Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1992 AHSME Problems/Problem 17"
Line 25: Line 25:
 
==Solution 3 (Longer, but less knowledge of modular arithmetic needed)==
 
==Solution 3 (Longer, but less knowledge of modular arithmetic needed)==
 
We know that a number is divisible by <math>3</math> if the sum of its digits is divisible by <math>3</math>. Observe that the sum of all digits in the given integer would be:  
 
We know that a number is divisible by <math>3</math> if the sum of its digits is divisible by <math>3</math>. Observe that the sum of all digits in the given integer would be:  
−
<cmath>8 \cdot \sum_{n=1}^{9} n</cmath> for all units digits of the integers <math>20</math> through <math>90</math> (we exclude <math>0</math> from the sum since it adds nothing to the total)  
+
<cmath>7 \cdot \sum_{n=1}^{9} n</cmath> for all units digits of the integers <math>20</math> through <math>90</math> (we exclude <math>0</math> from the sum since it adds nothing to the total)  
−
<cmath>\sum_{n=2}^{9} n</cmath>  for every digit in the <math>10</math>'s place of the integers <math>20</math> through <math>90</math>.  
+
<cmath>10+10 \cdot \sum_{n=2}^{8} n</cmath>  for every digit in the <math>10</math>'s place of the integers <math>20</math> through 89<math>.  
−
This excludes the digits from <math>19</math>, <math>91</math>, or <math>92</math>, so we compute their sum with <cmath>1+9+9+1+9+2</cmath>
+
This excludes the digits from </math>19<math>, </math>90<math> (excluded from the second sum, adds nothing to the first), </math>91<math>, </math>92<math>, so we compute their sum with <cmath>1+9+9+9+1+9+2</cmath>
  
 
Now, let's compute all of our sums  
 
Now, let's compute all of our sums  
−
<cmath>8 \cdot \sum_{n=1}^{9} n = 8 \cdot 45 = 360</cmath>  
+
<cmath>7 \cdot \sum_{n=1}^{9} n = 7 \cdot 45 = 315</cmath>  
−
<cmath>\sum_{n=2}^{9} n = -1 +\sum_{n=1}^{9} n =44</cmath>  
+
<cmath>10+10 \cdot \sum_{n=2}^{8} n = 10+ 10*35= 10+350=360</cmath>  
−
and <math>1+9+9+1+9+2</math> is regrouped to <cmath> 3 \cdot 9+4 = 31</cmath>
+
and </math>1+9+9+1+9+2<math> is regrouped to <cmath> 4 \cdot 9+4 = 40</cmath>
 
Now, our total digit sum should be:  
 
Now, our total digit sum should be:  
−
<cmath>360+44+31 = 404+31 =435</cmath>
+
<cmath>315+360+40 = 675+40 =705</cmath>
−
Now that we have the total digit sum, we can tell it's divisible by 3, since <math>4+3+5 = 12</math>  
+
Now that we have the total digit sum, we can tell it's divisible by 3, since </math>7+5 = 12<math>  
 
<cmath>12 \equiv 0 \pmod{3}</cmath>  
 
<cmath>12 \equiv 0 \pmod{3}</cmath>  
−
So <math>435 \equiv 0 \pmod{3}</math>  
+
So </math>705 \equiv 0 \pmod{3}<math>  
  
−
Let's try to factor out groups of <math>3</math>  
+
Let's try to factor out groups of </math>3<math>  
−
<cmath>435 = 3 \cdot 145</cmath>  
+
<cmath>705 = 3 \cdot 235</cmath>  
−
Now, to pull another <math>3</math> out from <math>145</math>, <math>145</math> must be divisible by 3.  
+
Now, to pull another </math>3<math> out from </math>235<math>, </math>235<math> must be divisible by 3.  
−
<cmath>1+4+5 = 10</cmath> so  
+
<cmath>2+3+5 = 10</cmath> so  
−
<cmath>145 \equiv 1 \pmod{3}</cmath>  
+
<cmath>235 \equiv 1 \pmod{3}</cmath>  
−
<math>145</math> is not divisible by <math>3</math>.
+
</math>235<math> is not divisible by </math>3<math>.
  
−
Therefore, the greatest power of <math>3</math> the given integer is divisible by is <math>3 = 3^1</math>  
+
Therefore, the greatest power of </math>3<math> the given integer is divisible by is </math>3 = 3^1<math>  
  
−
Thus, <math>k=1 \implies \boxed{B}</math>
+
Thus, </math>k=1 \implies \boxed{B}$
  
 
~shockfront99
 
~shockfront99

Revision as of 17:50, 29 October 2025

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


Solution 3 (Longer, but less knowledge of modular arithmetic needed)

We know that a number is divisible by $3$ if the sum of its digits is divisible by $3$. Observe that the sum of all digits in the given integer would be: \[7 \cdot \sum_{n=1}^{9} n\] for all units digits of the integers $20$ through $90$ (we exclude $0$ from the sum since it adds nothing to the total) \[10+10 \cdot \sum_{n=2}^{8} n\] for every digit in the $10$'s place of the integers $20$ through 89$. This excludes the digits from$19$,$90$(excluded from the second sum, adds nothing to the first),$91$,$92$, so we compute their sum with <cmath>1+9+9+9+1+9+2</cmath>

Now, let's compute all of our sums <cmath>7 \cdot \sum_{n=1}^{9} n = 7 \cdot 45 = 315</cmath> <cmath>10+10 \cdot \sum_{n=2}^{8} n = 10+ 10*35= 10+350=360</cmath> and$ (Error compiling LaTeX. Unknown error_msg)1+9+9+1+9+2$is regrouped to <cmath> 4 \cdot 9+4 = 40</cmath> Now, our total digit sum should be: <cmath>315+360+40 = 675+40 =705</cmath> Now that we have the total digit sum, we can tell it's divisible by 3, since$7+5 = 12$<cmath>12 \equiv 0 \pmod{3}</cmath> So$705 \equiv 0 \pmod{3}$Let's try to factor out groups of$3$<cmath>705 = 3 \cdot 235</cmath> Now, to pull another$3$out from$235$,$235$must be divisible by 3. <cmath>2+3+5 = 10</cmath> so <cmath>235 \equiv 1 \pmod{3}</cmath>$235$is not divisible by$3$.

Therefore, the greatest power of$ (Error compiling LaTeX. Unknown error_msg)3$the given integer is divisible by is$3 = 3^1$Thus,$k=1 \implies \boxed{B}$

~shockfront99

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1992_AHSME_Problems/Problem_17&oldid=264419"