Maths Olympiad Prep

Library / /5 of 377

Combinatorics Difficulty 4.0 AIME Prove it United States

Problem:
The product of the digits of a 5-digit number is 180180. How many such numbers exist?

Solution

Solution:
Let the digits be a,b,c,d,ea, b, c, d, e. Then abcde=180=22325a b c d e = 180 = 2^{2} \cdot 3^{2} \cdot 5. We observe that there are 6 ways to factor 180180 into digits a,b,c,d,ea, b, c, d, e (ignoring differences in ordering):
180=11459=11566=12259=12356=13345=22335. 180 = 1 \cdot 1 \cdot 4 \cdot 5 \cdot 9 = 1 \cdot 1 \cdot 5 \cdot 6 \cdot 6 = 1 \cdot 2 \cdot 2 \cdot 5 \cdot 9 = 1 \cdot 2 \cdot 3 \cdot 5 \cdot 6 = 1 \cdot 3 \cdot 3 \cdot 4 \cdot 5 = 2 \cdot 2 \cdot 3 \cdot 3 \cdot 5.
There are (respectively) 60,30,60,120,6060, 30, 60, 120, 60, and 3030 permutations of these breakdowns, for a total of 360360 numbers.

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.