Difference between revisions of "2013 AMC 10A Problems/Problem 13"
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When <math>x=9</math>, <math>18+y<20</math> so <math>y<2</math>. This yields <math>2</math> more numbers. | When <math>x=9</math>, <math>18+y<20</math> so <math>y<2</math>. This yields <math>2</math> more numbers. | ||
| − | Summing, we get <math>40 + 8 + 6 + 4 + 2 = | + | Summing, we get <math>40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}</math> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2013|ab=A|num-b=12|num-a=14}} | {{AMC10 box|year=2013|ab=A|num-b=12|num-a=14}} | ||
Revision as of 14:26, 8 February 2013
Problem
How many three-digit numbers are not divisible by 
, have digits that sum to less than 
, and have the first digit equal to the third digit?
Solution
These three digit numbers are of the form 
. We see that 
and 
, as 
does not yield a three-digit integer and 
yields a number divisible by 5.
The second condition is that the sum 
. When 
is 
, 
, 
, or 
, y can be any digit from 
to 
, as 
. This yields 
numbers.
When 
, we see that 
so 
. This yields 
more numbers.
When 
, 
so 
. This yields 
more numbers.
When 
, 
so 
. This yields more numbers.
When 
, 
so 
. This yields more numbers.
Summing, we get 
See Also
| 2013 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 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 | ||
