Difference between revisions of "2017 AMC 10B Problems/Problem 23"
Stormersyle (talk | contribs) |
Stormersyle (talk | contribs) (→Solution 3) |
||
Line 15: | Line 15: | ||
==Solution 3== | ==Solution 3== | ||
− | Because a number is equivalent to the sum of its digits modulo 9, we have that <math>N\equiv 1+2+3+4+5+...+44\equiv \frac{44 | + | Because a number is equivalent to the sum of its digits modulo 9, we have that <math>N\equiv 1+2+3+4+5+...+44\equiv \frac{44\times 45}{2}\equiv 0\pmod{9}</math>. Furthermore, we see that <math>N-9</math> ends in the digit 5 and thus is divisible by 5, so <math>N-9</math> is divisible by 45, meaning the remainder is <math>\boxed{\textbf{(C) }9}</math> |
-Stormersyle | -Stormersyle | ||
Revision as of 09:47, 19 February 2019
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, by inspection, the answer is .
Note: the sum of the digits of is .
Solution 2
Noting the solution above, we try to find the sum of the digits to figure out its remainder when divided by . From thru , the sum is . thru , the sum is , thru is , and thru is . Thus the sum of the digits is , and thus is divisible by . Now, refer to the above solution. and . From this information, we can conclude that and . Therefore, and so the remainder 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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
yeet