Difference between revisions of "2017 AMC 12B Problems/Problem 19"
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final += i | final += i | ||
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N = int(createNumberString(44)) | N = int(createNumberString(44)) | ||
remainder = N % 45 | remainder = N % 45 |
Revision as of 13:44, 26 September 2020
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
We will consider this number and . By looking at the last digit, it is obvious that the number is . To calculate the number , note that
so it is equivalent to
Let be the remainder when this number is divided by . We know that and , so by the Chinese remainder theorem, since , , or . So the answer is .
Solution 2
We know that this number is divisible by because the sum of the digits is , which is divisible by . If we subtracted from the integer we would get , which is also divisible by and by . Thus the remainder is , or .
Solution 3 (Beginner's Method)
To find the sum of digits of our number, we break it up into cases, starting with , , , , or .
Case 1: ,
Case 2: (We add 10 to the previous cases, as we are in the next ten's place)
Case 3: ,
Case 4: ,
Case 5: ,
Thus the sum of the digits is , so the number is divisible by . We notice that the number ends in "", which is more than a multiple of . Thus if we subtracted from our number it would be divisible by , and . (Multiple of n - n = Multiple of n)
So our remainder is , the value we need to add to the multiple of to get to our number.
Solution 4
This Python code works:
def createNumberString(n): final = "" for i in range (1, n+1) final += i return final N = int(createNumberString(44)) remainder = N % 45 print(remainder)
See Also
2017 AMC 12B (Problems • Answer Key • Resources) | |
Preceded by Problem 18 |
Followed by Problem 20 |
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 12 Problems and Solutions |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.