Difference between revisions of "2017 AMC 10B Problems/Problem 23"
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− | ==Problem | + | ==Problem== |
Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>? | Let <math>N=123456789101112\dots4344</math> be the <math>79</math>-digit number that is formed by writing the integers from <math>1</math> to <math>44</math> in order, one after the other. What is the remainder when <math>N</math> is divided by <math>45</math>? | ||
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math> | <math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 4\qquad\textbf{(C)}\ 9\qquad\textbf{(D)}\ 18\qquad\textbf{(E)}\ 44</math> | ||
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==Solution 1== | ==Solution 1== | ||
We only need to find the remainders of N when divided by 5 and 9 to determine the answer. | We only need to find the remainders of N when divided by 5 and 9 to determine the answer. | ||
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. Solving these modular congruence using | + | 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 the [[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=== | ||
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==Solution 3 (Clever way using divisibility rules)== | ==Solution 3 (Clever way using divisibility rules)== | ||
− | We know that <math>45 = 5 \ | + | We know that <math>45 = 5 \cdot 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+\cdots + 44 = \frac{44 \cdot 45}{2}</math>. We can quickly see this is a multiple of <math>9</math> because <math>\frac{44}{2} \cdot 45 = 22 \cdot 45</math>. We know <math>123 \ldots 4344</math> is not a multiple of <math>5</math> because the units digit isn't <math>5</math> or <math>0</math>. We can just subtract by 9 until we get a number whose units digit is 5 or 0. |
− | We have <math>123 \ldots | + | We have <math>123 \ldots 4344</math> is divisible by <math>9</math>, so we can subtract by <math>9</math> to get <math>123 \ldots 4335</math> and we know that this is divisible by 5. So our answer is <math>\boxed{\textbf{(C) } 9}</math> |
~Arcticturn | ~Arcticturn | ||
+ | ==Solution 4== | ||
+ | Since <math>N</math> ends with the number <math>4</math>, <math>N\equiv 4\pmod 5</math>. Also, the sum of the digits of <math>N</math> are divisible by <math>9</math>, so <math>N</math> must be divisible by <math>9</math>. Therefore, we have the system of equations: | ||
+ | <math>N\equiv 4\pmod 5</math> | ||
+ | <math>N\equiv 0\pmod 9</math> | ||
+ | According to the second equation, <math>N\equiv \{0, 9, 18, 27, 36\} \pmod {45}</math>. The only one of these solutions that is <math>4\pmod 5</math> is <math>\boxed{\textbf{(C) } 9}</math> | ||
+ | |||
+ | ~[https://artofproblemsolving.com/wiki/index.php/User:Sid2012 sid2012] | ||
==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}} |
Latest revision as of 19:25, 4 November 2024
Contents
[hide]Problem
Let be the -digit number that is formed by writing the integers from to in order, one after the other. What is the remainder when is divided by ?
Solution 1
We only need to find the remainders of N when divided by 5 and 9 to determine the answer. By inspection, . The remainder when is divided by is , but since , we can also write this as , which has a remainder of 0 mod 9. Solving these modular congruence using the Chinese Remainder Theorem we get the remainder to be . Therefore, the answer is .
Alternative Ending to Solution 1
Once we find our 2 modular congruences, we can narrow our options down to and because the remainder when is divided by should be a multiple of 9 by our modular congruence that states has a remainder of when divided by . Also, our other modular congruence states that the remainder when divided by should have a remainder of when divided by . Out of options and , only satisfies that the remainder when is divided by .
Solution 2
Realize that for all positive integers .
Apply this on the expanded form of :
Solution 3 (Clever way using divisibility rules)
We know that , so we can apply our restrictions to that. We know that the units digit must be or , and the digits must add up to a multiple of . . We can quickly see this is a multiple of because . We know is not a multiple of because the units digit isn't or . We can just subtract by 9 until we get a number whose units digit is 5 or 0.
We have is divisible by , so we can subtract by to get and we know that this is divisible by 5. So our answer is
~Arcticturn
Solution 4
Since ends with the number , . Also, the sum of the digits of are divisible by , so must be divisible by . Therefore, we have the system of equations:
According to the second equation, . The only one of these solutions that is is
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 |
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