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. Thus multiplying our specific 2-digit number by 100 doesn't change its modulus value. As we can see, for every 3 numbers (beginning with 0 mod 3 for convenience) our modulus total goes up by 3. From 21 to 92, we have 72 numbers, for which there are the same number of 0 mod 3's, 1 mod 3's, etc. We also have 19 and 20, which add 3 to the modular total. Thus we have a number congruent to 75 mod 3 congruent to 0 mod 3. However, for k > 1, 75 must be divisible by 3^2 = 9. It is not, 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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