2017 AMC 10 B
Complete problem set with solutions and individual problem pages
Mary thought of a positive twodigit number. She multiplied it by and added . Then she switched the digits of the result, obtaining a number between and , inclusive. What was Mary's number? (2017 AMC 10B Problem, Question#1)
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Let her digit number be . Multiplying by makes it a multiple of , meaning that the sum of its digits is divisible by . Adding on increases the sum of the digits by , and reversing the digits keeps the sum of the digits the same; this means that the resulting number must be more than a multiple of . There are two such numbers between and : and .Now that we have narrowed down the choices, we can simply test the answers to see which one will provide a twodigit number when the steps are reversed:For ,we reverse the digits, resulting in Subtracting , we get .We can already see that dividing this by will not be a two-digit number, so does not meet our requirements.Therefore, the answer must be the reversed steps applied to .We have the following: Therefore, our answer is .
Working backwards, we reverse the digits of each number from and subtract from each, so we have ,,,,The only numbers from this list that are divisible by are and . We divide both by , yielding and . Since is not a twodigit number, the answer is .
You can just plug in the numbers to see which one works. When you get to , you multiply by and add to get . When you reverse the digits of , you get , which is within the given range. Thus, the answer is .
