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nick1816
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For how many integers n, such that 2≤n≤80, [m][fraction](n-1)*n*(n+1)/ 26 May 2019, 17:47
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D
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Difficulty:

85% (hard)

Question Stats:

45% (02:25) correct 55% (02:24) wrong based on 97 sessions

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For how many integers n, such that 2≤n≤80, \(\frac{(n-1)*n*(n+1)}{8}\) is an integer.

A. 9
B. 29
C. 40
D. 49
E. 59
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D
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For how many integers n, such that 2≤n≤80, [m][fraction](n-1)*n*(n+1)/ 26 May 2019, 18:09
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nick1816
For how many integers n, such that 2≤n≤80, \(\frac{(n-1)*n*(n+1)}{8}\) is an integer.

A. 9
B. 29
C. 40
D. 49
E. 59

I started putting random numbers to find a pattern so that I can extrapolate.

I observed that for every (n-1)*n*(n+1) where n is a multiple of 8 the function f(n) is perfectly divisible by 8.
Also, n that is odd, f(n) is also perfectly divisible by 8.

So, there are 10 multiples of 8 between [2,80].
And there are 39 odd numbers between [2,80].
So possible values of n (Integers) 49, IMO D
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For how many integers n, such that 2≤n≤80, [m][fraction](n-1)*n*(n+1)/ 26 May 2019, 19:48
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For how many integers n, such that 2≤n≤80, \(\frac{(n-1)*n*(n+1)}{8}\) is an integer.

A. 9
B. 29
C. 40
D. 49
E. 59

(n-1)*n*(n+1) is a product of 3 consecutive integers.

So, it can be of 2 types
1. even*odd*even
2. Odd*even*odd

Case1: even*odd*even
If one even number is a multiple of 2, the next even number will definitely be a multiple of 4.
Eg: 2*3*4, 10*11*12 etc..

So, number if possible values of n are
3, 5, . . . . 79.

To find number of terms of Arithmetic Progression, use Last term
79 = 3 + (n-1)2
n = 39.

Case2: Odd*even*odd
In this case, the even number SHOULD be a multiple of 8 as the other 2 numbers are odd.
So, favourable values are just the number of multiples of 8 from 2≤n≤80 which are 10 values.

So, total values of n = 39 + 10 = 49

IMO D.

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For how many integers n, such that 2≤n≤80, [m][fraction](n-1)*n*(n+1)/ 26 May 2019, 22:52
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(n-1)n(n+1)/8=Integer , where 2<=n=<80

(n^2-1)n/8=Integer ,True only when
(1) n=odd. e.g. (7^2-1)(7)/8=Integer
(2) (n^2-1)n is a multiple of 8 e.g. (8^2-1)8/8 =Integer

Odd numbers =(80-2)/2 =39
Multiples of 8 = (80-8)/8 +1 =10
Therefore total number of integers is 49
Answer D

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For how many integers n, such that 2≤n≤80, [m][fraction](n-1)*n*(n+1)/ 22 Jun 2024, 21:30
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For how many integers n, such that 2≤n≤80, \(\frac{(n-1)*n*(n+1)}{8}\) is an integer.

If n-1 is even; n is odd; One of (n-1) and (n+1) is multiple of 2 and other is multiple of 4; (n-1)(n+1) is multiple of 8; \(\frac{(n-1)*n*(n+1)}{8}\) is an integer.
n = {3, 5, 7, .... , 79}: Number of integers = (79-3)/2 + 1 = 39

If n is a multiple of 8; \(\frac{(n-1)*n*(n+1)}{8}\) is an integer.
n = {8,16, ... 80}; Number of integers = (80-8)/8 + 1 = 10

Number of integers n, such that 2≤n≤80, \(\frac{(n-1)*n*(n+1)}{8}\) is an integer = 39 + 10 = 49

IMO D
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