Grades 11-12 Video Solutions 2021-video 2021 11-12/13

Video Transcription

Problem number 13 states, how many three-digit natural numbers have the property that when their digits are written in reverse order, the result is a three-digit number, which  is 99 more than the original number.  So let's say that we have a three-digit number ABC. Its value in terms of its digits would be 100 times A plus 10 times B plus C. If we reverse it, we would have a number CBA.  Its value in terms of its digits would be 100C plus 10B plus A. The problem tells us that CBA is 99 greater than ABC, so we would explain it mathematically  like 99 plus ABC equals CBA. Looking at the values, this is how it would look. Now let's start combining terms. Subtract CA and 10B from each side.  Let's bring the 99A to the other side and factor out the 99. Now if we divide both sides by 99, we get C minus A is equal to 1.  So the last digit must be 1 greater than the first digit. So 1B2 would be an example of a number that works. 2B3 would work, and so on through 8B9.  We cannot do 0B1 because that number would be a two-digit number. So we have eight forms of numbers, but each form can have any value of B that it wants.  It does not matter, so we have 10 options for any of the forms. So in total, we get 80 numbers.  So the question asks us, how many three-digit natural numbers have the property that when their digits are written in reverse order, the result is a three-digit number, which  is 99 more than the original number? The answer is 80.

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