1992 AHSME Problems/Problem 17

Revision as of 13:14, 4 January 2021 by Mathpro12345 (talk | contribs) (Solution 2)

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

As our first trial with 3, We can say that $N\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\pmod{3}$. 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 $N\equiv 19+20+21+22+...+91+92\pmod{3}$, and adding that up using the rainbow strategy, we get $N\equiv (111\times37)\pmod{3}$. 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 $\boxed{B}$

~mathpro12345

See also

1992 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 16
Followed by
Problem 18
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS