amparvardi wrote: Find four positive integers each not exceeding $70000$ and each having more than $100$ divisors. I dont know how to solve this problem in a simple way. Attempting an approach with real numbers, and considering the number as $\prod p_i^{x_i-1}$, we have to minimize $\prod p_i^{x_i-1}$ under the constraint $\prod x_i=100$ It's [rather] easy to see that this imply $x_i=\frac{(100\prod\ln(p_i))^{\frac 1n}}{\ln(p_i)}$ and $n(100\prod\ln(p_i))^{\frac 1n}\le \ln(70000)+\sum\ln(p_i)$ Numeric calculus gives then $n\ge 4$ and an easy computation gives too $n\le 6$ Above formulas give then real values for $x_i$ allowing some tests and trials for integer values. And this gives three solutions : $2^5\cdot 3^2\cdot 5^2\cdot 7=50400$ and $108$ divisors $2^6\cdot 3^3\cdot 5 \cdot 7=60480$ and $112$ divisors $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11=55440$ and $120$ divisors We get a nearly solution : $2^4\cdot 3^4\cdot 5\cdot 7=45360$ with $100$ divisors (but the request is "more than $100$" so So I got only three numbers and with heavy calculus. I wonder you we can find an olympiad-level solution to this problem. :

Hi Mr. pco. Official solution : There are five such numbers: \[69300 = 2^2 \cdot 3^2 \cdot 5^2 \cdot 7 \cdot 11 \quad : \quad 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 108 \text{ divisors};\]\[50400 = 2^5 \cdot 3^2 \cdot 5^2 \cdot 7 \quad : \quad 6 \cdot 3 \cdot 3 \cdot 2 = 108 \text{ divisors};\]\[60480 = 2^6 \cdot 3^3 \cdot 5 \cdot 7 \quad : \quad 7 \cdot 4 \cdot 2 \cdot 2 = 112 \text{ divisors};\]\[55440 = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 11 \quad : \quad 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 = 120 \text{ divisors};\]\[65520 = 2^4 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13 \quad : \quad 5 \cdot 3 \cdot 2\cdot 2 \cdot 2 = 120 \text{ divisors}.\]

amparvardi wrote: Hi Mr. pco. Official solution : There are five such numbers: \[69300 = 22 \cdot 32 \cdot 52 \cdot 7 \cdot 11 \quad : \quad 3 \cdot 3 \cdot 3 \cdot 2 \cdot 2 = 108 \text{ divisors};\] \[50400 = 25 \cdot 32 \cdot 52 \cdot 7 \quad : \quad 6 \cdot 3 \cdot 3 \cdot 2 = 108 \text{ divisors};\] \[60480 = 26 \cdot 33 \cdot 5 \cdot 7 \quad : \quad 7 \cdot 4 \cdot 2 \cdot 2 = 112 \text{ divisors};\] \[55440 = 24 \cdot 32 \cdot 5 \cdot 7 \cdot 11 \quad : \quad 5 \cdot 3 \cdot 2 \cdot 2 \cdot 2 = 120 \text{ divisors};\] \[65520 = 24 \cdot 32 \cdot 5 \cdot 7 \cdot 13 \quad : \quad 5 \cdot 3 \cdot 2\cdot 2 \cdot 2 = 120 \text{ divisors}.\] And how does the official solution suggest that students find these numbers ?

I wrote exactly the solution. There isn't anything more here, I attached the solution. This problem is sixth problem. [Solution from the book IMO Compendium.]

Attachments:
Solution.pdf (97kb)

50400 55440 60480 65520.

megarnie wrote: 50400 55440 60480 65520. good job !