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

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== Solution ==
 
== Solution ==
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<math>\fbox{B}</math>
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Note that 100 is congruent to 1 mod 3. Thus multiplying our specific 2-digit number by 100 doesn't change its modulus value. As we can see, for every 3 numbers (beginning with 0 mod 3 for convenience) our modulus total goes up by 3. From 21 to 92, we have 72 numbers, for which there are the same number of 0 mod 3's, 1 mod 3's, etc. We also have 19 and 20, which add 3 to the modular total. Thus we have a number congruent to 75 mod 3 congruent to 0 mod 3. However, for k > 1, 75 must be divisible by 3^2 = 9. It is not, thus our number is only divisible by 3 and k = 1 <math>\fbox{B}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 19:51, 11 January 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

Note that 100 is congruent to 1 mod 3. Thus multiplying our specific 2-digit number by 100 doesn't change its modulus value. As we can see, for every 3 numbers (beginning with 0 mod 3 for convenience) our modulus total goes up by 3. From 21 to 92, we have 72 numbers, for which there are the same number of 0 mod 3's, 1 mod 3's, etc. We also have 19 and 20, which add 3 to the modular total. Thus we have a number congruent to 75 mod 3 congruent to 0 mod 3. However, for k > 1, 75 must be divisible by 3^2 = 9. It is not, thus our number is only divisible by 3 and k = 1 $\fbox{B}$.

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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