Art of Problem Solving

2016 AMC 10A Problems/Problem 6: Difference between revisions

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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>
== 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 $1$ through $30$ once. Emilio copies Ximena's numbers, replacing each occurrence of the digit $2$ by the digit $1$. Ximena adds her numbers and Emilio adds his numbers. How much larger is Ximena's sum than Emilio's?

$\textbf{(A)}\ 13\qquad\textbf{(B)}\ 26\qquad\textbf{(C)}\ 102\qquad\textbf{(D)}\ 103\qquad\textbf{(E)}\ 110$

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 $10\cdot 10+1\cdot 3=\boxed{\textbf{(D) }103.}$

See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
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

These problems are copyrighted © by the Mathematical Association of America.