Grades 9-10 Video Solutions 2023-2023_9-10

Video Transcription

How many three-digit positive integers x are there such that subtracting the sum of digits of x from x gives a three-digit number whose digits are all  the same. So let's let the three-digit number x be represented as a b and c where a b and c are all digits. Then we know that we can write x as 100 a plus  10 b plus c. Then calculating the difference between x and the sum of its digits we find that 100 a plus 10 b plus c minus a minus b minus c is 99 a plus  plus 9 b. Clearly this is divisible by 9 so we want a three-digit number whose digits are all the same and is divisible by 9. Using the divisibility  rules for 9 we can find that this number can only be 3 3 3 or 6 6 6. Now when a equals 3 and b equals 4 we find that 3 times 3 times 99 plus 4 times 9  is 333 and because we have no limit on what c can be then any 3 4 0 3 4 1 all the way to 3 4 9 would work. Similarly we find that a equals 6 and b equals 8  means that 99 a plus 9 b would give 6 6 6. Then because there's no bounds on c 6 8 0 6 8 1 all the way to 6 8 9 would be valid as x. Then we found a total of  d20 numbers that can be x and satisfy our properties.

Video Summary

There are 20 three-digit positive integers \( x \) that satisfy the given conditions. Integers \( x \) are represented as 340 to 349 and 680 to 689. The process involves subtracting the sum of the digits of \( x \) from \( x \) to yield another three-digit number whose digits are all the same. The possible results of this subtraction are 333 or 666. By systematically choosing feasible \( a \) and \( b \) values (where \( a = 3, b = 4 \) and \( a = 6, b = 8 \)), each scenario provides 10 possible \( x \) values, resulting in 20 solutions overall.

Keywords

three-digit integers

digit subtraction

333

666

solutions