Difference between revisions of "2017 AMC 10B Problems/Problem 23"
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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=== |
Revision as of 08:28, 21 August 2022
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 this is not a multiple of because the units digit doesn't end in 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
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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