Maths Olympiad Prep

Track / Stage 4 / 184 of 340 #444 of 1964

Problem 444

AMC 12 late, AIME early
Combinatorics Difficulty 4.9 Find the answer

7.2. Find the sum of all three-digit natural numbers that do not contain the digit 0 or the digit 5.

A number or a short expression. Spacing, $ signs and \frac vs / are all fine.

Official solution

Answer: 284160. Solution. We will add the numbers in a column. Each last digit appears in the units place as many times as there are three-digit numbers ending with this digit. Therefore, it will appear 88=648 \cdot 8=64 times (since a total of 8 digits are used for the hundreds and tens places). Thus, the sum of the digits in the last place is 64(1+2+3+4+6+7+8+9)=256064 \cdot(1+2+3+4+6+7+8+9)=2560. Similarly, in the tens and hundreds places, we get the same sum. In the end, we get 2560100+256010+2560=2560111=2841602560 \cdot 100+2560 \cdot 10+2560=2560 \cdot 111=284160.

Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic, difficulty and ordering added by this site.