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 9: / Line 9: @@
 == Solution ==
+=== Solution 1===
-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 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 <math>3</math> or <math>9</math> by summing the digits. Looking at the one's place, we can start out with <math>0, 1, 2, 3, 4, 5, 6, 7, 8, 9</math> and continue cycling though the numbers from <math>0</math> through <math>9</math>. For each one of these cycles, we add <math>0 + 1 + ... + 9 = 45</math>. This is divisible by <math>9</math>, thus we can ignore the sum. However, this excludes <math>19</math>, <math>90</math>, <math>91</math> and <math>92</math>. These remaining units digits sum up to <math>9 + 1 + 2 = 12</math>, which means our units sum is <math>3 \pmod 9</math>. As for the tens digits, for <math>2, 3, 4, \cdots , 8</math> we have <math>10</math> sets of those: <cmath>\frac{8 \cdot 9}{2} - 1 = 35,</cmath> which is congruent to <math>8 \pmod 9</math>. We again have <math>19, 90, 91</math> and <math>92</math>, so we must add <cmath>1 + 9 \cdot 3 = 28</cmath> to our total. <math>28</math> is congruent to <math>1 \pmod 9</math>. Thus our sum is congruent to <math>3 \pmod 9</math>, and <math>k = 1
+\implies \boxed{B}</math>.
+==Solution 2==
+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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