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 3)
(→‎Solution 3 (Clever way using divisibility rules))
Line 22: Line 22:
  
 
==Solution 3 (Clever way using divisibility rules)==
 
==Solution 3 (Clever way using divisibility rules)==
−
We know that <math>45</math> = <math>5</math> x <math>9</math>, so we can apply our restrictions to that. We know that the units digit must be <math>5</math> or <math>0</math>, and the digits must add up to a multiple of <math>9</math>. <math>1+2+3+4... + 44</math> = <math>44*45/2</math>. We can quickly see this is a multiple of <math>9</math> because <math>44/2 * 45</math> = <math>22*45</math>. We know this is not a multiple of <math>5</math> because the units digit doesn't end in <math>5</math> or <math>10</math>. We can just subtract by 9 until we get a number whose units digit is 5 or 0!
+
We know that <math>45</math> = <math>5</math> x <math>9</math>, so we can apply our restrictions to that. We know that the units digit must be <math>5</math> or <math>0</math>, and the digits must add up to a multiple of <math>9</math>. <math>1+2+3+4... + 44</math> = <math>44*45/2</math>. We can quickly see this is a multiple of <math>9</math> because <math>44/2 * 45</math> = <math>22*45</math>. We know this is not a multiple of <math>5</math> because the units digit doesn't end in <math>5</math> or <math>10</math>. We can just subtract by 9 until we get a number whose units digit is 5 or 0.
  
 
We have <math>123...44</math> is divisible by <math>9</math>, so we can subtract by <math>9</math> to get <math>123...35</math> and we know that this is divisible by 5. So our answer is <math>\boxed{\textbf{(C) } 9}</math>
 
We have <math>123...44</math> is divisible by <math>9</math>, so we can subtract by <math>9</math> to get <math>123...35</math> and we know that this is divisible by 5. So our answer is <math>\boxed{\textbf{(C) } 9}</math>

Revision as of 14:50, 30 July 2021

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 1

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. Solving these modular congruence using CRT(Chinese Remainder Theorem) we get the remainder to be $9 \pmod{45}$. Therefore, the answer is $\boxed{\textbf{(C) } 9}$.

Alternative Ending to Solution 1

Once we find our 2 modular congruences, we can narrow our options down to ${C}$ and ${D}$ because the remainder when $N$ is divided by $45$ should be a multiple of 9 by our modular congruence that states $N$ has a remainder of $0$ when divided by $9$. Also, our other modular congruence states that the remainder when divided by $45$ should have a remainder of $4$ when divided by $5$. Out of options $C$ and $D$, only $\boxed{\textbf{(C) } 9}$ satisfies that the remainder when $N$ is divided by $45 \equiv 4 \text{ (mod 5)}$.

Solution 2

Realize that $10 \equiv 10 \cdot 10 \equiv 10^{k} \pmod{45}$ for all positive integers $k$.

Apply this on the expanded form of $N$: \[N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv\] \[10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv\] \[10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}\]

Solution 3 (Clever way using divisibility rules)

We know that $45$ = $5$ x $9$, so we can apply our restrictions to that. We know that the units digit must be $5$ or $0$, and the digits must add up to a multiple of $9$. $1+2+3+4... + 44$ = $44*45/2$. We can quickly see this is a multiple of $9$ because $44/2 * 45$ = $22*45$. We know this is not a multiple of $5$ because the units digit doesn't end in $5$ or $10$. We can just subtract by 9 until we get a number whose units digit is 5 or 0.

We have $123...44$ is divisible by $9$, so we can subtract by $9$ to get $123...35$ and we know that this is divisible by 5. So our answer is $\boxed{\textbf{(C) } 9}$

~Arcticturn

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=159290"