Question

The repeating decimals $0.abab\overline{ab}$ and $0.abcabc\overline{abc}$ satisfy $0.abab\overline{ab} + 0.abcabc\overline{abc} = \frac{33}{37},$ where $a,$ $b,$ and $c$ are (not necessarily distinct) digits. Find the three-digit number $abc.$

答案

Solution:

Writing $ab$ and $abc$ for the two- and three-digit numbers, the decimals equal $\frac{ab}{99}$ and $\frac{abc}{999}.$ Since $99 = 9 \cdot 11$ and $999 = 27 \cdot 37,$ the common denominator is $27 \cdot 37 \cdot 11 = 10989,$ and multiplying the equation by it gives $111 \cdot ab + 11 \cdot abc = \frac{33}{37} \cdot 10989 = 9801.$

Modulo $11,$ since $9801 = 11 \cdot 891$ and $111 \equiv 1,$ this forces $11 \mid ab,$ so $a = b.$ Then $ab = 11a,$ and dividing the equation by $11$ gives $111a + abc = 891.$ Since $abc = 110a + c,$ this is $221a + c = 891,$ which requires $a = 4$ and $c = 7.$

Thus $a = b = 4,$ $c = 7,$ and the three-digit number $abc$ is $447.$

2014 AIME II Problem 4

Solution:

Problem 4 in Other Years