Maths Olympiad Prep

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Number theory Difficulty 5.0 AIME, harder Find the answer

The product of the digits of a 5 -digit number is 180 . How many such numbers exist?

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

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 180 into digits a,b,c,d,ea, b, c, d, e (ignoring differences in ordering): 180=180= 11459=11566=12259=12356=13345=223351 \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 30 permutations of these breakdowns, for a total of 360 numbers.

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Source: Omni-MATH, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.