Difference between revisions of "2013 AIME I Problems/Problem 2"
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== Solution == | == Solution == | ||
The number takes a form of <math>5\text{x,y,z}5</math>, in which <math>5|x+y+z</math>. Let <math>x</math> and <math>y</math> be arbitrary digits. For each pair of <math>x,y</math>, there are exactly two values of <math>z</math> that satisfy the condition of <math>5|x+y+z</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math> | The number takes a form of <math>5\text{x,y,z}5</math>, in which <math>5|x+y+z</math>. Let <math>x</math> and <math>y</math> be arbitrary digits. For each pair of <math>x,y</math>, there are exactly two values of <math>z</math> that satisfy the condition of <math>5|x+y+z</math>. Therefore, the answer is <math>10\times10\times2=\boxed{200}</math> | ||
+ | |||
+ | == 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>. | ||
== 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 22:28, 5 September 2018
Contents
Problem 2
Find the number of five-digit positive integers, , that satisfy the following conditions:
-
(a) the number
![$n$](http://latex.artofproblemsolving.com/1/7/4/174fadd07fd54c9afe288e96558c92e0c1da733a.png)
![$5,$](http://latex.artofproblemsolving.com/2/1/4/21415a2eb558eb7c199a0382fcd50cbeda9299fb.png)
-
(b) the first and last digits of
![$n$](http://latex.artofproblemsolving.com/1/7/4/174fadd07fd54c9afe288e96558c92e0c1da733a.png)
-
(c) the sum of the digits of
![$n$](http://latex.artofproblemsolving.com/1/7/4/174fadd07fd54c9afe288e96558c92e0c1da733a.png)
![$5.$](http://latex.artofproblemsolving.com/b/8/3/b835f2c6c592edc4583ce996f86bcc0d07ca8da5.png)
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.