Difference between revisions of "2003 AMC 12A Problems/Problem 18"
(→Solution 2) |
m |
||
(One intermediate revision by one other user not shown) | |||
Line 21: | Line 21: | ||
== Solution 3 == | == Solution 3 == | ||
− | Let <math>abcde</math> be the five digits of <math>n</math>. Then <math>q = abc</math> and <math>r = de</math>. By the divisibility rules of <math>11</math>, <math>q = a | + | Let <math>abcde</math> be the five digits of <math>n</math>. Then <math>q = abc</math> and <math>r = de</math>. By the divisibility rules of <math>11</math>, <math>q = a - b + c \pmod{11}</math> and <math>r = -d + e \pmod{11}</math>, so <math>q + r = a - b + c - d + e = abcde = n \pmod{11}</math>. Thus, <math>n</math> must be divisble by <math>11</math>. There are <math>\frac{99990 - 10010}{11} + 1 = 8181</math> five-digit multiples of <math>11</math>, so the answer is <math>\boxed{\textbf{(B) }8181}</math>. |
+ | ==Video Solution 1== | ||
+ | https://youtu.be/OpGHj-B0_hg?t=672 | ||
+ | |||
+ | ~IceMatrix | ||
==See Also== | ==See Also== |
Latest revision as of 17:20, 18 July 2022
Problem
Let be a -digit number, and let and be the quotient and the remainder, respectively, when is divided by . For how many values of is divisible by ?
Solution 1
When a -digit number is divided by , the first digits become the quotient, , and the last digits become the remainder, .
Therefore, can be any integer from to inclusive, and can be any integer from to inclusive.
For each of the possible values of , there are at least possible values of such that .
Since there is "extra" possible value of that is congruent to , each of the values of that are congruent to have more possible value of such that .
Therefore, the number of possible values of such that is .
Solution 2
Notice that . This means that any number whose quotient and remainder sum is divisible by 11 must also be divisible by 11. Therefore, there are possible values. The answer is .
Solution 3
Let be the five digits of . Then and . By the divisibility rules of , and , so . Thus, must be divisble by . There are five-digit multiples of , so the answer is .
Video Solution 1
https://youtu.be/OpGHj-B0_hg?t=672
~IceMatrix
See Also
2003 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Question | |
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 |
2003 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.