Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

If x is the product of the positive integers from 1 to 8, [#permalink]
25 Apr 2007, 21:50

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 ^.