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

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

papa I wrote this second one

Solution 3 (Longer, but less knowledge of modular arithmetic needed)

We know that a number is divisible by $3$ if the sum of its digits is divisible by $3$ . Observe that the sum of all digits in the given integer would be: $7 \cdot \sum_{n=1}^{9} n$ for all units digits of the integers $20$ through $90$ (we exclude $0$ from the sum since it adds nothing to the total) $10+10 \cdot \sum_{n=2}^{8} n$ for every digit in the $10$ 's place of the integers $20$ through 89 $. This excludes the digits from$ 19 $,$ 90 $(excluded from the second sum, adds nothing to the first),$ 91 $,$ 92$, so we compute their sum with <cmath>1+9+9+9+1+9+2</cmath>

Now, let's compute all of our sums <cmath>7 \cdot \sum_{n=1}^{9} n = 7 \cdot 45 = 315</cmath> <cmath>10+10 \cdot \sum_{n=2}^{8} n = 10+ 10*35= 10+350=360</cmath> and$ (Error compiling LaTeX. Unknown error_msg)1+9+9+1+9+2 $is regrouped to <cmath> 4 \cdot 9+4 = 40</cmath> Now, our total digit sum should be: <cmath>315+360+40 = 675+40 =705</cmath> Now that we have the total digit sum, we can tell it's divisible by 3, since$ 7+5 = 12 $<cmath>12 \equiv 0 \pmod{3}</cmath> So$ 705 \equiv 0 \pmod{3} $Let's try to factor out groups of$ 3 $<cmath>705 = 3 \cdot 235</cmath> Now, to pull another$ 3 $out from$ 235 $,$ 235 $must be divisible by 3. <cmath>2+3+5 = 10</cmath> so <cmath>235 \equiv 1 \pmod{3}</cmath>$ 235 $is not divisible by$ 3$.

Therefore, the greatest power of$ (Error compiling LaTeX. Unknown error_msg)3 $the given integer is divisible by is$ 3 = 3^1 $Thus,$ k=1 \implies \boxed{B}$

~shockfront99

1992 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 16	Followed by Problem 18
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