1992 AHSME Problems/Problem 17
Problem
The 2-digit integers from 19 to 92 are written consecutively to form the integer . Suppose that is the highest power of 3 that is a factor of . What is ?
Solution
Solution 1
We can determine if our number is divisible by or by summing the digits. Looking at the one's place, we can start out with and continue cycling though the numbers from through . For each one of these cycles, we add . This is divisible by , thus we can ignore the sum. However, this excludes , , and . These remaining units digits sum up to , which means our units sum is . As for the tens digits, for we have sets of those: which is congruent to . We again have and , so we must add to our total. is congruent to . Thus our sum is congruent to , and .
Solution 2
Every number is congruent to its digit sum mod , so but applying the result in reverse, , , etc., so the sum just become We can simplify this using the formula for the sum of an arithmetic series, giving , which is congruent to , as before. Hence our answer is .
See also
1992 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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