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Divisibility and Remainders on the GMAT
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07 Jun 2018, 14:18
Bunuel wrote: Divisibility Applied to Remainders Part II Let's continue on our endeavor to understand divisibility and remainders in this post. Last week’s post focused on situations where the remainders were equal. Today, let’s see how to deal with situations where the remainders are different. Question: When positive integer n is divided by 3, the remainder is 2. When n is divided by 7, the remainder is 5. How many values less than 100 can n take?(A) 0 (B) 2 (C) 3 (D) 4 (E) 5 Solution: So n is a number 2 greater than a multiple of 3 (or we can say, it is 1 less than the next multiple of 3). It is also 5 greater than a multiple of 7 (or we can say it is 2 less than the next multiple of 7) \(n = 3a + 2 = 3x – 1\) \(n = 7b + 5 = 7y – 2\) No common remainder! When we have a common remainder,the smallest value of n would be the common remainder. Say, if n were of the forms: (3a + 1) and (7b + 1), the smallest number of both these forms is 1. When 1 is divided by 3, the quotient is 0 and the remainder is 1. When 1 is divided by 7, the quotient is 0 and the remainder is 1. But that is not the case here. So then, what do we do now? Let’s try and work with some trial and error now. n belongs to both the lists given below: Numbers of the form (3a+2): 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50… Numbers of the form (7b + 5): 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75… Which numbers are common to both the lists? 5, 26, 47 and there should be more. Do you see some link between these numbers? Let me show you some connections: – 26 is 21 more than 5. – 47 is 21 more than 26. – 21 is the LCM of 3 and 7. How do we explain these? Say, we identified that the smallest positive number which gives a remainder of 2 when divided by 3 and a remainder of 5 when divided by 7 is 5 (note here that when we divide 5 by 7, the quotient is 0 and the remainder is 5). What will be the next such number? Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… away from 5 i.e. it will be a multiple of 3 and a multiple of 7 away from 5. The smallest such multiple is obviously the LCM (lowest common multiple) of 3 and 7 i.e. 21. Hence the next such number will be 21 away from 5. We get 26. Use the same logic to get the next such number. It will be another 21 away from 26 so we get 47. By the same logic, the next few such numbers will be 68, 89, 110 etc. How many such numbers will be less than 100? 5, 26, 47, 68, 89 i.e. 5 such numbers. So, does this mean that when the remainders are not equal, you will need to make the lists given above. Well yes, in a way, but you can do it mentally. Let me explain. In the first 100 numbers, there will be many more numbers of the form (3a+2) than (7b + 5). Since n should be of both the forms, let’s start checking in the smaller list first. Say b = 0, the number is 5. Is 5 of the form (3a + 2). Yes! Your list has served its purpose. Now all we need to do is keep adding 21 (the LCM of 3 and 7) to get the next few numbers! This turned out to be an easy example. Let’s change the values a little to make it a little more cumbersome. This question is discussed HERE. Hi pushpitkc can you explain following: what are these numbers in red ?... quotients ? or what? "Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… " why there will be many more numbers of the form (3a+2) than (7b + 5) ? how can i figure that out that there are more numbers of the form (3a+2) Numbers of the form (3a+2): 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50… here, do we add divisor to remainder ? Numbers of the form (7b + 5): 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75… here, do we add divisor to remainder ? What is the point of rewriting \(n = 3a + 2\) AS \(3x – 1\) and not using it in solution above ? And \(n = 7b + 5\) AS \(7y – 2\) same here From other solutions i see that people equate \(3a + 2 = 7b + 5\) why ? in which cases can we equate ? thank you:)



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Re: Divisibility and Remainders on the GMAT
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07 Jun 2018, 22:43
dave13 wrote: Hi pushpitkc can you explain following: what are these numbers in red ?... quotients ? or what? "Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… " why there will be many more numbers of the form (3a+2) than (7b + 5) ? how can i figure that out that there are more numbers of the form (3a+2) Numbers of the form (3a+2): 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50… here, do we add divisor to remainder ? Numbers of the form (7b + 5): 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75… here, do we add divisor to remainder ? What is the point of rewriting \(n = 3a + 2\) AS \(3x – 1\) and not using it in solution above ? And \(n = 7b + 5\) AS \(7y – 2\) same here From other solutions i see that people equate \(3a + 2 = 7b + 5\) why ? in which cases can we equate ? thank you:) Hey dave13First things first, the first list of numbers are of form 3x and 7x, where x takes integer values starting from 1. Every time you add 2 to 3x, you will get a number of the form 3x + 2. Similarly, we get 7x + 5 when you add 5 to 7x. 3x + 2 > 2,5,8,11,14,17,20(7 numbers) 7x + 5 > 5,12,19(3 numbers) So, you can extrapolate and find out that there are more numbers of the form 3x + 2 than of the form 7x + 5. I think the reason we rewrite 3a+2 as 3x–1 and 7b+5 as 7b2 is to explain that they yield a common list. Here, we need to find the numbers which are present in both the lists. So, you can't equate in this case. Hope this helps you!
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Re: Divisibility and Remainders on the GMAT
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08 Jun 2018, 00:23
pushpitkc wrote: dave13 wrote: Hi pushpitkc can you explain following: what are these numbers in red ?... quotients ? or what? "Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… " why there will be many more numbers of the form (3a+2) than (7b + 5) ? how can i figure that out that there are more numbers of the form (3a+2) Numbers of the form (3a+2): 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50… here, do we add divisor to remainder ? Numbers of the form (7b + 5): 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75… here, do we add divisor to remainder ? What is the point of rewriting \(n = 3a + 2\) AS \(3x – 1\) and not using it in solution above ? And \(n = 7b + 5\) AS \(7y – 2\) same here From other solutions i see that people equate \(3a + 2 = 7b + 5\) why ? in which cases can we equate ? thank you:) Hey dave13First things first, the first list of numbers are of form 3x and 7x, where x takes integer values starting from 1. Every time you add 2 to 3x, you will get a number of the form 3x + 2. Similarly, we get 7x + 5 when you add 5 to 7x. 3x + 2 > 2,5,8,11,14,17,20(7 numbers) 7x + 5 > 5,12,19(3 numbers) So, you can extrapolate and find out that there are more numbers of the form 3x + 2 than of the form 7x + 5. I think the reason we rewrite 3a+2 as 3x–1 and 7b+5 as 7b2 is to explain that they yield a common list. Here, we need to find the numbers which are present in both the lists. So, you can't equate in this case. Hope this helps you! pushpitkc thank you for your nice explanation one question: what are these numbers in red (quotients ?) and how do we get them ?... "Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… " thanks and have a great day



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Divisibility and Remainders on the GMAT
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08 Jun 2018, 01:14
dave13 wrote: pushpitkc wrote: dave13 wrote: Hi pushpitkc can you explain following: what are these numbers in red ?... quotients ? or what? "Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… " why there will be many more numbers of the form (3a+2) than (7b + 5) ? how can i figure that out that there are more numbers of the form (3a+2) Numbers of the form (3a+2): 2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50… here, do we add divisor to remainder ? Numbers of the form (7b + 5): 5, 12, 19, 26, 33, 40, 47, 54, 61, 68, 75… here, do we add divisor to remainder ? What is the point of rewriting \(n = 3a + 2\) AS \(3x – 1\) and not using it in solution above ? And \(n = 7b + 5\) AS \(7y – 2\) same here From other solutions i see that people equate \(3a + 2 = 7b + 5\) why ? in which cases can we equate ? thank you:) Hey dave13First things first, the first list of numbers are of form 3x and 7x, where x takes integer values starting from 1. Every time you add 2 to 3x, you will get a number of the form 3x + 2. Similarly, we get 7x + 5 when you add 5 to 7x. 3x + 2 > 2,5,8,11,14,17,20(7 numbers) 7x + 5 > 5,12,19(3 numbers) So, you can extrapolate and find out that there are more numbers of the form 3x + 2 than of the form 7x + 5. I think the reason we rewrite 3a+2 as 3x–1 and 7b+5 as 7b2 is to explain that they yield a common list. Here, we need to find the numbers which are present in both the lists. So, you can't equate in this case. Hope this helps you! pushpitkc thank you for your nice explanation one question: what are these numbers in red (quotients ?) and how do we get them ?... "Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… " thanks and have a great day dave13This is the part of the solution that you seem to have a problem with Say, we identified that the smallest positive number which gives a remainder of 2 when divided by 3 and a remainder of 5 when divided by 7 is 5 (note here that when we divide 5 by 7, the quotient is 0 and the remainder is 5). What will be the next such number? Since the next number will also belong to both the lists above so it will be 3/6/9/12/15/18/21… away from 5 and it will also be 7/14/21/28/35/42… away from 5 i.e. it will be a multiple of 3 and a multiple of 7 away from 5. The smallest such multiple is obviously the LCM (lowest common multiple) of 3 and 7 i.e. 21. Hence the next such number will be 21 away from 5. We get 26. Use the same logic to get the next such number. It will be another 21 away from 26 so we get 47. By the same logic, the next few such numbers will be 68, 89, 110 etc. How many such numbers will be less than 100? 5, 26, 47, 68, 89 i.e. 5 such numbers. Once we identify the smallest number(in this case 5) which gives a remainder of 2 when divided by 3. In order to identify the next such number(which gives a remainder of 2 when divided by 3), we need to add 3/6/9/12/15/18/21 to 5. This list contains numbers which when added to 5, give us numbers when divided by 3, give us a remainder of 2:{5,5+3,5+6,5+9,5+12,5+15,5+18,5+21,5+24,5+27,5+30...} or {5,8,11,14,17,20,23,26,29,32,35... } Similarly, to identify the next such number(which gives a remainder of 5 when divided by 7), you will need to add 7/14/21/28/35/42 to 5. The list is {5,5+7,5+14,5+21,5+28,5+35...} or {5,12,19,26,33,40...} The next step is finding the common numbers in both the list which is {5, 26, 47, 68, 89} Hope this helps clear your confusion!
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Re: Divisibility and Remainders on the GMAT
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22 Jan 2019, 21:49
globaldesi wrote: Bunuel wrote: https://gmatclub.com/forum/newunitsdi ... 68569.html is no longer valid. Bunuel am I looking at something wrong? ___________________ Fixed that. Thank you.
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Re: Divisibility and Remainders on the GMAT
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