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

Note that 100 is congruent to 1 mod 3 and 1 mod 9 (this will be useful later on). Thus multiplying our specific 2-digit number by 100 doesn't change its modulus value, so we can write our number, preserving its value mod 3 or mod 9 as 19 + 20 + 21 + ... + 92. As we can see, for every 3 numbers (beginning with 0 mod 3 for convenience) our modulus total goes up by 3. Thus, if we define a 3-cycle to be 3 numbers that satisfy 0 mod 3 to 2 mod 3, than a 3-cycle of numbers adds up to 3 mod 3 or 0 mod 3. From 21 to 92, we have an integer number of 3-cycles, and thus can ignore them as they don't change our modulus sum. 19 and 20 add 1 + 2 = 3 to the modulus sum, so our number is divisible by 3. To calculate our number mod 9, we define a 9-cycle in the same manner as before; we have numbers congruent to 1 + 2 + 3 + ... + 8 = 36, which is 0 mod 9. Thus, we can ignore complete 9-cycles with or without the first number (as that contributes 0 to the modulus sum). Starting at 19, we have some integer number of 9-sets. The last complete 9-set terminates in 89, thus our number mod 9 is 0 + 1 + 2 = 3. Thus our number is only divisible by 3 and k = 1 .

Solution 2

Let us sum the digits: these yields 92(93)/2 - 18(19)/2. If this is divisible by 3, k >= 1. Since 3 | 18, 92 it is. Now we check if this number is divisible by 9. 92 is congruent to 2 mod 9, 93 congruent to 3 mod 9 thus resulting in the first term being congruent to 3 mod 9. Thus, our second term must be congruent to 6 mod 9 for our expression to be divisible by 9. However, 9 | 18. thus our number is not divisible by 9 and k = 1 .

See also

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.