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# If k is a positive integer and 150!/24^k is a positive integer, what

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If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 00:52
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If k is a positive integer and $$\frac{150!}{24^k}$$ is a positive integer, what is the greatest possible value of k ?

A. 46
B. 47
C. 48
D. 49
E. 50

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Re: If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 01:19
Let's do the prime factorizing of 24 -> $$2^3 * 3$$

So, the maximum power of 3 that can satisfy will be -
$$\frac{150}{3^1}$$ + $$\frac{150}{3^2}$$ + $$\frac{150}{3^3}$$ + $$\frac{150}{3^4}$$ = 50 + 16 + 5 + 1 = 72

So, the maximum power of 2 that can satisfy will be -
$$\frac{150}{2^1}$$ + $$\frac{150}{2^2}$$ + $$\frac{150}{2^3}$$ + $$\frac{150}{2^4}$$ + $$\frac{150}{2^5}$$ + $$\frac{150}{2^6}$$ + $$\frac{150}{2^7}$$ = 75 + 37 + 18 + 9 + 4 + 2 + 1 = 146

So, the maximum power of 8 ($$2^3 = 8$$) that can satisfy will be -> $$\frac{146}{3}$$ and get = 48

We take the minimum power, hence '48 = 'C' is the winner
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Re: If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 01:38
1
Quote:
If k is a positive integer and $$150!/24^k$$ is a positive integer, what is the greatest possible value of k ?

A. 46
B. 47
C. 48
D. 49
E. 50

CONCEPT: Power of any Prime Number in any factorial can be calculated by following understanding  Power of prime x in n! = [n/x] + [n/x^2] + [n/x^3] + [n/x^4] + ... and so on Where,

[n/x] = No. of Integers that are multiple of x from 1 to n

[n/x^2] = No. of Integers that are multiple of x^2 from 1 to n whose first power has been counted in previous step and second is being counted at this step

[n/x^3] = No. of Integers that are multiple of x^3 from 1 to n whose first two powers have been counted in previous two step and third power is counted at this step
And so on.....

Where [n/x] is greatest Integer value of (n/x) less than or equal to (n/x) i.e. [100/3] = [33.33] = 33 i.e. [100/9] = [11.11] = 11 etc.

So $$24 = 2^3*3$$

Now, Power of prime 2 in 150! = [150/2] + [150/2^2] + [150/2^3] + [150/2^4] + ... = 75+37+18+9+4+2+1 = 146
Now, Power of prime 3 in 150! = [150/3] + [150/3^2] + [150/3^3] + [150/3^4] + ... = 50+16+5+1 = 72

i.e. $$150! = 2^{146}*3^{72}*...$$

i.e. $$150! = (2^3)^{48}*3^{72}*2^2*... = (24)^{48}*2^2*3^{24}*...$$

i.e. We get maximum of 48 pairs of $$2^3*3$$

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Re: If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 07:05
Quote:
If k is a positive integer and 150!/24^k is a positive integer, what is the greatest possible value of k ?

A. 46
B. 47
C. 48
D. 49
E. 50

150!/24^k=150!/(8*3)^k=150!/(2^3*3)^k

two's in 150!: sum of quotient = 150/2^(1…2…3…4…n)
two's in 150!: 150/{2,4,8,16,32,64,128}=75+37+18+9+4+2+1=146
eight's in 150!: quotient 146/3=48

Since there are more three's than eight's, then it is safe to assume that largest k is 48.

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Re: If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 13:04
If k is a positive integer and $$\frac{150!}{24^{k}}$$ is a positive integer, what is the greatest possible value of k ?

$$24^{k}=(8∗3)^{k}=2^{3k}∗3^{k}$$
--> In order to get the greatest possible value of k, we need to find the number of 2s and 3s in 150!:

$$[\frac{150}{2}]+[\frac{150}{2^{2}}]+[\frac{150}{2^{3}}]+[\frac{150}{2^{4}}]+[\frac{150}{2^{5}}]+[\frac{150}{2^{6}}]+[\frac{150}{2^{7}}]+...=$$

=$$75+37+18+9+4+2+1= 146$$ --> $$3k= 146$$ --> $$k = 48 \frac{2}{3}$$

$$[\frac{150}{3}]+[\frac{150}{3^{2}}]+[\frac{150}{3^{3}}]+[\frac{150}{3^{4}}]+...=$$

$$=50+16+5+1=71$$
-->$$k=71$$

Well, The greatest possible value of $$k$$ is $$48$$.

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Re: If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 13:27
The greatest possible value of k is 48

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Re: If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 16:38
1
$$150!/(2^{3k}.3^k)$$

The greatest value of positive integer k can be determined by following:

--> The maximum number for $$2^{3k}$$: $$[150/2+150/(2^2)+150/(2^3)+150/(2^4)+150/(2^5)+150/(2^6)+150/(2^7)]*{1/3}=(75+37+18+9+4+2+1)/3=146/3=48$$ with remainder of 2

--> The maximum number for $$3^{k}$$: $$[150/3+150/(3^2)+150/(3^3)+150/(3^4)]=(50+16+5+1)=72$$

The greatest value of positive integer k is min (48, 72) = 48.

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Re: If k is a positive integer and 150!/24^k is a positive integer, what  [#permalink]

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26 Feb 2020, 23:03
24 = 2^3*3

so we have to calculate the no.of 8's in 150!

|150/2|=75
|75/2|=37
|37/2|=18
|18/2|=9
|9/2|=4
|4/2|=2
|2/2|=1

sum = 146

so there are 146 2's in 1501

but we need the no.of 8's,......which we can get by |146/3| = 48

there are 48 8's in 150!

OA:C
Re: If k is a positive integer and 150!/24^k is a positive integer, what   [#permalink] 26 Feb 2020, 23:03
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