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If x is the product of the positive integers from 1 to 8,

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If x is the product of the positive integers from 1 to 8, [#permalink] New post 25 Apr 2007, 21:50
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]
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 [#permalink] New post 25 Apr 2007, 22:45
are you sure the question is correct?

the way i see it i*k*m*p = 4*6*8

but the greatest answer is 12, and the greatest product of 4 numbers that adds to 12 is 81, which is less than 4*6*8

im a bit rusty with this though
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 [#permalink] New post 25 Apr 2007, 22:49
Ok, I am boldy going to say that the answer choices are messed up :)

Let's see -

x = 8!

Also, x = 2i3k5m7p => ikmp = 192

now, factoring 192 = 2 x 2 x 2 x 2 x 2 x 2 x 3

We see that one of ikmp is definitely 3 and remaining are even numbers

this implies i + k + m + p = ODD

So from given answer choices , only B and D hold water.

Also, the minimum sum = 15 (3 + 4 + 4 + 4)

IF, there was no condition that i,k,m and p are positive, then choice B is the answer
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 [#permalink] New post 25 Apr 2007, 23:18
x= 8! = 8*7*6*5*4*3*2*1 = (2^7)(3^2)(5)(7)

So i = 6, k=3, m=1, p=1 (if 2i3k5m7p meant 2*i*3*k*5*m*7*p)

Then i+k+m+p = 11
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 [#permalink] New post 25 Apr 2007, 23:56
well,

in other words,

2*i*3*k*5*m*7*p = (2^7)(3^2)(5)(7) ----- (1)

Cancelling out common terms,

i * k * m * p = (2^6)(3) = 192 ----- (2)

taking values of i = 6 / k=3 / m,p = 1 does not satify the (2)
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hi [#permalink] New post 26 Apr 2007, 00:36
I agree with grand_mba.. I used the same approach and got 15 as the answer.. Since I don't have the answer with me I think 15 IS the correct answer..
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 [#permalink] New post 26 Apr 2007, 18:25
ywilfred wrote:
x= 8! = 8*7*6*5*4*3*2*1 = (2^7)(3^2)(5)(7)

So i = 6, k=3, m=1, p=1 (if 2i3k5m7p meant 2*i*3*k*5*m*7*p)

Then i+k+m+p = 11


Yes, D.

I think the problem meant 2^i + 3^k + 5^m + 7^p.

not 2xi + 3xk and so on...
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 [#permalink] New post 26 Apr 2007, 22:22
ricokevin wrote:
ywilfred wrote:
x= 8! = 8*7*6*5*4*3*2*1 = (2^7)(3^2)(5)(7)

So i = 6, k=3, m=1, p=1 (if 2i3k5m7p meant 2*i*3*k*5*m*7*p)

Then i+k+m+p = 11


Yes, D.

I think the problem meant 2^i + 3^k + 5^m + 7^p.

not 2xi + 3xk and so on...


I agree - excellent thinking Ricokevin ! :-D
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 [#permalink] New post 30 Apr 2007, 00:56
ricokevin wrote:
ywilfred wrote:
x= 8! = 8*7*6*5*4*3*2*1 = (2^7)(3^2)(5)(7)

So i = 6, k=3, m=1, p=1 (if 2i3k5m7p meant 2*i*3*k*5*m*7*p)

Then i+k+m+p = 11


Yes, D.

I think the problem meant 2^i + 3^k + 5^m + 7^p.

not 2xi + 3xk and so on...


if (2^7)(3^2)(5)(7) = (2^i )( 3^k )( 5^m )(7^p)
can yu pls elaborate how yu got the values as i=6, k= 3, m=1 , p = 1
isn't i=7, k=2, m=1, p=1 ..
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 [#permalink] New post 30 Apr 2007, 01:03
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 .
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 [#permalink] New post 30 Apr 2007, 01:22
ywilfred wrote:
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 :cry:
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 [#permalink] New post 30 Apr 2007, 10:11
JyotiBharadwaj wrote:
ywilfred wrote:
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 :cry:


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 ^.
  [#permalink] 30 Apr 2007, 10:11
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