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

Revision as of 10:47, 19 February 2019

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 solution. $N \equiv 4 \text{ (mod 5)}$ 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, $N \equiv 54 \text{ (mod 45)}$ and $N \equiv 9 \text{ (mod 45)}$ so the remainder is $\boxed{\textbf{(C) }9}$

Solution 3

Because a number is equivalent to the sum of its digits modulo 9, we have that $N\equiv 1+2+3+4+5+...+44\equiv \frac{44}{45}{2}\equiv 0\pmod{9}$ . Furthermore, we see that $N-9$ ends in the digit 5 and thus is divisible by 5, so $N-9$ is divisible by 45, meaning the remainder is $\boxed{\textbf{(C) }9}$ -Stormersyle

@@ Line 13: / Line 13: @@
 ==Solution 2==
 Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by <math>9</math>. From <math>1</math> thru <math>9</math>, the sum is <math>45</math>. <math>10</math> thru <math>19</math>, the sum is <math>55</math>, <math>20</math> thru <math>29</math> is <math>65</math>, and <math>30</math> thru <math>39</math> is <math>75</math>. Thus the sum of the digits is <math>45+55+65+75+4+5+6+7+8 = 240+30 = 270</math>, and thus <math>N</math> is divisible by <math>9</math>. Now, refer to the above solution. <math>N \equiv 4 \text{ (mod 5)}</math> and <math>N \equiv 0 \text{ (mod 9)}</math>. From this information, we can conclude that <math>N \equiv 54 \text{ (mod 5)}</math> and <math>N \equiv 54 \text{ (mod 9)}</math>. Therefore, <math>N \equiv 54 \text{ (mod 45)}</math> and <math>N \equiv 9 \text{ (mod 45)}</math> so the remainder is <math>\boxed{\textbf{(C) }9}</math>
+==Solution 3==
+Because a number is equivalent to the sum of its digits modulo 9, we have that <math>N\equiv 1+2+3+4+5+...+44\equiv \frac{44}{45}{2}\equiv 0\pmod{9}</math>. Furthermore, we see that <math>N-9</math> ends in the digit 5 and thus is divisible by 5, so <math>N-9</math> is divisible by 45, meaning the remainder is <math>\boxed{\textbf{(C) }9}</math>
+-Stormersyle
 ==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
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