Difference between revisions of "2016 AMC 10A Problems/Problem 6"
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| + | == Problem == | ||
Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's? | Ximena lists the whole numbers <math>1</math> through <math>30</math> once. Emilio copies Ximena's numbers, replacing each occurrence of the digit <math>2</math> by the digit <math>1</math>. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's? | ||
<math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math> | <math>\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110</math> | ||
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| + | == Solution == | ||
| + | |||
| + | For every tens digit 2, we subtract 10, and for every units digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, the answer is <math>10\cdot 10+1\cdot 3=\boxed{\textbf{(D) }103.}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2016|ab=A|num-b=5|num-a=7}} | ||
| + | {{MAA Notice}} | ||
Revision as of 18:34, 3 February 2016
Problem
Ximena lists the whole numbers 
through 
once. Emilio copies Ximena's numbers, replacing each occurrence of the digit 
by the digit . Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?
Solution
For every tens digit 2, we subtract 10, and for every units digit 2, we subtract 1. Because 2 appears 10 times as a tens digit and 2 appears 3 times as a units digit, the answer is 
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 5 |
Followed by Problem 7 | |
| 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 | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
