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04 Mar 2015, 04:49
| FROM Veritas Prep Blog: Advanced Number Properties on GMAT - Part I |
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Don’t worry, we are not going to discuss (Even + Even = Even) and (Odd + Odd = Even) type of basic number properties in this post. What we have in mind for today is something based on this but far more advanced. Often, people complain that they thoroughly understand the theory but have difficulties applying it and hence are stuck at a score of 600. They look for practice questions and tend to ignore concepts since they already “know” them. We often ask them to go back to concepts since we believe that a strong foundation of concepts is necessary for ‘score increase’. Mind you, when we do that, we don’t mean to ask them to review the basic concepts again, we mean to ask them to deduce and work on advanced concepts. Let’s show you with the help of a question. Question: If two integers are chosen at random out of first 5 positive integers, what is the probability that their product will be of the form a^2 – b^2, where a and b are both positive integers? A. 2/5 B. 3/5 C. 7/10 D. 4/5 E. 9/10 Solution: This might look like a probability question but isn’t. Questions like these are the reason we ask you to go through basics of every topic including probability. If you do not know probability at all, you may skip this question even though it needs very basic knowledge of probability. Probability will tell you that Required probability = Favorable cases/Total cases Total cases are very easy to find: 5C2 = 10 or 5*4/2 = 10 whatever you prefer. This is the number of ways in which you select any 2 distinct numbers out of the given 5 distinct numbers. Number of favorable cases is the challenge here. That is why it is a number properties question and not so much a probability question. Let’s focus on the main part of the question: First five positive integers: 1, 2, 3, 4, 5 We need to select two integers such that their product is of the form a^2 – b^2. What does a^2 – b^2 remind you of? It reminds me of (a + b)(a – b). So the product needs to be of the form (a + b)(a – b). So is it necessary that of the two numbers we pick, one must be of the form (a + b) and the other must be (a – b)? No. Note that we should be able to write the product in this form. It is not necessary that the numbers must be of this form only. But first let’s focus on numbers which are already of the form (a + b) and (a – b). Say you pick two numbers, 2 and 5. Are they of the form (a + b) and (a – b) such that a and b are integers? No. 5 = 3.5 + 1.5 2 = 3.5 – 1.5 So a = 3.5, b = 1.5. a and b are not integers. What about numbers such as 3 and 5? Are they of the form (a + b) and (a – b) such that a and b are integers? Yes. 5 = 4 + 1 3 = 4 – 1 Note that whenever the average of the numbers will be an integer, we will be able to write them as a+b and a – b because one number will be some number more than the average and the other will be the same number less than average. So a will be the average and the amount more or less will be b. When will the average of two numbers (Number1 + Number2)/2 be an integer? When the sum of the two numbers is even! When is the sum of two numbers even? It is when both the numbers are even or when both are odd. So then does the question boil down to “favorable cases are when we select both numbers even or both numbers odd?” Yes and No. When we select both even numbers or both odd numbers, the product can be written as a^2 – b^2. But are those the only cases when the product can be written as a^2 – b^2? The question is not so much as whether both the numbers are even or both are odd as whether the product of the numbers can be written as product of two even numbers or two odd numbers. We need to be able to write the product (whatever we obtain) as product of two even or two odd numbers. To explain this, let’s say we pick two numbers 4 and 5 4*5 = 20 Can we write 20 as product of two even numbers? Yes 2*10. So even though, 4 is even and 5 is odd, their product can be written as product of two even numbers. So in which all cases will this happen? - Whenever you have at least 4 in the product, you can write it as product of two even numbers: give one 2 to one number and the other 2 to the other number to make both even. If the product is even but not a multiple of 4, it cannot be written as product of two even numbers or product of two odd numbers. It can only be written as product of one even and one odd number. If the product is odd, it can always be written as product of two odd numbers. Let’s go back to our question: We have 5 numbers: 1, 2, 3, 4, 5 Our favorable cases constitute those in which either both numbers are odd or the product has 4 as a factor. 3 Odd numbers: 1, 3, 5 2 Even numbers: 2, 4 Number of cases when both numbers are odd = 3C2 = 3 (select 2 of the 3 odd numbers) Number of cases when 4 is a factor of the product = Number of cases such that we select 4 and any other number = 1*4C1 = 4 Total number of favorable cases = 3 + 4 = 7 Note that this includes the case where we take both even numbers. Had there been more even numbers such as 6, we would have included more cases where we pick both even numbers such as 2 and 6 since their product would have 4 as a factor. Required Probability = 7/10 Answer (C) Takeaway: When can we write a number as difference of squares? - When the number is odd or - When the number has 4 as a factor Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
04 Mar 2015, 04:59
| FROM Veritas Prep Admissions Blog: Advanced Number Properties on the GMAT - Part V |
 Today, let’s look in detail at a relation between arithmetic mean and geometric mean of two numbers. It is one of those properties which make sense the moment someone explains to us but are very hard to arrive on our own. When two positive numbers are equal, their Arithmetic Mean = Geometric Mean = The number itself Say, the two numbers are m and n (and are equal). Their arithmetic mean = (m+n)/2 = 2m/2 = m Their geometric mean = sqrt(m*n) = sqrt(m^2) = m (the numbers are positive so |m| = m) We also know that Arithmetic Mean >= Geometric Mean So when arithmetic mean is equal to geometric mean, it means the arithmetic mean is taking its minimum value. So when (m+n)/2 is minimum, it implies (m+n) is minimum. Therefore, sum of numbers takes its minimum value when the numbers are equal. When geometric mean is equal to arithmetic mean, it means the geometric mean is taking its maximum value. So when sqrt(m*n) is maximum, it means m*n is maximum. Therefore, product of numbers takes its maximum value when the numbers are equal. Let’s see how to solve a difficult question using this concept. Question: If x and y are positive, is x^2 + y^2 > 100? Statement 1: 2xy < 100 Statement 2: (x + y)^2 > 200 Solution: We need to find whether x^2 + y^2 must be greater than 100. Statement 1: 2xy < 100 Plug in some easy values to see that this is not sufficient alone. If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100 If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100 So x^2 + y^2 may be less than or greater than 100. Statement 2: (x + y)^2 > 200 There are two ways to deal with this statement. One is the algebra way which is easier to understand but far less intuitive. Another is using the concept we discussed above. Let’s look at both: Algebra solution: We know that (x – y)^2 >= 0 because a square is never negative. So x^2 + y^2 – 2xy >= 0 x^2 + y^2 >= 2xy This will be true for all values of x and y. Now, statement 2 gives us x^2 + y^2 + 2xy > 200. The left hand side is greater than 200. If on the left we substitute 2xy with (x^2 + y^2), the left hand side will either become greater than or same as before. So in any case, the left hand side will remain greater than 200. x^2 + y^2 + (x^2 + y^2) > 200 2(x^2 + y^2) > 200 x^2 + y^2 > 100 This statement alone is sufficient to say that x^2 + y^2 will be greater than 100. But, we agree that the first step where we start with (x – y)^2 is not intuitive. It may not hit you at all. Hence, here is another way to analyze this statement. Logical solution: Let’s try to find the minimum value of x^2 + y^2. It will take minimum value when x^2 = y^2 i.e. when x = y (x and y are both positive) We are given that (x+y)^2 > 200 (x+x)^2 > 200 x > sqrt(50) So x^2 + y^2 will take a value greater than [sqrt(50)]^2 + [sqrt(50)]^2 = 100. So in any case, x^2 + y^2 will be greater than 100. This statement alone is sufficient to answer the question. Answer (B) Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
04 Mar 2015, 04:54
| FROM Veritas Prep Blog: Advanced Number Properties on the GMAT - Part III |
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Continuing our discussion on number properties, today we will discuss how factorials affect the behavior of odd and even integers. Since we are going to deal with factorials, positive integers will be our concern. Using a question, we will see how factorials are divided. Question: If x and y are positive integers, is y odd? Statement 1: (y+2)!/x! = odd Statement 2: (y+2)!/x! is greater than 2 Solution: The question stem doesn’t give us much information – just that x and y are positive integers. Question: Is y odd? Statement 1: (y+2)!/x! = odd Note that odd and even are identified only for integers. Since (y+2)!/x! is odd, it must be a positive integer. This means that x! must be equal to or less than (y+2)! Now think, how are y and y+2 related? If y+2 is odd, y+1 is even and hence y is odd. If y+2 is even, by the same logic, y is even. (y+2)! = 1*2*3*4*…*y*(y+1)*(y+2) x! = 1*2*3*4*…*x Note that (y+2)! and x! have common factors starting from 1. Since x! is less than or equal to (y+2)!, x will be less than or equal to (y+2). So all factors in the denominator, from 1 to x will be there in the numerator too and will get canceled leaving us with the last few factors of (y+2)! To explain this, let us take a few examples: Example 1: Say, y+2 = 6, x = 6 (y+2)!/x! = 6!/6! = 1 Example 2: Say, y+2 = 7, x = 6 (y+2)!/x! = 7!/6! = (1*2*3*4*5*6*7)/(1*2*3*4*5*6) = 7 (only one leftover factor) Example 3: Say, y+2 = 6, x = 4 (y+2)!/x! = 6!/4! = (1*2*3*4*5*6)/(1*2*3*4) = 5*6 (two leftover factors) If the division of two factorials is an integer, the factorial in the numerator must be larger than or equal to the factorial in the denominator. So what does (y+2)!/x! is odd imply? It means that the leftover factors must be all odd. But the leftover factors will be consecutive integers. So after one odd factor, there will be an even factor. If we want (y+2)!/x! to be odd, we must have either no leftover factors (such that (y+2)!/x! = 1) or only one leftover factor and that too odd. If we have no leftover factor, it doesn’t matter what y+2 is as long as it is equal to x. It could be odd or even. If there is one leftover factor, then y+2 must be odd and hence y must be odd. Hence y could be odd or even. This statement alone is not sufficient. Statement 2: (y+2)!/x! is greater than 2 This tells us that y+2 is not equal to x since (y+2)!/x! is not 1. But all we know is that it is greater than 2. It could be anything as seen in examples 2 and 3 above. This statement alone is not sufficient. Both statements together tell us that y+2 is greater than x such that (y+2)!/x! is odd. So there must be only one leftover factor and it must be odd. The leftover factor will be the last factor i.e. y+2. This tells us that y+2 must be odd. Hence y must be odd too. Answer (C) Takeaways: Assuming a and b are positive integers, - a!/b! will be an integer only if a >= b - a!/b! will be an odd integer whenever a = b or a is odd and a = b+1 - a!/b! will be an even integer whenever a is even and a = b+1 or a > b+1 Think about this: what happens when we put 0 in the mix? Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
04 Mar 2015, 04:50
| FROM Veritas Prep Blog: Advanced Number Properties on the GMAT - Part II |
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Before we get started, be sure to take a look at Part I of this article. Number properties concepts come across as pretty easy, theoretically, but they have some of the toughest questions. Today let’s take a look at some properties of prime numbers and their sum. Note that don’t try to “learn” all the takeaways you come across for number properties – it will be very stressful. Instead, try to understand why the properties are such so that if you get a question related to some such properties, you can replicate the results effortlessly. To start off, we would like to take up a simple question and then using the takeaway derived from it, we will look at a harder problem. Question 1: Which of the following CANNOT be the sum of two prime numbers? (A) 19 (B) 45 (C) 68 (D) 79 (E) 88 Solution: What do we know about sum of two prime numbers? e.g. 3 + 5 = 8 5 + 11 = 16 5 + 17 = 22 23 + 41 = 64 Do you notice something? The sum is even in all these cases. Why? Because most prime numbers are odd. When we add two odd numbers, we get an even sum. We have only 1 even prime number and that is 2. Hence to obtain an odd sum, one number must be 2 and the other must be odd. 2 + 3 = 5 2 + 7 = 9 2 + 17 = 19 Look at the options given in the question. Three of them are odd which means they must be of the form 2 + Another Prime Number. Let’s check the odd options first: (A) 19 = 2 + 17 (Both Prime. Can be written as sum of two prime numbers.) (B) 45 = 2 + 43 (Both Prime. Can be written as sum of two prime numbers.) (D) 79 = 2 + 77 (77 is not prime.) 79 cannot be written as sum of two prime numbers. Note that 79 cannot be written as sum of two primes in any other way. One prime number has to be 2 to get a sum of 79. Hence there is no way in which we can obtain 79 by adding two prime numbers. (D) is the answer. Now think what happens if instead of 79, we had 81? 81 = 2 + 79 Both numbers are prime hence all three odd options can be written as sum of two prime numbers. Then we would have had to check the even options too (at least one of which would be different from the given even options). Think, how would we find which even numbers can be written as sum of two primes? We will give the solution of that next week. So the takeaway here is that if you get an odd sum on adding two prime numbers, one of the numbers must be 2. Question 2: If m, n and p are positive integers such that m < n < p, is m a factor of the odd integer p? Statement 1: m and n are prime numbers such that (m + n) is a factor of 119. Statement 2: p is a factor of 119. Solution: First of all, we are dealing with positive integers here – good. No negative numbers, 0 or fraction complications. Let’s move on. The question stem tells us that p is an odd integer. Also, m < n < p. Question: Is m a factor of p? There isn’t much information in the question stem for us to process so let’s jump on to the statements directly. Statement 1: m and n are prime numbers such that (m + n) is a factor of 119. Write down the factors of 119 first to get the feasible range of values. 119 = 1, 7, 17, 119 All factors of 119 are odd numbers. So (m + n), a sum of two primes must be odd. This means one of m and n is 2. There are many possible values of m and n e.g. 2 and 5 (to give sum 7) or 2 and 15 (to give sum 17) or 2 and 117 (to give sum 119). Note that we also have m < n. This means that in each case, m must be 2 and n must be the other number of the pair. So now we know that m is 2. We also know that p is an odd integer. Is m a factor of p? No. Odd integers are those which do not have 2 as a factor. Since m is 2, p does not have m as a factor. This statement alone is sufficient to answer the question! Statement 2: p is a factor of 119 This tells us that p is one of 7, 17 and 119. p cannot be 1 because m < n < p where all are positive integers. But it tells us nothing about m. All we know is that it is less than p. For example, if p is 7, m could be 1 and hence a factor of p or it could be 5 and not a factor of p. Hence this statement alone is not sufficient. Answer (A) Something to think about: In this question, if you are given that m is not 1, does it change our answer? Key Takeaways: - When two distinct prime numbers are added, their sum is usually even. If their sum is odd, one of the numbers must be 2. - Think what happens in case you add three distinct prime numbers. The sum will be usually odd. In case the sum is even, one number must be 2. - Remember the special position 2 occupies among prime numbers – it is the only even prime number. Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
04 Mar 2015, 04:57
| FROM Veritas Prep Blog: Advanced Number Properties on the GMAT - Part IV |
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As pointed out by a reader, we need to complete the discussion on a question discussed in our previous ‘Advanced Number Properties’ posts so let’s do that today. Note that the discussion that follows doesn’t fall in the purview of GMAT and you needn’t know it. You will be able to solve any question without taking this post into account but that has never stopped us from letting loose our curiosity so here goes… Question 1: Which of the following CANNOT be the sum of two prime numbers? (A) 19 (B) 45 (C) 58 (D) 79 (E) 88 Solution: We discussed in that post that the sum of two prime numbers is usually even because prime numbers are usually odd. We also discussed that if the sum of two prime numbers is odd, it means one of the prime numbers is certainly 2 – the only even prime number. For example: 2 + 3 = 5 2 + 7 = 9 2 + 17 = 19 Then it makes perfect sense to first look at the options which are odd. To be sum of two prime numbers, the sum must be of the form 2 + Another Prime Number. We saw that (D) 79 = 2 + 77 (77 is not prime.) and hence we got (D) as our answer. Now the question we raised there was: What happens if instead of 79, we had 81? 81 = 2 + 79 Then all three odd options would have been sum of two prime numbers and we would have needed to check the even options too. How do you figure whether an even number can be written as the sum of two prime numbers? This is where Goldbach’s Conjecture comes into play (you don’t really need to know it. We are doing it for intellectual purposes. GMAC will never put you in this fix). It says “Every even integer greater than 2 can be expressed as the sum of two primes.” Mind you, it’s a conjecture i.e. it hasn’t been proven for all even numbers (only for even numbers till 4 * 10^{18}) but it does seem to hold. For example: 4 = 2 + 2 6 = 3 + 3 8 = 3 + 5 10 = 3 + 7 = 5 + 5 12 = 5 + 7 and so on… So given any even sum greater than 2, you can say that it CAN be written as sum of two prime numbers, for all practical purposes. In fact, and here we are going into really geeky territory, we expect that every large even integer has not just one representation as the sum of two primes, but in fact has very many such representations. For all we know, 6 may be the only even number greater than 2 which cannot be written as the sum of two distinct prime numbers. Coming back to our original question, we will actually check only odd numbers to see whether they can be written as sum of two primes. One of them has to be such that it cannot be written as sum of two primes and finding that is very simple! (as discussed in the previous post) So all in all, the question that seemed very tedious turned out to be very simple! Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
04 Mar 2015, 05:05
16 May 2017, 12:23
| FROM Veritas Prep Blog: Advanced Number Properties on the GMAT - Part II |
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Before we get started, be sure to take a look at Part I of this article. Number properties concepts come across as pretty easy, theoretically, but they have some of the toughest questions. Today let’s take a look at some properties of prime numbers and their sum. Note that don’t try to “learn” all the takeaways you come across for number properties – it will be very stressful. Instead, try to understand why the properties are such so that if you get a question related to some such properties, you can replicate the results effortlessly. To start off, we would like to take up a simple question and then using the takeaway derived from it, we will look at a harder problem. Question 1: Which of the following CANNOT be the sum of two prime numbers? (A) 19 (B) 45 (C) 68 (D) 79 (E) 88 Solution: What do we know about sum of two prime numbers? e.g. 3 + 5 = 8 5 + 11 = 16 5 + 17 = 22 23 + 41 = 64 Do you notice something? The sum is even in all these cases. Why? Because most prime numbers are odd. When we add two odd numbers, we get an even sum. We have only 1 even prime number and that is 2. Hence to obtain an odd sum, one number must be 2 and the other must be odd. 2 + 3 = 5 2 + 7 = 9 2 + 17 = 19 Look at the options given in the question. Three of them are odd which means they must be of the form 2 + Another Prime Number. Let’s check the odd options first: (A) 19 = 2 + 17 (Both Prime. Can be written as sum of two prime numbers.) (B) 45 = 2 + 43 (Both Prime. Can be written as sum of two prime numbers.) (D) 79 = 2 + 77 (77 is not prime.) 79 cannot be written as sum of two prime numbers. Note that 79 cannot be written as sum of two primes in any other way. One prime number has to be 2 to get a sum of 79. Hence there is no way in which we can obtain 79 by adding two prime numbers. (D) is the answer. Now think what happens if instead of 79, we had 81? 81 = 2 + 79 Both numbers are prime hence all three odd options can be written as sum of two prime numbers. Then we would have had to check the even options too (at least one of which would be different from the given even options). Think, how would we find which even numbers can be written as sum of two primes? We will give the solution of that next week. So the takeaway here is that if you get an odd sum on adding two prime numbers, one of the numbers must be 2. Question 2: If m, n and p are positive integers such that m < n < p, is m a factor of the odd integer p? Statement 1: m and n are prime numbers such that (m + n) is a factor of 119. Statement 2: p is a factor of 119. Solution: First of all, we are dealing with positive integers here – good. No negative numbers, 0 or fraction complications. Let’s move on. The question stem tells us that p is an odd integer. Also, m < n < p. Question: Is m a factor of p? There isn’t much information in the question stem for us to process so let’s jump on to the statements directly. Statement 1: m and n are prime numbers such that (m + n) is a factor of 119. Write down the factors of 119 first to get the feasible range of values. 119 = 1, 7, 17, 119 All factors of 119 are odd numbers. So (m + n), a sum of two primes must be odd. This means one of m and n is 2. There are many possible values of m and n e.g. 2 and 5 (to give sum 7) or 2 and 15 (to give sum 17) or 2 and 117 (to give sum 119).St 1 already mentions that m and n are prime then how can we consider the pair 2 and 15 (to give sum 17) or 2 and 117 (to give sum 119) for m and n. Bunuel please correct me if i am wrong. Note that we also have m < n. This means that in each case, m must be 2 and n must be the other number of the pair. So now we know that m is 2. We also know that p is an odd integer. Is m a factor of p? No. Odd integers are those which do not have 2 as a factor. Since m is 2, p does not have m as a factor. This statement alone is sufficient to answer the question! Statement 2: p is a factor of 119 This tells us that p is one of 7, 17 and 119. p cannot be 1 because m < n < p where all are positive integers. But it tells us nothing about m. All we know is that it is less than p. For example, if p is 7, m could be 1 and hence a factor of p or it could be 5 and not a factor of p. Hence this statement alone is not sufficient. Answer (A) Something to think about: In this question, if you are given that m is not 1, does it change our answer? Key Takeaways: - When two distinct prime numbers are added, their sum is usually even. If their sum is odd, one of the numbers must be 2. - Think what happens in case you add three distinct prime numbers. The sum will be usually odd. In case the sum is even, one number must be 2. - Remember the special position 2 occupies among prime numbers – it is the only even prime number. Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
02 Aug 2017, 02:35
| FROM Veritas Prep Blog: Advanced Number Properties on the GMAT - Part II |
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Before we get started, be sure to take a look at Part I of this article. Number properties concepts come across as pretty easy, theoretically, but they have some of the toughest questions. Today let’s take a look at some properties of prime numbers and their sum. Note that don’t try to “learn” all the takeaways you come across for number properties – it will be very stressful. Instead, try to understand why the properties are such so that if you get a question related to some such properties, you can replicate the results effortlessly. To start off, we would like to take up a simple question and then using the takeaway derived from it, we will look at a harder problem. Question 1: Which of the following CANNOT be the sum of two prime numbers? (A) 19 (B) 45 (C) 68 (D) 79 (E) 88 Solution: What do we know about sum of two prime numbers? e.g. 3 + 5 = 8 5 + 11 = 16 5 + 17 = 22 23 + 41 = 64 Do you notice something? The sum is even in all these cases. Why? Because most prime numbers are odd. When we add two odd numbers, we get an even sum. We have only 1 even prime number and that is 2. Hence to obtain an odd sum, one number must be 2 and the other must be odd. 2 + 3 = 5 2 + 7 = 9 2 + 17 = 19 Look at the options given in the question. Three of them are odd which means they must be of the form 2 + Another Prime Number. Let’s check the odd options first: (A) 19 = 2 + 17 (Both Prime. Can be written as sum of two prime numbers.) (B) 45 = 2 + 43 (Both Prime. Can be written as sum of two prime numbers.) (D) 79 = 2 + 77 (77 is not prime.) 79 cannot be written as sum of two prime numbers. Note that 79 cannot be written as sum of two primes in any other way. One prime number has to be 2 to get a sum of 79. Hence there is no way in which we can obtain 79 by adding two prime numbers. (D) is the answer. Now think what happens if instead of 79, we had 81? 81 = 2 + 79 Both numbers are prime hence all three odd options can be written as sum of two prime numbers. Then we would have had to check the even options too (at least one of which would be different from the given even options). Think, how would we find which even numbers can be written as sum of two primes? We will give the solution of that next week. So the takeaway here is that if you get an odd sum on adding two prime numbers, one of the numbers must be 2. Question 2: If m, n and p are positive integers such that m < n < p, is m a factor of the odd integer p? Statement 1: m and n are prime numbers such that (m + n) is a factor of 119. Statement 2: p is a factor of 119. Solution: First of all, we are dealing with positive integers here – good. No negative numbers, 0 or fraction complications. Let’s move on. The question stem tells us that p is an odd integer. Also, m < n < p. Question: Is m a factor of p? There isn’t much information in the question stem for us to process so let’s jump on to the statements directly. Statement 1: m and n are prime numbers such that (m + n) is a factor of 119. Write down the factors of 119 first to get the feasible range of values. 119 = 1, 7, 17, 119 All factors of 119 are odd numbers. So (m + n), a sum of two primes must be odd. This means one of m and n is 2. There are many possible values of m and n e.g. 2 and 5 (to give sum 7) or 2 and 15 (to give sum 17) or 2 and 117 (to give sum 119). Note that we also have m < n. This means that in each case, m must be 2 and n must be the other number of the pair. So now we know that m is 2. We also know that p is an odd integer. Is m a factor of p? No. Odd integers are those which do not have 2 as a factor. Since m is 2, p does not have m as a factor. This statement alone is sufficient to answer the question! Statement 2: p is a factor of 119 This tells us that p is one of 7, 17 and 119. p cannot be 1 because m < n < p where all are positive integers. But it tells us nothing about m. All we know is that it is less than p. For example, if p is 7, m could be 1 and hence a factor of p or it could be 5 and not a factor of p. Hence this statement alone is not sufficient. Answer (A) Something to think about: In this question, if you are given that m is not 1, does it change our answer? Key Takeaways: - When two distinct prime numbers are added, their sum is usually even. If their sum is odd, one of the numbers must be 2. - Think what happens in case you add three distinct prime numbers. The sum will be usually odd. In case the sum is even, one number must be 2. - Remember the special position 2 occupies among prime numbers – it is the only even prime number. Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
24 Aug 2017, 10:34
| FROM Veritas Prep Admissions Blog: Advanced Number Properties on the GMAT - Part V |
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Today, let’s look in detail at a relation between arithmetic mean and geometric mean of two numbers. It is one of those properties which make sense the moment someone explains to us but are very hard to arrive on our own. When two positive numbers are equal, their Arithmetic Mean = Geometric Mean = The number itself Say, the two numbers are m and n (and are equal). Their arithmetic mean = (m+n)/2 = 2m/2 = m Their geometric mean = sqrt(m*n) = sqrt(m^2) = m (the numbers are positive so |m| = m) We also know that Arithmetic Mean >= Geometric Mean So when arithmetic mean is equal to geometric mean, it means the arithmetic mean is taking its minimum value. So when (m+n)/2 is minimum, it implies (m+n) is minimum. Therefore, sum of numbers takes its minimum value when the numbers are equal. When geometric mean is equal to arithmetic mean, it means the geometric mean is taking its maximum value. So when sqrt(m*n) is maximum, it means m*n is maximum. Therefore, product of numbers takes its maximum value when the numbers are equal. Let’s see how to solve a difficult question using this concept. Question: If x and y are positive, is x^2 + y^2 > 100? Statement 1: 2xy < 100 Statement 2: (x + y)^2 > 200 Solution: We need to find whether x^2 + y^2 must be greater than 100. Statement 1: 2xy < 100 Plug in some easy values to see that this is not sufficient alone. If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100 If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100 So x^2 + y^2 may be less than or greater than 100. Statement 2: (x + y)^2 > 200 There are two ways to deal with this statement. One is the algebra way which is easier to understand but far less intuitive. Another is using the concept we discussed above. Let’s look at both: Algebra solution: We know that (x – y)^2 >= 0 because a square is never negative. So x^2 + y^2 – 2xy >= 0 x^2 + y^2 >= 2xy This will be true for all values of x and y. Now, statement 2 gives us x^2 + y^2 + 2xy > 200. The left hand side is greater than 200. If on the left we substitute 2xy with (x^2 + y^2), the left hand side will either become greater than or same as before. So in any case, the left hand side will remain greater than 200. x^2 + y^2 + (x^2 + y^2) > 200 2(x^2 + y^2) > 200 x^2 + y^2 > 100 This statement alone is sufficient to say that x^2 + y^2 will be greater than 100. But, we agree that the first step where we start with (x – y)^2 is not intuitive. It may not hit you at all. Hence, here is another way to analyze this statement. Logical solution: Let’s try to find the minimum value of x^2 + y^2. It will take minimum value when x^2 = y^2 i.e. when x = y (x and y are both positive) We are given that (x+y)^2 > 200 (x+x)^2 > 200 x > sqrt(50) So x^2 + y^2 will take a value greater than [sqrt(50)]^2 + [sqrt(50)]^2 = 100. So in any case, x^2 + y^2 will be greater than 100. This statement alone is sufficient to answer the question. Answer (B) Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
24 Aug 2017, 10:52
| FROM Veritas Prep Admissions Blog: Advanced Number Properties on the GMAT - Part V |
|
Today, let’s look in detail at a relation between arithmetic mean and geometric mean of two numbers. It is one of those properties which make sense the moment someone explains to us but are very hard to arrive on our own. When two positive numbers are equal, their Arithmetic Mean = Geometric Mean = The number itself Say, the two numbers are m and n (and are equal). Their arithmetic mean = (m+n)/2 = 2m/2 = m Their geometric mean = sqrt(m*n) = sqrt(m^2) = m (the numbers are positive so |m| = m) We also know that Arithmetic Mean >= Geometric Mean So when arithmetic mean is equal to geometric mean, it means the arithmetic mean is taking its minimum value. So when (m+n)/2 is minimum, it implies (m+n) is minimum. Therefore, sum of numbers takes its minimum value when the numbers are equal. When geometric mean is equal to arithmetic mean, it means the geometric mean is taking its maximum value. So when sqrt(m*n) is maximum, it means m*n is maximum. Therefore, product of numbers takes its maximum value when the numbers are equal. Let’s see how to solve a difficult question using this concept. Question: If x and y are positive, is x^2 + y^2 > 100? Statement 1: 2xy < 100 Statement 2: (x + y)^2 > 200 Solution: We need to find whether x^2 + y^2 must be greater than 100. Statement 1: 2xy < 100 Plug in some easy values to see that this is not sufficient alone. If x = 0 and y = 0, 2xy < 100 and x^2 + y^2 < 100 If x = 40 and y = 1, 2xy < 100 but x^2 + y^2 > 100 So x^2 + y^2 may be less than or greater than 100. Statement 2: (x + y)^2 > 200 There are two ways to deal with this statement. One is the algebra way which is easier to understand but far less intuitive. Another is using the concept we discussed above. Let’s look at both: Algebra solution: We know that (x – y)^2 >= 0 because a square is never negative. So x^2 + y^2 – 2xy >= 0 x^2 + y^2 >= 2xy This will be true for all values of x and y. Now, statement 2 gives us x^2 + y^2 + 2xy > 200. The left hand side is greater than 200. If on the left we substitute 2xy with (x^2 + y^2), the left hand side will either become greater than or same as before. So in any case, the left hand side will remain greater than 200. x^2 + y^2 + (x^2 + y^2) > 200 2(x^2 + y^2) > 200 x^2 + y^2 > 100 This statement alone is sufficient to say that x^2 + y^2 will be greater than 100. But, we agree that the first step where we start with (x – y)^2 is not intuitive. It may not hit you at all. Hence, here is another way to analyze this statement. Logical solution: Let’s try to find the minimum value of x^2 + y^2. It will take minimum value when x^2 = y^2 i.e. when x = y (x and y are both positive) We are given that (x+y)^2 > 200 (x+x)^2 > 200 x > sqrt(50) So x^2 + y^2 will take a value greater than [sqrt(50)]^2 + [sqrt(50)]^2 = 100. So in any case, x^2 + y^2 will be greater than 100. This statement alone is sufficient to answer the question. Answer (B) Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog! |
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Hi Generic [Bot],
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