Factors 2

How many factors of 40 ! 40! are perfect cubes ?

Notation :
! ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × × 8 8! = 1\times2\times3\times\cdots\times8 .


The answer is 2912.

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1 solution

Margaret Yu
Jun 3, 2016

40 ! = 2 38 × 3 18 × 5 9 × 7 5 × 1 1 3 × 1 3 3 × 1 7 2 × 1 9 2 × 23 × 29 × 31 × 37 40! = 2^{38} \times 3^{18} \times 5^{9} \times 7^{5} \times 11^{3} \times 13^{3} \times 17^{2} \times 19^{2} \times 23 \times 29 \times 31 \times 37

40 ! = 2 36 × 2 2 × 3 18 × 5 9 × 7 3 × 7 2 × 1 1 3 × 1 3 3 × 1 7 2 × 1 9 2 × 23 × 29 × 31 × 37 40! = 2^{36} \times 2^{2} \times 3^{18} \times 5^{9} \times 7^{3} \times 7^{2} \times 11^{3} \times 13^{3} \times 17^{2} \times 19^{2} \times 23 \times 29 \times 31 \times 37

40 ! = ( 2 3 ) 12 × ( 3 3 ) 6 × ( 5 3 ) 3 × ( 7 3 ) 1 × ( 1 1 3 ) 1 × ( 1 3 3 ) 1 × 2 2 × 7 2 × 1 7 2 × 1 9 2 × 23 × 29 × 31 × 37 40! = (2^{3})^{12} \times (3^{3})^{6} \times (5^{3})^{3} \times (7^{3})^{1} \times (11^{3})^{1} \times (13^{3})^{1} \times 2^{2} \times 7^{2} \times 17^{2} \times 19^{2} \times 23 \times 29 \times 31 \times 37

( 12 + 1 ) × ( 6 + 1 ) × ( 3 + 1 ) × ( 1 + 1 ) × ( 1 + 1 ) × ( 1 + 1 ) (12+1) \times (6+1) \times (3+1) \times (1+1) \times (1+1) \times (1+1)

13 × 7 × 4 × 2 × 2 × 2 = 2912 13 \times 7 \times 4 \times 2 \times 2 \times 2 = \boxed{2912}

good approach..+1

Ayush G Rai - 5 years ago

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