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
We can determine if our number is divisible by 3 or 9 by summing the digits. Looking at the one's place, we can start out with 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 and continue cycling though the numbers from 0 through 9. For each one of these cycles, we add 0 + 1 + ... + 9 => 45. This is divisible by 9, thus we can ignore the sum. However, this excludes 19, 90, 91 and 92. These remaining units digits sum up to 9 + 1 + 2 => 12, which means our units sum is 3 mod 9. As for the tens digits, for 2, 3, 4, ..., 8 we have 10 sets of those -> 8(9)/2 - 1 => 35, which is congruent to 8 mod 9. We again have 19, 90, 91 and 92, so we must add 1 + 9 * 3 => 28 to our total. 28 is congruent to 1 mod 9. Thus our sum is congruent to 3 mod 9, and k = 1 
.
Alternate Solution
Every number is congruent to its digit sum mod 
, so is congruent to 
mod , but applying the result in reverse, 
is congruent to 
, 
is congruent to 
, etc., so the sum just become 
mod . We can simplify this using the formula for the sum of an arithmetic series, giving 
, which is congruent to 
mod , as before.
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
| 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 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
