2017 AMC 10B Problems/Problem 1
Contents
Problem
Mary thought of a positive two-digit 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?
Solution
Solution 1
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 (we can ignore numbers such as that do not satisfy this since the numbers must be between and ) 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 two-digit 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 .
Solution 2
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 two-digit number, the answer is .
Solution 3
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 .
Video Solution
~savannahsolver
Video Solution by TheBeautyofMath
With whiteboard at home: https://www.youtube.com/watch?v=zTGuz6EoBWY
~IceMatrix
See Also
2017 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by First Problem |
Followed by Problem 2 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.