Difference between revisions of "2013 AIME I Problems/Problem 2"
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== Solution 2 == | == Solution 2 == | ||
| − | For a number to be divisible by <math>5</math>, the last digit of the number must be <math>5</math> or <math>0</math>. However, since the first digit is the same as the last one, the last (and first) digits can not be <math>0</math>, so the number must be in the form <math>\overline{5abc5}</math>, where <math>a+b+c</math> is divisible by 5. This gives us a answer of <math>10\times10\times10\times\frac{1}{5}=\boxed{200}</math> | + | For a number to be divisible by <math>5</math>, the last digit of the number must be <math>5</math> or <math>0</math>. However, since the first digit is the same as the last one, the last (and first) digits can not be <math>0</math>, so the number must be in the form <math>\overline{5abc5}</math>, where <math>a+b+c</math> is divisible by 5. This gives us a answer of <math>10\times10\times10\times\frac{1}{5}=\boxed{200}</math> |
== See also == | == See also == | ||
{{AIME box|year=2013|n=I|num-b=1|num-a=3}} | {{AIME box|year=2013|n=I|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 10:00, 15 September 2018
Contents
Problem 2
Find the number of five-digit positive integers, 
, that satisfy the following conditions:
-
(a) the number
is divisible by 
-
(b) the first and last digits of
are equal, and
-
(c) the sum of the digits of
is divisible by 
Solution
The number takes a form of 
, in which 
. Let 
and 
be arbitrary digits. For each pair of 
, there are exactly two values of 
that satisfy the condition of . Therefore, the answer is 
Solution 2
For a number to be divisible by 
, the last digit of the number must be or 
. However, since the first digit is the same as the last one, the last (and first) digits can not be , so the number must be in the form 
, where 
is divisible by 5. This gives us a answer of 
See also
| 2013 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
