Difference between revisions of "2004 AMC 8 Problems/Problem 8"
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Find the number of two-digit positive integers whose digits total <math>7</math>. | Find the number of two-digit positive integers whose digits total <math>7</math>. | ||
| − | <math> \ | + | <math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math> |
== Solution == | == Solution == | ||
| − | The numbers are <math>16, 25, 34, 43, 52, 61, 70</math> which gives us a total of | + | The numbers are <math>16, 25, 34, 43, 52, 61, 70</math> which gives us a total of <math>\boxed{\textbf{(B)}\ 7}</math>. |
| + | |||
| + | ==See Also== | ||
| + | {{AMC8 box|year=2004|num-b=7|num-a=9}} | ||
| + | {{MAA Notice}} | ||
Latest revision as of 00:55, 5 July 2013
Problem
Find the number of two-digit positive integers whose digits total 
.
Solution
The numbers are 
which gives us a total of 
.
See Also
| 2004 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 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 AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
