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

m (Solution 2)
m (Solution 2)
Line 12: Line 12:
  
 
==Solution 2==
 
==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 or below solutions. <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>
+
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>
  
 
==See Also==
 
==See Also==
 
{{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}}
 
{{AMC10 box|year=2017|ab=B|num-b=22|num-a=24}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 00:08, 25 January 2018

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 $\boxed{\textbf{(C) }9}$

See Also

2017 AMC 10B (ProblemsAnswer KeyResources)
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

Invalid username
Login to AoPS