Difference between revisions of "2017 AMC 10B Problems/Problem 23"

Revision as of 21:05, 27 November 2020

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 $N$ : $N = 1(10)^{78} + 2(10)^{77} + \cdots + 9(10)^{70} + 10(10)^{68} + 11(10)^{66} + \cdots + 43(10)^{2} + 44 \equiv$ $10(1 + 2 + \cdots + 43) + 44 \equiv 10 \left (\frac{43 \cdot 44}2 \right ) + 44 \equiv$ $10 \left (\frac{-2 \cdot -1}2 \right ) - 1 \equiv \boxed{\textbf{(C) } 9} \pmod{45}$

@@ Line 10: / Line 10: @@
 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. Solving these modular congruence using CRT(Chinese Remainder Theorem) we get the remainder to be <math>9 \pmod{45}</math>. Therefore, the answer is <math>\boxed{\textbf{(C) } 9}</math>.
-==Alternative Ending to Solution 1==
+===Alternative Ending to Solution 1===
 Once we find our 2 modular congruences, we can narrow our options down to <math>{C}</math> and <math>{D}</math> because the remainder when <math>N</math> is divided by <math>45</math> should be a multiple of 9 by our modular congruence that states <math>N</math> has a remainder of <math>0</math> when divided by <math>9</math>. Also, our other modular congruence states that the remainder when divided by <math>45</math> should have a remainder of <math>4</math> when divided by <math>5</math>. Out of options <math>C</math> and <math>D</math>, only <math>\boxed{\textbf{(C) } 9}</math> satisfies that the remainder when <math>N</math> is divided by <math>45 \equiv 4 \text{ (mod 5)}</math>.

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