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 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 
.
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 16 |
Followed by Problem 18 | |
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