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Ainat [17]
3 years ago
9

If the integers a and n are greater than 1 and the product of the first 8 positive integers is a multiple of a^n, what is the va

lue of a?(1) a^n = 64(2) n=6
Mathematics
1 answer:
daser333 [38]3 years ago
3 0

Answer:

a=2

Step-by-step explanation:

Well, first let us find product of the first 8 positive integers, which is 1*2*3*4*5*6*7*8=40320. It can also be written as 40320=2^{7}*3²*5*7. From the formula above, it can be extracted that if a=2, n can be any number from 3 to 7 (considering that a and n are not equal and greater than 1).

In (1) a^n=64 and in (2) n=6 so it can be written as 2^6=64 so a=2 and n=6. So the value of a is 2. a=2

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