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If x is the product of the positive integers from 1 to 9, inclusive,

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If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]

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25 May 2016, 03:27
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If x is the product of the positive integers from 1 to 9, 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. 13
[Reveal] Spoiler: OA

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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]

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25 May 2016, 03:54
x= 9!
= 9* 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= (3^2) * (2^3) * 7 * (2*3) * 5 * 2^2 * 3 * 2 * 1
= 2^7 * 3^4 * 5 * 7
= 2^i * 3^k * 5^m * 7^p

i+k+m+p = 7+4+1+1
= 13
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If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]

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31 May 2016, 14:27
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Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$

Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1

Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$

Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$

Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$

Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$

i + k + m + p = 6 + 4 + 2 + 1 = 13

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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]

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31 May 2016, 19:42
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fantaisie wrote:
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$

Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1

Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$

Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$

Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$

Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$

i + k + m + p = 6 + 4 + 2 + 1 = 13

Hi fantaisie ,
It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !!

$$\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7$$

$$\frac{9}{3}+\frac{9}{9}= 3+1 = 4$$

$$\frac{9}{5} = 1$$

$$\frac{9}{7} = 1$$

Finding the powers of a prime number p, in the n!

The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]

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01 Jun 2016, 01:56
Skywalker18 wrote:
fantaisie wrote:
Given x = 9! = $$2^i * 3^k * 5^m * 7^p$$

Find the number of powers of prime numbers: $$n / i^1 + n / i^2 ...$$ until i^z <n, z any positive number z=>1

Number of 2s in expression 9!: i = $$\frac{9}{2} + \frac{9}{8}≈ 5 + 1 = 6$$

Number of 3s in expression 9!: k = $$\frac{9}{3} + \frac{9}{9} ≈ 3 + 1 = 4$$

Number of 5s in expression 9!: m = $$\frac{9}{5} ≈ 2$$

Number of 7s in expression 9!: p = $$\frac{9}{7} ≈ 1$$

i + k + m + p = 6 + 4 + 2 + 1 = 13

Hi fantaisie ,
It seems that you have missed 2^2 while trying to find powers of 2 in 9! . Also you have rounded up in a few cases . Hope this helps !!

$$\frac{9}{2}+\frac{9}{4}+\frac{9}{8}= 4+2+1 = 7$$

$$\frac{9}{3}+\frac{9}{9}= 3+1 = 4$$

$$\frac{9}{5} = 1$$

$$\frac{9}{7} = 1$$

Finding the powers of a prime number p, in the n!

The formula is:
$$\frac{n}{p}+\frac{n}{p^2}+\frac{n}{p^3}+...+\frac{n}{p^k}$$, where $$k$$ must be chosen such that $$p^k\leq{n}$$

Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? $$\frac{9}{5}$$ = 1.8, do we only take into account actual integers? Thank you!
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]

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01 Jun 2016, 02:03
fantaisie wrote:

Thank you very much for pointing out my mistake! I have one question: what's the rule when it comes to rounding in this formula? $$\frac{9}{5}$$ = 1.8, do we only take into account actual integers? Thank you!

Hi,

rounding of is always done to the lower integer...
Reason --
say we are looking for number of 5 in 9! -
9/5 + 9/5^2 = 1.8 + 0.3 = 1+0 = 1..
so 1 in 1.8 talks of integer 5 in the product....
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Re: If x is the product of the positive integers from 1 to 9, inclusive, [#permalink]

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09 Sep 2017, 04:06
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Re: If x is the product of the positive integers from 1 to 9, inclusive,   [#permalink] 09 Sep 2017, 04:06
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