number 21. What is the sum of all positive integers x smaller than 100 such that x squared minus 81 is a multiple of 100? Our expression x squared minus 81 is a perfect square so we can factor it into x minus 9 and x plus 9 and to require that this expression is a multiple of 100 means that x minus 9 times x plus 9 when divided by 100 is an integer. So now let's look at the factors x plus 9 and x minus 9 more carefully. We know that they multiply to an even number and when we subtract them we obtain 18. So we have the difference of x plus 9 and x minus 9 is 18. So any number dividing both factors divides also 18. We also know that since 18 is even any two numbers that subtract to an even number and multiply to an even number both have to be even. So now with this information let's look at the possible factorizations of 100 into two numbers. We have 1 times 100, 2 times 50, 4 times 25, 5 times 20 and finally 10 times 10 and we have to choose two even numbers here but one of them has to divide 18 so that cannot be the choice 10 times 10 we choose 2 times 50. What this tells us is that one of x plus 9 or x minus 9 is a multiple of 50 and we choose our values of x accordingly. So x could be equal to 9 for example in which case x minus 9 is equal to 0. X could be equal to 41 in which case x plus 9 is equal to 50. X could be equal to 59 in which case x minus 9 is equal to 50 and finally the last choice is x is equal to 91 in which case x plus 9 is equal to 100. Those are the only values of x that are allowed so we have 0, 2 50s and a 100 as possible factors and these sum up to 200 and so our answer here is A.

positive integers

multiple of 100

factor pairs

sum

solution