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If k is a positive integer and 150!/24^k is a positive integer, what 26 Feb 2020, 01:52
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

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Question Stats:

61% (02:39) correct 39% (02:32) wrong based on 220 sessions

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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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C
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If k is a positive integer and 150!/24^k is a positive integer, what 26 Feb 2020, 02:19
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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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If k is a positive integer and 150!/24^k is a positive integer, what 26 Feb 2020, 02:38
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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
Show more

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 \(2^3*3\)

Answer: Option C
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If k is a positive integer and 150!/24^k is a positive integer, what 26 Feb 2020, 08:05
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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
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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.

Ans (C)
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If k is a positive integer and 150!/24^k is a positive integer, what 26 Feb 2020, 14:04
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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 ?

\(24^{k}=(8∗3)^{k}=2^{3k}∗3^{k}\\
\)
--> 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 \(48\).

The answer is C.
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If k is a positive integer and 150!/24^k is a positive integer, what 26 Feb 2020, 14:27
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The greatest possible value of k is 48

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If k is a positive integer and 150!/24^k is a positive integer, what 26 Feb 2020, 17:38
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\(150!/(2^{3k}.3^k)\)

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.

FINAL ANSWER IS (C)
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If k is a positive integer and 150!/24^k is a positive integer, what 27 Feb 2020, 00:03
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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
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If k is a positive integer and 150!/24^k is a positive integer, what 05 May 2023, 06:06
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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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Exactly the correct way - just a shortcut for the number of powers. You don't have to actually square/cube/etc. the denominator, you can just take the previous value and ignore the remainder:

For 3's:
150/3 = 50
50/3 = 16
16/3 = 5
5/3 = 1

50+16+5+1 = 72

For 2's:
150/2 = 75
75/2 =37
37/2 =18
18/2 =9
9/2 = 4
4/2 = 2
2/2 = 1

75+ 37 + 18 + 9 + 4 + 2 + 1 = 142
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If k is a positive integer and 150!/24^k is a positive integer, what 27 Feb 2026, 05:12
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