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

Revision as of 00:47, 2 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, 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 $\fbox{B}$ .

1992 AHSME (Problems • Answer Key • Resources)
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@@ Line 8: / Line 8: @@
 \text{(E) more than } 3</math>
-== Solution (without using sum-of-digits formula) ==
+== Solution ==
-Note that 100 is congruent to 1 mod 3 and 1 mod 9 (this will be useful later on). Thus multiplying our specific 2-digit number by 100 doesn't change its modulus value, so we can write our number, preserving its value mod 3 or mod 9 as 19 + 20 + 21 + ... + 92. As we can see, for every 3 numbers (beginning with 0 mod 3 for convenience) our modulus total goes up by 3. Thus, if we define a 3-cycle to be 3 numbers that satisfy 0 mod 3 to 2 mod 3, than a 3-cycle of numbers adds up to 3 mod 3 or 0 mod 3. From 21 to 92, we have an integer number of 3-cycles, and thus can ignore them as they don't change our modulus sum. 19 and 20 add 1 + 2 = 3 to the modulus sum, so our number is divisible by 3. To calculate our number mod 9, we define a 9-cycle in the same manner as before; we have numbers congruent to 1 + 2 + 3 + ... + 8 = 36, which is 0 mod 9. Thus, we can ignore complete 9-cycles with or without the first number (as that contributes 0 to the modulus sum). Starting at 19, we have some integer number of 9-sets. The last complete 9-set terminates in 89, thus our number mod 9 is 0 + 1 + 2 = 3. Thus our number is only divisible by 3 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, 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>.
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Difference between revisions of "1992 AHSME Problems/Problem 17"

Revision as of 00:47, 2 February 2018

Problem

Solution

See also