115500 needs positive divisor

The total number of proper positive divisors of 115500 is

Note: The term "proper divisors" excludes the number itself.


The answer is 95.

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1 solution

Aditya Raut
Jul 13, 2014

The number given is

115500 = 100 × 11 × 105 = 2 2 . 3. 5 3 . 7.11 115500 = 100\times 11 \times 105 = 2^2.3.5^3.7. 11

Thus the number of factors is 3 × 2 × 4 × 2 × 2 = 96 3\times 2\times 4\times 2\times 2 = 96 and this includes the number itself. But in PROPER divisors, we have to exclude the number itself, hence the answer is 95 \boxed{95}

Where do you get the 3 × 2 × 4 × 2 × 2 = 96 3\times 2 \times 4 \times 2\times 2=96 ? :o

Astro Enthusiast - 6 years, 10 months ago

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Precious: Notice that the factor 2 can be included 0, 1 or 2 times, which is 3 possibilities. Likewise, the factor 3 can be included 0 or 1 times, which is 2 possibilities. 5 can be included from 0 to 3 times, which is 4 possibilities. And so on.

Bill Bell - 6 years, 10 months ago

R e s p o n s e Response to A s t r o Astro E n t h u s i a s t Enthusiast : Actually there is a formula that for every positive composite number N N and its distinct prime factors a , b , c , d . . . . . . . . . a,b,c,d......... with p , q , r , s . . . . . . . . . . . . . p,q,r,s............. being the indices of a , b , c , d . . . . . . . . a,b,c,d........ .

N = a p × b q × c r × d s × . . . . . . . . . . . . N=a^{p} \times b^{q} \times c^{r} \times d^{s} \times ............

Then total factors of N = ( p + 1 ) ( q + 1 ) ( r + 1 ) ( s + 1 ) . . . . . . . . . . . . . . . . . . . N=(p+1)(q+1)(r+1)(s+1)...................

Abhisek Mohanty - 6 years, 2 months ago

Proper divisors only excludes the number itself, but include 1. So the answer should be 96-1=95.

I've updated the answer to 95.

Calvin Lin Staff - 6 years, 11 months ago

number of factors and number Proper divisors are equal, isn't??

Arun AR - 6 years, 10 months ago

The correct answer is 94

Hariom Yadav - 3 years, 5 months ago

Yes the answer should be 96

akhilesh agrawal - 6 years, 11 months ago

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