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

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== Solution ==
 
== Solution ==
  
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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, 91and 92. These remaining units digits sum up to 9 + 1 + 2 => 12, which means our units sum is 3 mod 9. As for the tens digits, for 2, 3, 4, ..., 8 we have 10 sets of those -> 8(9)/2 - 1 => 35, which is congruent to 8 mod 9. We again have 19, 90, 91 and 92, so we must add 1 + 9 * 3 => 28 to our total. 28 is congruent to 1 mod 9. Thus our sum is congruent to 3 mod 9, and k = 1 <math>\fbox{B}</math>.
+
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 mod 9. As for the tens digits, for 2, 3, 4, ..., 8 we have 10 sets of those -> 8(9)/2 - 1 => 35, which is congruent to 8 mod 9. We again have 19, 90, 91 and 92, so we must add 1 + 9 * 3 => 28 to our total. 28 is congruent to 1 mod 9. Thus our sum is congruent to 3 mod 9, and k = 1 <math>\fbox{B}</math>.
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== Alternate Solution ==
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Every number is congruent to its digit sum mod <math>9</math>, so <math>N</math> is congruent to <math>1+9+2+0+...+9+2</math> mod <math>9</math>, but applying the result in reverse, <math>1+9</math> is congruent to <math>19</math>, <math>2+0</math> is congruent to <math>20</math>, etc., so the sum just become <math>19+20+...+92</math> mod <math>9</math>. 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</math> mod <math>9</math>, as before.
  
 
== See also ==
 
== See also ==

Revision as of 01:58, 20 February 2018

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

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 mod 9. As for the tens digits, for 2, 3, 4, ..., 8 we have 10 sets of those -> 8(9)/2 - 1 => 35, which is congruent to 8 mod 9. We again have 19, 90, 91 and 92, so we must add 1 + 9 * 3 => 28 to our total. 28 is congruent to 1 mod 9. Thus our sum is congruent to 3 mod 9, and k = 1 $\fbox{B}$.

Alternate Solution

Every number is congruent to its digit sum mod $9$, so $N$ is congruent to $1+9+2+0+...+9+2$ mod $9$, but applying the result in reverse, $1+9$ is congruent to $19$, $2+0$ is congruent to $20$, etc., so the sum just become $19+20+...+92$ mod $9$. 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 $1 \times 3 = 3$ mod $9$, as before.

See also

1992 AHSME (Problems • Answer Key • Resources)
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

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