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: ![\[\frac{8 \cdot 9}{2} - 1 = 35,\]](http://latex.artofproblemsolving.com/6/0/0/60098bf737199b312d181dbded419c56aa2b826d.png)
which is congruent to 
. We again have 
and , so we must add ![\[1 + 9 \cdot 3 = 28\]](http://latex.artofproblemsolving.com/9/e/9/9e9c9db2bcf0c94cf1af3e2f7149dadd87b85bc3.png)
to our total. 
is congruent to 
. Thus our sum is congruent to , and 
.
Solution 2
As our first trial with 3, We can say that 
. Since if the sum of the digits of a number is divisible by 3, then the number is also divisible by 3, we reverse that logic to say that 
, and adding that up using the rainbow strategy, we get 
. We can see that since 3 divides 111, the original number is divisible by 3. But, since 111 only has one 3 and 37 has no 3's, it is not divisble by 9. Thus, the answer is 
~mathpro12345
papa I wrote this second one
See also
| 1992 AHSME (Problems • Answer Key • Resources) | ||
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Followed by Problem 18 | |
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