Difference between revisions of "1992 AHSME Problems/Problem 17"

Latest revision as of 12:19, 4 January 2021

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

@@ Line 14: / Line 14: @@
 \implies \boxed{B}</math>.
-=== Solution 2===
+==Solution 2==
-Every number is congruent to its digit sum mod <math>9</math>, so <cmath>N \equiv 1+9+2+0+...+9+2 \pmod 9,</cmath> but applying the result in reverse, <math>1+9 \equiv 19</math>, <math>2+0 \equiv 20</math>, etc., so the sum just become <cmath>19+20+...+92 \pmod 9.</cmath> We can simplify this using the formula for the sum of an arithmetic series, giving <math>\frac{1}{2} \times 74 \times (19+92) = 37 \times 111</math>, which is congruent to <math>1 \times 3 = 3 \pmod 9</math>, as before. Hence our answer is <math>\boxed{\text{(B) } 1}</math>.
+As our first trial with 3, We can say that <math>N\equiv 1+9+2+0+2+1+2+2+...+9+1+9+2\pmod{3}</math>. 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 <math>N\equiv 19+20+21+22+...+91+92\pmod{3}</math>, and adding that up using the rainbow strategy, we get <math>N\equiv (111\times37)\pmod{3}</math>. 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 <math>\boxed{B}</math>
+~mathpro12345
+papa I wrote this second one
 == See also ==

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