1992 AHSME Problems/Problem 17

Revision as of 18:14, 13 January 2018 by Johnhankock (talk | contribs) (Solution)

Problem

The 2-digit integers from 19 to 92 are written consecutively to form the integer $N=192021\cdots9192$. Suppose that $3^k$ is the highest power of 3 that is a factor of $N$. What is $k$?

$\text{(A) } 0\quad \text{(B) } 1\quad \text{(C) } 2\quad \text{(D) } 3\quad \text{(E) more than } 3$

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. 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. To calculate our number mod 9, we notice that for every complete set of 9 numbers starting at a multiple of 9, we have numbers congruent to 1 + 2 + 3 + ... + 8 = 36, which is 0 mod 9. Thus, we can ignore complete sets of 9 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 $\fbox{B}$.

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
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