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If P is the product of all the integers from 1 to 36 inclusive.

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Kaushik786
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If P is the product of all the integers from 1 to 36 inclusive. [#permalink] New post   09 Mar 2021, 02:33
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If P is the product of all the integers from 1 to 36 inclusive, what is the greatest integer i for which 12^i is a factor of P ?

A. 17
B. 18
C. 19
D. 20
E. 21
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Re: If P is the product of all the integers from 1 to 36 inclusive. [#permalink] New post   09 Mar 2021, 02:55
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Kaushik786 wrote:
If P is the product of all the integers from 1 to 36 inclusive, what is the greatest integer i for which 12^i is a factor of P ?

A. 17
B. 18
C. 19
D. 20
E. 21


P = 36!
We need to find the highest power of 12 in 36!
12 = 2² x 3

Highest power of 2 in 36! is:
36/2 = 18
18/2 = 9
9/2 = 4.5 => 4
4/2 = 2
2/2 = 1
So we have: 18+9+4+2+1 = 34
So, highest power of 2² = 34/2 = 17

Highest power of 3 in 36! is:
36/3 = 12
12/3 = 4
4/3 = 1.33 => 1
So we have: 12+4+1 = 17
So, highest power of 3 = 17

Hence, highest power of 12 in 36! is also 17

Answer A

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If P is the product of all the integers from 1 to 36 inclusive. [#permalink] New post   09 Mar 2021, 07:57
Kaushik786 wrote:
If P is the product of all the integers from 1 to 36 inclusive, what is the greatest integer i for which 12^i is a factor of P ?

A. 17
B. 18
C. 19
D. 20
E. 21


the formula is:
\(\frac{n}{k}+\frac{n}{k^2}+\frac{n}{k^3}+ ... till n>k^2\)

Expressing 12 as \(2^2 * 3\) and taking individual cases:

Greatest integer of \(2^2\) in 36! = Greatest integer of 3 in 36! = 17 ------------ option A
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Re: If P is the product of all the integers from 1 to 36 inclusive. [#permalink] New post   11 Mar 2021, 10:28
sujoykrdatta wrote:
Kaushik786 wrote:
If P is the product of all the integers from 1 to 36 inclusive, what is the greatest integer i for which 12^i is a factor of P ?

A. 17
B. 18
C. 19
D. 20
E. 21


P = 36!
We need to find the highest power of 12 in 36!
12 = 2² x 3

Highest power of 2 in 36! is:
36/2 = 18
18/2 = 9
9/2 = 4.5 => 4
4/2 = 2
2/2 = 1
So we have: 18+9+4+2+1 = 34
So, highest power of 2² = 34/2 = 17

Highest power of 3 in 36! is:
36/3 = 12
12/3 = 4
4/3 = 1.33 => 1
So we have: 12+4+1 = 17
So, highest power of 3 = 17

Hence, highest power of 12 in 36! is also 17

Answer A

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Request your kind explanation on how your method is accounting for all the powers of 2 in 36!

Thank You in advance.

Shashank Singh
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If P is the product of all the integers from 1 to 36 inclusive. [#permalink] New post   11 Mar 2021, 10:47
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s1804singh wrote:
sujoykrdatta wrote:
Kaushik786 wrote:
If P is the product of all the integers from 1 to 36 inclusive, what is the greatest integer i for which 12^i is a factor of P ?

A. 17
B. 18
C. 19
D. 20
E. 21


P = 36!
We need to find the highest power of 12 in 36!
12 = 2² x 3

Highest power of 2 in 36! is:
36/2 = 18
18/2 = 9
9/2 = 4.5 => 4
4/2 = 2
2/2 = 1
So we have: 18+9+4+2+1 = 34
So, highest power of 2² = 34/2 = 17

Highest power of 3 in 36! is:
36/3 = 12
12/3 = 4
4/3 = 1.33 => 1
So we have: 12+4+1 = 17
So, highest power of 3 = 17

Hence, highest power of 12 in 36! is also 17

Answer A

Posted from my mobile device



Request your kind explanation on how your method is accounting for all the powers of 2 in 36!

Thank You in advance.

Shashank Singh


s1804singh
I am taking the liberty to explain
36/2 = 18
18/2 ( same as 36/2^2) = 9
So shortcut is to get the quotient from the previous power ,in this case 18 and divide by 2 and keep repeating by dividing the previous quotient by 2 till you get quotient 1
9/2 (same as 36/2^3) = 4.5 => 4
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Re: If P is the product of all the integers from 1 to 36 inclusive. [#permalink] New post   23 May 2023, 12:30
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