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The number n is the product of the first 49 natural numbers. What is

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The number n is the product of the first 49 natural numbers. What is  [#permalink]

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Updated on: 18 Feb 2019, 01:51
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53% (02:58) correct 47% (02:51) wrong based on 150 sessions

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Question of the Week #1

The number n is the product of the first 49 natural numbers. What is the maximum possible value of p + q such that both $$\frac{n}{{(24)}^p}$$ and $$\frac{n}{{(36)}^q}$$ are integers?

A. 11
B. 15
C. 20
D. 26
E. 30

To access all the questions: Question of the Week: Consolidated List

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Originally posted by EgmatQuantExpert on 01 Jun 2018, 11:57.
Last edited by EgmatQuantExpert on 18 Feb 2019, 01:51, edited 5 times in total.
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Re: The number n is the product of the first 49 natural numbers. What is  [#permalink]

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05 Jun 2018, 11:59

Solution

Given:
• The number n is the product of the first 49 natural numbers
• The numbers $$\frac{n}{(24)^p}$$ and $$\frac{n}{(36)^q}$$ are integers

To find:
• The maximum possible value of p + q

Approach and Working:
• When it is given that $$\frac{n}{(24)^p}$$ is an integer, it necessarily means p is the highest power of 24 that can divide the number n
• In a similar way we can say that, if $$\frac{n}{(36)^q}$$ is an integer, then q is the highest power of 36 that can divide the number n

Now, n is defined as the product of the first 49 natural numbers – means n = 49!

So, effectively we are trying to find out the highest power of 24 (which is p) and 36 (which is q) respectively which can divide 49!

• Now, as the numbers 24 and 36 are composite numbers, to find the highest power of them, we need to express them in terms of the prime factors and then figure out the individual instances of those prime factors

If we factorise the numbers 24 and 36, we get
• $$24 = 2^3 * 3^1$$
• $$36 = 2^2 * 3^2$$

As the numbers 24 and 36 both consist of powers of 2 and 3 only, first we will find out the instances of 2 and 3 individually in 49!

• The number of 2s present in $$49! = \frac{49}{2} + \frac{49}{2^2} + \frac{49}{2^3} + \frac{49}{2^4} + \frac{49}{2^5} = 24 + 12 + 6 + 3 + 1 = 46$$
• The number of 3s present in $$49! = \frac{49}{3} + \frac{49}{3^2} + \frac{49}{3^3} = 16 + 5 + 1 = 22$$

Considering the number 24, as it is equal to $$2^3 * 3^1$$
• Number of $$2^3$$s present = $$\frac{46}{3}$$ = 15
• Number of $$3^1$$s present = $$\frac{22}{1}$$ = 22
• Hence, number of combinations possible for $$2^3 * 3^1 = 15$$

Therefore, we can say highest power of 24 present in 49! = max (p) = 15

In a similar way, considering the number 36, as it is equal to $$2^2 * 3^2$$
• Number of $$2^2$$s present = $$\frac{46}{2}$$ = 23
• Number of $$3^2$$s present = $$\frac{22}{2}$$ = 11
• Hence, number of combinations possible for $$2^2 * 3^2 = 11$$

Therefore, we can say highest power of 36 present in 49! = max (q) = 11

As, we have the maximum values of p and q respectively, we can say
• Max (p + q) = 15 + 11 = 26

Hence, the correct answer is option D.

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Re: The number n is the product of the first 49 natural numbers. What is  [#permalink]

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01 Jun 2018, 12:20
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3
EgmatQuantExpert wrote:
Question of the Week #1

The number n is the product of the first 49 natural numbers. What is the maximum possible value of p + q such that both $$\frac{n}{{24}^p}$$ and $$\frac{n}{{36}^q}$$ are integers?

A. 11
B. 15
C. 20
D. 26
E. 30

The number n is the product of 49 natural numbers(49!)

n will contain $$[\frac{49}{2^1}]+[\frac{49}{2^2}]+[\frac{49}{2^3}]+[\frac{49}{2^4}]+[\frac{49}{2^5}] = 24+12+6+3+1 = 46$$ 2's.

n will contain $$[\frac{49}{3^1}]+[\frac{49}{3^2}]+[\frac{49}{3^3}] = 16+5+1 = 22$$ 3's.

The reason we find out the details of the prime number 2 and 3 in 49! is
because $$24 = 2^3 * 3$$ and $$36 = 2^2 * 3^2$$ (when prime-factorized)

n which contains the product of $$2^{46} * 3^{22}$$ and we know that $$\frac{n}{{24}^p}$$ is an integer | Max value(p) = 15.
n which contains the product of $$2^{46} * 3^{22}$$ and we know that $$\frac{n}{{36}^q}$$ is an integer | Max value(q) = 11.

Therefore, the maximum possible value for the sum of p and q is 15+11 = 26(Option D)

P.S The exponent of any prime number p in n! = $$[\frac{n}{p^1}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]$$+.. where [ ] is the greatest integer function.
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Re: The number n is the product of the first 49 natural numbers. What is  [#permalink]

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01 Jun 2018, 14:36
EgmatQuantExpert wrote:
Question of the Week #1

The number n is the product of the first 49 natural numbers. What is the maximum possible value of p + q such that both $$\frac{n}{{24}^p}$$ and $$\frac{n}{{36}^q}$$ are integers?

A. 11
B. 15
C. 20
D. 26
E. 30

My question is... are these 24 raise to the power p and 36 raise to the power q or 24*p and 36*q
because in a simple look it looks like it is a multiplication not powers.
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Re: The number n is the product of the first 49 natural numbers. What is  [#permalink]

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01 Jun 2018, 17:54
dkumar2012 wrote:
EgmatQuantExpert wrote:
Question of the Week #1

The number n is the product of the first 49 natural numbers. What is the maximum possible value of p + q such that both $$\frac{n}{{24}^p}$$ and $$\frac{n}{{36}^q}$$ are integers?

A. 11
B. 15
C. 20
D. 26
E. 30

My question is... are these 24 raise to the power p and 36 raise to the power q or 24*p and 36*q
because in a simple look it looks like it is a multiplication not powers.

Hey dkumar2012,
We have got powers in the question.
Multiplication comes with a "*" starmark sign or with equal font size.
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Re: The number n is the product of the first 49 natural numbers. What is  [#permalink]

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30 Mar 2019, 10:33
1
Please could you explain why we sum up 2^1, 2^2, 2^3, 2^4 and 2^5s? And the same thing for 3s?
And then I did not understand why we divided the sum by 3, not by 6?
Would be grateful to you for the explanation.
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Re: The number n is the product of the first 49 natural numbers. What is  [#permalink]

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30 Mar 2019, 13:37
1
RusskiyLev wrote:
Please could you explain why we sum up 2^1, 2^2, 2^3, 2^4 and 2^5s? And the same thing for 3s?
And then I did not understand why we divided the sum by 3, not by 6?
Would be grateful to you for the explanation.

Hey RusskiyLev,
Request you to go through this following article. It will surely help you to understand the logic:

Variations in Factorial Manipulations
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Re: The number n is the product of the first 49 natural numbers. What is  [#permalink]

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31 Mar 2019, 00:10
given N= 49!
so 24= 2^3*3^1
powers 49!/24 = 49!/2^3*3^1
49!/2+ 49!/4+49!/8+49!/16+49!/32 = 24+12+6+3+1 ; 46
similarly for
49!/3; 16+5+1=22
highest power is 2^3 ; 46/3 = 15 and 3^1 ; 22
P(max)= 15
now for
N=49!/36
36= 2^2 * 3^2
solve similarly for
highest power ; 46/2 " for 2 power" = 23 and 22/2"for 3 power" = 11

P(max) = 11
max p+maxq = 15+11; 26 IMO D

EgmatQuantExpert wrote:
Question of the Week #1

The number n is the product of the first 49 natural numbers. What is the maximum possible value of p + q such that both $$\frac{n}{{(24)}^p}$$ and $$\frac{n}{{(36)}^q}$$ are integers?

A. 11
B. 15
C. 20
D. 26
E. 30

To access all the questions: Question of the Week: Consolidated List

Re: The number n is the product of the first 49 natural numbers. What is   [#permalink] 31 Mar 2019, 00:10
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