Matematické Fórum

stuart clark

(A) Total number of divisers of $10!$ which are is in the form of $3n$

(B) Total number of divisers of $10!$ which are is in the form of $3n+1$

(C) Total number of divisers of $11!$ which are is in the form of $4n+2$

(D) Total number of divisers of $11!$ which are is in the form of $3n+2$

where $n\in \mathbb{N}+\{0\}$

vanok · 19. 09. 2013 14:52

Hi ↑ stuart clark:,
To solve proposed problems one can use the function $\tau$ , number of divisors.
http://2000clicks.com/mathhelp/NumberTh … rsTau.aspx

I solve that the first problem (the others have the similar solution)

First of all $10!=2^8.3^4.5^2.7$
I notice that $k=\tau (2^8.5^2.7)=(1+8)(1+2)(1+1)=9.3.2=54$
So we have $k$ numbers of the form $3n$ , with n not divisible by 3
and $k$ numbers of the form $3^2n$ , with n not divisible by 3
$k$ numbers of the form $3^3n$ , with n not divisible by 3
$k$ numbers of the form $3^4n$ , with n not divisible by 3

So we have $4 \cdot 54 = 216$ such numbers.

stuart clark · 23. 09. 2013 06:37

↑ vanok: Thanks

Got it.

Matematické Fórum

#1 18. 09. 2013 09:08 — Editoval stuart clark (18. 09. 2013 09:10)

number of divisers

#2 19. 09. 2013 14:52

Re: number of divisers

#3 23. 09. 2013 06:37

Re: number of divisers

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