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## Competency in Focus: Number Theory

This problem from American Mathematics Contest 10B (AMC 10B, 2019) is based on calculation of number theory. It is Question no. 19 of the AMC 10B 2019 Problem series.

## Next understand the problem

Let $S$ be the set of all positive integer divisors of $100,000.$ How many numbers are the product of two distinct elements of $S?$ $\textbf{(A) }98\qquad\textbf{(B) }100\qquad\textbf{(C) }117\qquad\textbf{(D) }119\qquad\textbf{(E) }121$
##### Source of the problem
American Mathematical Contest 2019, AMC 10B Problem 19

### Number Theory

4/10
##### Suggested Book

Challenges and Thrills in Pre College Mathematics Excursion Of Mathematics

Do you really need a hint? Try it first!
Any number is divisible by all of its factors. For eaxmple 50 is divisible by $2,5,10$ and $25$ out of these their are some prime numbers called Prime factors.
The prime factor of 100,000 are only 2 and 5, the rest of them are not the prime factor, they are composite factor. Also The prime factorization of $100,000$ is $2^5 \cdot 5^5$.
Any Number which divides 100,000 must be multiple of 2 and (or) 5. So it can be 10=5×2 or $200=2^{3} 5^{2}$.
Since prime factorization of $100,000$ is $2^5 \cdot 5^5$. Thus We can find possible value of a,b,c and d being between 0 and 5.

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