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MeghaP
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Bunuel
If n is an integer and (9! + 8! +7!)/3^n is an integer, what is the largest possible value of n?

A. 6
B. 5
C. 4
D. 3
E. 2


Hi all,
As I had mentioned in a similar type of Q, do NOT find the minimum of three in any single term and take the answer..
here in that case we will have 3s least in 7! or 8! so 7/3 =2.. ans would be 2.. BUT we have to add and take out COMMON factor to see if by addition is there any extra 3s being added..

so (9! + 8! +7!) = 7!(9*8+8+1) = 7!*81 ..
so number of 3s in 7!*18 = 2+4 = 6------ 2 in 7! and 4 in 81
ans 6..
A

Abhishek009 refer that healthy discssion on 49! and 50!, where there was a post that take the lowest as it was an easy approach...

How did you get 18??? (red part above) I tried solving this question by finding the maximum nos. of 3 in any one. Can you please explain this? Thank you.!

It's a typo should be 81.
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(9! + 8! +7!) = 7! (9 x 8 + 8 + 1) = 7! (72 + 8 + 1) = 7! x 81 = 7 x 6 x 5 x 4 x 3 x 2 x 9 x 9
= 7 x 3 x 2 x 5 x 4 x 3 x 2 x 3 x 3 x 3 x 3

So 6, answer choice A
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7!(9*8+8+1)/3^n
7! *81 /3^n
to make this expression integer n must be 6

Ans A



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