1992 AHSME Problems/Problem 17

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

Solution 1

We can determine if our number is divisible by $3$ or $9$ by summing the digits. Looking at the one's place, we can start out with $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$ and continue cycling though the numbers from $0$ through $9$. For each one of these cycles, we add $0 + 1 + ... + 9 = 45$. This is divisible by $9$, thus we can ignore the sum. However, this excludes $19$, $90$, $91$ and $92$. These remaining units digits sum up to $9 + 1 + 2 = 12$, which means our units sum is $3 \pmod 9$. As for the tens digits, for $2, 3, 4, \cdots , 8$ we have $10$ sets of those: \[\frac{8 \cdot 9}{2} - 1 = 35,\] which is congruent to $8 \pmod 9$. We again have $19, 90, 91$ and $92$, so we must add \[1 + 9 \cdot 3 = 28\] to our total. $28$ is congruent to $1 \pmod 9$. Thus our sum is congruent to $3 \pmod 9$, and $k = 1  \implies \boxed{B}$.

Solution 2

Every number is congruent to its digit sum mod $9$, so \[N \equiv 1+9+2+0+...+9+2 \pmod 9,\] but applying the result in reverse, $1+9 \equiv 19$, $2+0 \equiv 20$, etc., so the sum just become \[19+20+...+92 \pmod 9.\] We can simplify this using the formula for the sum of an arithmetic series, giving $\frac{1}{2} \times 74 \times (19+92) = 37 \times 111$, which is congruent to $1 \times 3 = 3 \pmod 9$, as before. Hence our answer is $\boxed{\text{(B) } 1}$.

See also

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