duongthang wrote:integer a and n is greater than 1, the product of first 8 integers is multiple of a^n,
what is value of a
1, a^n=64
2, n=6
I think this is already discussed here, can't find the link though
I remember it read "the product of first 8 positive integers is multiple of a^n", if it's so, then we can safely assume
8! = k*a^n, where k is a positive integer
(1) If a^n = 64, with both a and n as integers greater than 1, a can take values like 2, 4, and 8; insufficient
(2) If n = 6, then 8! = k*a^6, and there's only one integer greater than 1, which has 6 or more appearances in the 8!, and that's [spoiler]
2[/spoiler]; [spoiler]
sufficient[/spoiler]
[spoiler]
B[/spoiler]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
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