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

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Joined: 17 Jul 2010
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If x is the product of the positive integers from 1 to 8, in  [#permalink]

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21 Jan 2013, 15:11
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78% (01:45) correct 22% (02:15) wrong based on 426 sessions

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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 = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 12

The OG Guide and MGMAT Guide both have different solutions, a bit long. Can someone tell me if I'm doing this incorrectly.

If I'm plugging #'s in, I'm getting 2+3+4+5, = 14, but not all can be added, because not all are prime, and some numbers are repeated right, so if I take the sum of all primes in 2+3+4+5, without repeats I'll get 1+3+2+5, then I get 11? is this correct? I know 1 is not prime, and the first 2, and 4 share the same primes, so do I use 1 as a digit for 2, and use 2 as a prime # for 4? to end up with 1+3+2+5?
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Posts: 58434
Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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21 Jan 2013, 15:29
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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 = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 12

Given that $$x=8!=2^7*3^2*5*7$$. Hence, $$x=2^7*3^2*5^1*7^1=2^i * 3^k * 5^m * 7^p$$, since i, k, m, and p are positive integers, then we can equate the exponents, so we have that $$i=7$$, $$k=2$$, $$m=1$$, and $$p=1$$.

Therefore, $$i + k + m + p=7+2+1+1=11$$.

Hope it's clear.
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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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21 Jan 2013, 20:56
3
1
laythesmack23 wrote:
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 = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 12

The OG Guide and MGMAT Guide both have different solutions, a bit long. Can someone tell me if I'm doing this incorrectly.

If I'm plugging #'s in, I'm getting 2+3+4+5, = 14, but not all can be added, because not all are prime, and some numbers are repeated right, so if I take the sum of all primes in 2+3+4+5, without repeats I'll get 1+3+2+5, then I get 11? is this correct? I know 1 is not prime, and the first 2, and 4 share the same primes, so do I use 1 as a digit for 2, and use 2 as a prime # for 4? to end up with 1+3+2+5?

You cannot plug in numbers. You need to find the values of i, k, m and p.

x = 1*2*3*4*5*6*7*8 = 8!

$$x = 2^i*3^k*5^m*7^p$$

To get the value of i, you need to find the number of 2s in x i.e. 8! (including the 2s you get in 4, 6 and 8). You can quickly count - one from 2, two from 4, one from 6 and three from 8 = total seven 2s are there in 8!

To get the value of k, you need to find the number of 3s in 8!. There are two 3s in 8! (one from 3 and another from 6)
It is easy to see that there is only one 5 and one 7 in 8!.

$$x = 2^7*3^2*5^1*7^1$$

So 7 + 2 + 1 + 1 = 11

Check this post for more on powers in factorials: http://www.veritasprep.com/blog/2011/06 ... actorials/
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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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11 Aug 2013, 01:51
Stiv wrote:
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 $$x = 2^i3^k5^m7^p$$, then i + k + m + p =
A 4
B 7
C 8
D 11
E 12

$$X= 1*2*3*4*5*6*7*8$$ OR =$$1*2*3*2^2*5*(2*3)*7*2^3$$ =$$1*2^7*3^2*5*7$$
THEREFORE $$i + k + m + p = 7+2+1+1 = 11$$

HENCE D
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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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11 Aug 2013, 02:40
If given any factorial .Then the maximum power of a prime number in that factorial can be obtained by using the following formula.

If given n! then the maximum power of a prime number p in that factorial is obtained by

[n/p] + [n/p^2] + [n/p^3] + ......

where [x] is a step function and gives the greatest integer less than or equal to X.

We have to continue the above formula until we get the value of some term zero.

For example take 8!.

To find the power of 2 in 8!. We apply this formula.

[8/2] + [8/4] + [8/8] + [8/16] +....

= 4 + 2 + 1 + 0 + 0 + ..... = 7
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Re: If x is the product of the positive integers from 1 to 8, in  [#permalink]

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22 Nov 2017, 13:32
1
feellikequitting wrote:
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 = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 12

Let’s break 8! into prime factors:

8 x 7 x 6 x 5 x 4 x 3 x 2

2^3 x 7 x 2 x 3 x 5 x 2^2 x 3 x 2

2^7 x 3^2 x 5^1 x 7^1

We see that i = 7, k = 2, m = 1, and p = 1.

Thus, i + k + m + p = 7 + 2 + 1 + 1 = 11.

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

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20 Mar 2018, 16:24
feellikequitting wrote:
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 = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 12

x is the product of the positive integers from 1 to 8 = 8! = (1) (2) (3) (4) (5) (6) (7) (8) = 2^7 * 3^2 * 5^1 * 7^1

i + k + m + p = 7 + 2+ 1+1 = 11

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

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30 Dec 2018, 10:44
Top Contributor
feellikequitting wrote:
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 = 2^i * 3^k * 5^m * 7^p, then i + k + m + p =

A. 4
B. 7
C. 8
D. 11
E. 12

GIVEN: x = (8)(7)(6)(5)(4)(3)(2)(1)
Let's find the prime factorization of x by rewriting each value as the product of primes.
We get: x = (2)(2)(2)(7)(2)(3)(5)(2)(2)(3)(2)
Rearrange to get: x = (2)(2)(2)(2)(2)(2)(2)(3)(3)(5)(7)
Rewrite as powers to get: $$x = (2^7)(3^2)(5^1)(7^1)$$

We're told that $$x = (2^i)(3^k)(5^m)(7^p)$$

This means: i = 7, k = 2, m = 1 and p = 1

So, i + k + m + p = 7 + 2 + 1 + 1 = 11

Cheers,
Brent
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Re: If x is the product of the positive integers from 1 to 8, in   [#permalink] 30 Dec 2018, 10:44
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