If the integers a and n are greater than 1 and the product of first 8 positive integers is a multiple of a^n , what is the value of a?
1) a^n = 64
2) n = 6
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nisagl750
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S1: a=2,n=6prat_agl wrote:If the integers a and n are greater than 1 and the product of first 8 positive integers is a multiple of a^n , what is the value of a?
1) a^n = 64
2) n = 6
a=4,n=3
a=8,n=2
Insuff:
S2: n=6, So a=2
Suff
IMO B
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eaakbari
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IMO C
(I) a could be 2,4,8 Hence Insuff
(II) n is 6 but but we have no idea about a^n or a. Hence Insuff.
(I) &(II)
a^6 = 64 only corresponds to a = 2
Hence Suff
Answer C
(I) a could be 2,4,8 Hence Insuff
(II) n is 6 but but we have no idea about a^n or a. Hence Insuff.
(I) &(II)
a^6 = 64 only corresponds to a = 2
Hence Suff
Answer C
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nisagl750
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When n=6, Only integer 2 can satisfy i.e only 2^6 will be a factor of multiples of first 8 integers.eaakbari wrote:
(II) n is 6 but but we have no idea about a^n or a. Hence Insuff.
First 8 multiples will include: 1x2x3x4x5x6x7x8
= 1*2*3*(2^2)*5*(2*3)*7*(2^3)
= 1* (2^7) * (3^3) * 5 * 7
So only sixth power of 2 will satisfy.... i.e a=2(remember a>1) Suff.
Hence B
