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

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By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.
 
By inspection, <math>N \equiv 4 \text{ (mod 5)}</math>.
 
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. Therefore, the answer is <math>\boxed{\textbf{(C) } 9}</math>.
 
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. Therefore, the answer is <math>\boxed{\textbf{(C) } 9}</math>.
 
==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==
 
==Solution 3==

Revision as of 16:45, 17 January 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

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, the answer is $\boxed{\textbf{(C) } 9}$.

Solution 3

Because a number is equivalent to the sum of its digits modulo 9, we have that $N\equiv 1+2+3+4+5+...+44\equiv \frac{44\times 45}{2}\equiv 0\pmod{9}$. 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

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

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