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25 Apr 2019, 06:35
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
52% (02:30) correct
48%
(02:35)
wrong
based on 27
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What is the greatest three-digit positive integer n for which the sum of the first n positive integers is not a divisor of the product of the first n positive integers?
(A) 995
(B) 996
(C) 997
(D) 998
(E) 999
(A) 995
(B) 996
(C) 997
(D) 998
(E) 999
25 Apr 2019, 09:54
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The sum of the first n positive integers is n(n+1)/2 (the average of the list 1, 2, 3, ... , n is just (n+1)/2, and we have n numbers in that list, so, since sum = average*n, the sum is n(n+1)/2 ).
Notice that n and n+1 are consecutive integers, so one of them is even, the other odd. If n is even, then n(n+1)/2 is equal to the product of the two integers n/2 and n+1, and if n is odd, then it is equal to the product of the two integers (n+1)/2 and n.
The product of the first n positive integers is just n!. Notice that if we write out n! in full, we see instantly that every integer n or smaller is a factor of n!.
If n is odd, then the sum of the first n positive integers is equal to (n+1)/2 times n, and both of those numbers are less than or equal to n. So they're both clearly factors of n!. So in the case where n is odd, n! will always be divisible by n(n+1)/2.
So if n(n+1)/2 is not a divisor of n! then n must be even. In that case, we're trying to divide n! by n/2, which we can certainly do, but also by the odd integer n+1. Often we can do that, if n+1 can be divided up into smaller numbers. But we will definitely not be able to divide n! by n+1 if n+1 is a prime number. So we're looking for an answer choice that gives us a prime number if we add 1 to it. The only two candidate answer choices are B and D, but 998+1 is 999 which is clearly divisible by 3 and 37, so is not prime. That only leaves B as a possibility, but proving that 997 is in fact a prime number is impossible to do in two minutes, and is not something you'd ever need to know for the GMAT, so this is not a realistic GMAT question. Where is it from?
Notice that n and n+1 are consecutive integers, so one of them is even, the other odd. If n is even, then n(n+1)/2 is equal to the product of the two integers n/2 and n+1, and if n is odd, then it is equal to the product of the two integers (n+1)/2 and n.
The product of the first n positive integers is just n!. Notice that if we write out n! in full, we see instantly that every integer n or smaller is a factor of n!.
If n is odd, then the sum of the first n positive integers is equal to (n+1)/2 times n, and both of those numbers are less than or equal to n. So they're both clearly factors of n!. So in the case where n is odd, n! will always be divisible by n(n+1)/2.
So if n(n+1)/2 is not a divisor of n! then n must be even. In that case, we're trying to divide n! by n/2, which we can certainly do, but also by the odd integer n+1. Often we can do that, if n+1 can be divided up into smaller numbers. But we will definitely not be able to divide n! by n+1 if n+1 is a prime number. So we're looking for an answer choice that gives us a prime number if we add 1 to it. The only two candidate answer choices are B and D, but 998+1 is 999 which is clearly divisible by 3 and 37, so is not prime. That only leaves B as a possibility, but proving that 997 is in fact a prime number is impossible to do in two minutes, and is not something you'd ever need to know for the GMAT, so this is not a realistic GMAT question. Where is it from?
