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 12B Problems/Problem 19"

m (→‎Solution 2)
m (→‎Solution)
Line 14: Line 14:
 
<cmath>\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.</cmath>
 
<cmath>\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.</cmath>
  
−
Let <math>x</math> be the remainder when this number is divided by <math>45</math>. We know that <math>x\equiv 0 \pmod {9}</math> and <math>x\equiv 4 \pmod {5}</math>, so by the Chinese remainder theorem, since <math>9(-1)\equiv 1 \pmod{5}</math>, <math>x\equiv 5(0)+9(-1)(4) \pmod {5\cdot 9}</math>, or <math>x\equiv -36 \equiv 9 \pmod {45}</math>. So the answer is <math>\boxed {\bold {(C)}}</math>.
+
Let <math>x</math> be the remainder when this number is divided by <math>45</math>. We know that <math>x\equiv 0 \pmod {9}</math> and <math>x\equiv 4 \pmod {5}</math>, so by the Chinese remainder theorem, since <math>9(-1)\equiv 1 \pmod{5}</math>, <math>x\equiv 5(0)+9(-1)(4) \pmod {5\cdot 9}</math>, or <math>x\equiv -36 \equiv 9 \pmod {45}</math>. So the answer is <math>\boxed {\textbf {(C)}}</math>.
  
 
==Solution 2==
 
==Solution 2==

Revision as of 15:05, 11 February 2019

Problem

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 will consider this number $\bmod\ 5$ and $\bmod\ 9$. By looking at the last digit, it is obvious that the number is $\equiv 4\bmod\ 5$. To calculate the number $\bmod\ 9$, note that

\[123456\cdots 4344 \equiv 1+2+3+4+5+6+7+8+9+(1+0)+(1+1)+\cdots+(4+3)+(4+4) \equiv 1+2+\cdots+44 \bmod\ 9,\]

so it is equivalent to

\[\frac{44\cdot 45}{2} = 22\cdot 45 \equiv 0\bmod\ 9.\]

Let $x$ be the remainder when this number is divided by $45$. We know that $x\equiv 0 \pmod {9}$ and $x\equiv 4 \pmod {5}$, so by the Chinese remainder theorem, since $9(-1)\equiv 1 \pmod{5}$, $x\equiv 5(0)+9(-1)(4) \pmod {5\cdot 9}$, or $x\equiv -36 \equiv 9 \pmod {45}$. So the answer is $\boxed {\textbf {(C)}}$.

Solution 2

We know that this number is divisible by $9$ because the sum of the digits is $270$, which is divisible by $9$. Subtracting 9 from the integer we get $1234 \cdots 4335$, which is also divisible by $5$, making it also divisible by $45$. Thus the remainder is $9$, or $\boxed{\bold{B}}$.

See Also

2017 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 18 		Followed by
Problem 20
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 12 Problems and Solutions
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_12B_Problems/Problem_19&oldid=101850"