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multiples of 3

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by Gopi » Sat Aug 08, 2009 1:01 am
Hi All,

This is a basic question, I am posting this problem as the answer given in my book is not matching with my answer. Request you guys to check and clarify.

Thanks in advance.
Gopi

For any integer n, such that n>3, n! denotes the product of all integers from 1 to n, inclusive. How many multiples of 3 are there in between n!-6 and n!+6, inclusive?

A) 1 B) 2 C) 3 D) 4 E) 5


Here is how I worked out this problem,

for any integer n, such that n>3 , n! will always be a multiple of 3.

=> n! is a multiple of 3
=> n!-3 is also a multiple of 3
Similarly n!-6, n!+3 and n!+6 are also multiples of 3 so in total there are 5 multiples of 3 in between n!-6 and n!+6, both inclusive.

hence my answer is E (5)


Request you guys to check this and clarify.
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Source: — Problem Solving |

by tohellandback » Sat Aug 08, 2009 1:33 am
substitute n=4,
answer is E.--5
The powers of two are bloody impolite!!
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by vittalgmat » Sat Aug 08, 2009 6:58 am
easy way is to plug in numbers.
choose n= 4
n!= 24
24-6 = 18; 24+6 = 30.
Number of multiples: 5.

E

HT helps
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