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>. | ||
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==Solution 3== | ==Solution 3== |
Revision as of 15:45, 17 January 2020
Contents
[hide]Problem 23
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
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. Therefore, the answer is .
Solution 3
Because a number is equivalent to the sum of its digits modulo 9, we have that . Furthermore, we see that ends in the digit 5 and thus is divisible by 5, so is divisible by 45, meaning the remainder is -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 | ||
All AMC 10 Problems and Solutions |
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