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04 Mar 2015, 04:49
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| 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! |
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04 Mar 2015, 04:59
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| 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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04 Mar 2015, 04:54
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| 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! |
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