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If x is the product of the positive integers from 1 to 8, [#permalink]

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25 Apr 2007, 22:50

This topic is locked. If you want to discuss this question please re-post it in the respective forum.

If x is the product of the positive integers from 1 to 8, inclusive, and if i, k, m, and p are positive integers such that x = 2i3k5m7 p, then i + k + m + p =

A. 4

B. 7

C. 8

D. 11

E. 12

PS: I do not have the OA. A friend of mine asked and he doesn have the OA either..
When I tried solving, I get 15 as the minimum answer..[/b]

If it's (2^7)(3^2)(5)(7) = (2^i )( 3^k )( 5^m )(7^p) , then yes, i=7.

But since you wrote 2i3k5m7p, that's how I ended up with i =6 .

yu r right...
but if (2^7)(3^2)(5)(7) = (2*i)(3*k)(5*m)(7*p)

I am still not able to figure out how i is 6
we can write the above equation as (2*64)(3*3)(5)(7) = (2*i)(3*k)(5*m)(7*p)
correct??
or I gues mt mind is not working

If it's (2^7)(3^2)(5)(7) = (2^i )( 3^k )( 5^m )(7^p) , then yes, i=7.

But since you wrote 2i3k5m7p, that's how I ended up with i =6 .

yu r right... but if (2^7)(3^2)(5)(7) = (2*i)(3*k)(5*m)(7*p)

I am still not able to figure out how i is 6 we can write the above equation as (2*64)(3*3)(5)(7) = (2*i)(3*k)(5*m)(7*p) correct?? or I gues mt mind is not working

ywilfred didnot say that i = 6 is wright. i = 7.

also * is multiplication. if you still use *, it confuses everybody.
for power, you can use ^.