Difference between revisions of "2009 AMC 12A Problems/Problem 12"
(→See Also) |
(→Solution) |
||
Line 18: | Line 18: | ||
We can write each integer between <math>1</math> and <math>999</math> inclusive as <math>\overline{abc}=100a+10b+c</math> where <math>a,b,c\in\{0,1,\dots,9\}</math> and <math>a+b+c>0</math>. | We can write each integer between <math>1</math> and <math>999</math> inclusive as <math>\overline{abc}=100a+10b+c</math> where <math>a,b,c\in\{0,1,\dots,9\}</math> and <math>a+b+c>0</math>. | ||
The sum of digits of this number is <math>a+b+c</math>, hence we get the equation <math>100a+10b+c = 6(a+b+c)</math>. This simplifies to <math>94a + 4b - 5c = 0</math>. Clearly for <math>a>0</math> there are no solutions, hence <math>a=0</math> and we get the equation <math>4b=5c</math>. This obviously has only one valid solution <math>(b,c)=(5,4)</math>, hence the only solution is the number <math>54</math>. | The sum of digits of this number is <math>a+b+c</math>, hence we get the equation <math>100a+10b+c = 6(a+b+c)</math>. This simplifies to <math>94a + 4b - 5c = 0</math>. Clearly for <math>a>0</math> there are no solutions, hence <math>a=0</math> and we get the equation <math>4b=5c</math>. This obviously has only one valid solution <math>(b,c)=(5,4)</math>, hence the only solution is the number <math>54</math>. | ||
+ | |||
+ | === Solution 3 === | ||
+ | |||
+ | The sum of the digits is at most <math>9+9+9=27</math>. Therefore the number is at most <math>6\cdot 27 = 162</math>. Since the number is <math>6</math> times the sum of its digits, it must be divisible by <math>6</math>, therefore also by <math>3</math>, therefore the sum of its digits must be divisible by <math>3</math>. Next we can conclude that the number must be divisible by <math>18</math>, not just by <math>6</math>. Since the number is divisible by <math>18</math>, it is also divisible by <math>9</math>, therefore the sum of its digits is divisible by <math>9</math>, therefore the number is divisible by <math>54</math>, which leaves us with <math>54</math>, <math>108</math> and <math>162</math>. Only <math>54</math> is <math>6</math> times its digits, hence the answer is <math>\boxed{1}</math>. | ||
== See Also == | == See Also == |
Revision as of 12:12, 30 December 2014
Problem
How many positive integers less than are
times the sum of their digits?
Solution
Solution 1
The sum of the digits is at most . Therefore the number is at most
. Out of the numbers
to
the one with the largest sum of digits is
, and the sum is
. Hence the sum of digits will be at most
.
Also, each number with this property is divisible by , therefore it is divisible by
, and thus also its sum of digits is divisible by
.
We only have six possibilities left for the sum of the digits: ,
,
,
,
, and
. These lead to the integers
,
,
,
,
, and
. But for
the sum of digits is
, which is not
, therefore
is not a solution. Similarly we can throw away
,
,
, and
, and we are left with just
solution: the number
.
Solution 2
We can write each integer between and
inclusive as
where
and
.
The sum of digits of this number is
, hence we get the equation
. This simplifies to
. Clearly for
there are no solutions, hence
and we get the equation
. This obviously has only one valid solution
, hence the only solution is the number
.
Solution 3
The sum of the digits is at most . Therefore the number is at most
. Since the number is
times the sum of its digits, it must be divisible by
, therefore also by
, therefore the sum of its digits must be divisible by
. Next we can conclude that the number must be divisible by
, not just by
. Since the number is divisible by
, it is also divisible by
, therefore the sum of its digits is divisible by
, therefore the number is divisible by
, which leaves us with
,
and
. Only
is
times its digits, hence the answer is
.
See Also
2009 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 11 |
Followed by Problem 13 |
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.