Difference between revisions of "2004 AMC 8 Problems/Problem 8"

Revision as of 03:36, 24 December 2012

Find the number of two-digit positive integers whose digits total $7$ .

$\textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10$

The numbers are $16, 25, 34, 43, 52, 61, 70$ which gives us a total of $\boxed{\textbf{(B)}\ 7}$ .

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

@@ Line 2: / Line 2: @@
 Find the number of two-digit positive integers whose digits total <math>7</math>.
-<math> \mathrm{(A)\ 6 }\qquad\mathrm{(B)\ 7 }\qquad\mathrm{(C)\ 8 }\qquad\mathrm{(D)\ 9 }\qquad\mathrm{(E)\ 10 } </math>
+<math> \textbf{(A)}\ 6\qquad\textbf{(B)}\ 7\qquad\textbf{(C)}\ 8\qquad\textbf{(D)}\ 9\qquad\textbf{(E)}\ 10</math>
 == Solution ==
-The numbers are <math>16, 25, 34, 43, 52, 61, 70</math> which gives us a total of <math>7</math>. <math>\boxed{\textbf{(B)}\ 7}</math>
+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}}