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 "2017 AMC 10B Problems/Problem 23"

(→‎Solution 2)
m (→‎Solution)
Line 7: Line 7:
 
We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
 
We only need to find the remainders of N when divided by 5 and 9 to determine the answer.
 
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.
 
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.
−
The remainder when <math>N</math> is divided by <math>9</math> is <math>1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which has a remainder of 0 mod 9. Therefore, by CRT, the answer is <math>\boxed{\textbf{(C) } 9}</math>.
+
The remainder when <math>N</math> is divided by <math>9</math> is <math>1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4</math>, but since <math>10 \equiv 1 \text{ (mod 9)}</math>, we can also write this as <math>1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45</math>, which has a remainder of 0 mod 9. Therefore, by inspection, the answer is <math>\boxed{\textbf{(C) } 9}</math>.
  
−
Note: the sum of the digits of <math>N</math> is <math>270</math>
+
Note: the sum of the digits of <math>N</math> is <math>270</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 18:38, 18 February 2017

Problem 23

Let $N=123456789101112\dots4344$ be the $79$-digit number that is formed by writing the integers from $1$ to $44$ in order, one after the other. What is the remainder when $N$ is divided by $45$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44$

Solution

We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, $N \equiv 4 \text{ (mod 5)}$. The remainder when $N$ is divided by $9$ is $1+2+3+4+ \cdots +1+0+1+1 +1+2 +\cdots + 4+3+4+4$, but since $10 \equiv 1 \text{ (mod 9)}$, we can also write this as $1+2+3 +\cdots +10+11+12+ \cdots 43 + 44 = \frac{44 \cdot 45}2 = 22 \cdot 45$, which has a remainder of 0 mod 9. Therefore, by inspection, the answer is $\boxed{\textbf{(C) } 9}$.

Note: the sum of the digits of $N$ is $270$.

Solution 2

Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by $9$. From $1$ thru $9$, the sum is $45$. $10$ thru $19$, the sum is $55$, $20$ thru $29$ is $65$, and $30$ thru $39$ is $75$. Thus the sum of the digits is $45+55+65+75+4+5+6+7+8 = 240+30 = 270$, and thus $N$ is divisible by $9$. Now, refer to the above or below solutions. $N \equiv 4 \text{ (mod 5)}$ and $N \equiv 0 \text{ (mod 9)}$. From this information, we can conclude that $N \equiv 54 \text{ (mod 5)}$ and $N \equiv 54 \text{ (mod 9)}$. Therefore, $N \equiv 54 \text{ (mod 45)}$. The remainder is thus $9$, so the answer is $\boxed{\textbf{(C) }9}$

See Also

2017 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 22 		Followed by
Problem 24
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
All AMC 10 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_10B_Problems/Problem_23&oldid=84092"