2 2 s in Divisor Product of 36000 36000

Let P n P_n be the product of all positive divisors of a positive integer n n . For example, P 1 = 1 P_1 = 1 and P 4 = 1 × 2 × 4 = 8. P_4 = 1 \times 2 \times 4 = 8.

If 2 m 2^m is the highest integer power of 2 2 that divides P 36000 P_{36000} , what is m m ?


The answer is 180.

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3 solutions

Note that 36000 = 2 5 3 2 5 3 36000 = 2^5 \cdot 3^2 \cdot 5^3 . Also, note that a divisor of 36000 36000 is of the form 2 a 3 b 5 c ( a , b , c N 0 ) 2^a \cdot 3^b \cdot 5^c (a, b, c \in \mathbb{N}_0) , where a 5 a \leq 5 , b 2 b \leq 2 and c 3 c \leq 3 . Then, for a fixed a a , 2 a 2^a appears 3 4 = 12 3 \cdot 4\ = 12 times among the divisors of 36000 36000 .

Knowing that, as a a takes values from 0 0 to 5 5 , the highest integer power of 2 2 that divides P 36000 P_{36000} is i = 0 5 ( 2 i ) 12 = ( 2 1 + 2 + 3 + 4 + 5 ) 12 = 2 180 \displaystyle\prod_{i=0}^{5} {(2^i)}^{12} = {(2^{1+2+3+4+5})}^{12} = 2^{180} .

Yes, that is the one!

Muhammad Rasel Parvej - 3 years, 3 months ago
Giorgos K.
Feb 28, 2018

Here is a Mathematica code that outputs the Prime factors of this product

FactorInteger[Times @@ Divisors@36000]

which returns

{{2, 180}, {3, 72}, {5, 108}}

so, the answer is 180

Which computer language is this?

Ong Zi Qian - 3 years, 3 months ago

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Mathematica ...

Giorgos K. - 3 years, 3 months ago

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Oops, sorry. I have not heard it before.

Ong Zi Qian - 3 years, 3 months ago

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@Ong Zi Qian you can test Mathematica online here https://sandbox.open.wolframcloud.com
Just paste the code and hit shift+enter to run
Also under the "Help" tab you will find the documentation which is an excellent guide to the features of this language

Giorgos K. - 3 years, 3 months ago

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@Giorgos K. Oh, I got it. Thank you!

Ong Zi Qian - 3 years, 3 months ago

@Giorgos K. But why do I cannot toggle your LaTeX?

Ong Zi Qian - 3 years, 3 months ago

First of all note that, 36000 = 2 5 × 3 2 × 5 3 36000= 2^5×3^2×5^3 . As we know that from number theory,

the product of positive divisors of n N = n τ ( n ) 2 n∈N =n^{\frac{τ(n)}{2}} where τ ( n ) = τ(n)= number of positive divisors of n N n∈N .

τ ( 36000 ) = τ ( 2 5 × 3 2 × 5 3 ) τ(36000) =τ( 2^5×3^2×5^3) = ( 5 + 1 ) × ( 2 + 1 ) × ( 3 + 1 ) = (5+1)×(2+1)×(3+1) = 72 = 72

This imply,

P 36000 = 3600 0 36 P_{36000}= 36000^{36} = ( 2 5 × 3 2 × 5 3 ) 36 =(2^5×3^2×5^3)^{36} = ( 2 5 ) 36 × ( 3 2 ) 36 × ( 5 3 ) 36 = (2^5)^{36} × (3^2)^{36} ×(5^3)^{36}

Which clearly imply the highest integer powers of 2 2 that divides P 36000 P_{36000} is ( 2 5 ) 36 = 2 180 (2^5)^{36}= 2^{180} this imply m = 180 m=180

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