Difference between revisions of "1987 AHSME Problems/Problem 16"
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\textbf{(C)}\ 82 \qquad | \textbf{(C)}\ 82 \qquad | ||
\textbf{(D)}\ 108 \qquad | \textbf{(D)}\ 108 \qquad | ||
− | \textbf{(E)}\ 113 </math> | + | \textbf{(E)}\ 113 </math> |
+ | |||
+ | == Solution == | ||
+ | Since <math>VYX + 1 = VVW</math>, i.e. adding <math>1</math> causes the "fives" digit to change, we must have <math>X = 4</math> and <math>W = 0</math>. Now since <math>VYZ + 1 = VYX</math>, we have <math>X = Z + 1 \implies Z = 4 - 1 = 3</math>. Finally, note that in <math>VYX + 1 = VVW</math>, adding <math>1</math> will cause the "fives" digit to change by <math>1</math> if it changes at all, so <math>V = Y + 1</math>, and thus since <math>1</math> and <math>2</math> are the only digits left (we already know which letters are assigned to <math>0</math>, <math>3</math>, and <math>4</math>), we must have <math>V = 2</math> and <math>Y = 1</math>. Thus <math>XYZ = 413_{5} = 4 \cdot 5^{2} + 1 \cdot 5 + 3 = 100 + 5 + 3 = 108</math>, which is answer <math>\boxed{\text{D}}</math>. | ||
== See also == | == See also == |
Latest revision as of 14:32, 1 March 2018
Problem
A cryptographer devises the following method for encoding positive integers. First, the integer is expressed in base . Second, a 1-to-1 correspondence is established between the digits that appear in the expressions in base and the elements of the set . Using this correspondence, the cryptographer finds that three consecutive integers in increasing order are coded as , respectively. What is the base- expression for the integer coded as ?
Solution
Since , i.e. adding causes the "fives" digit to change, we must have and . Now since , we have . Finally, note that in , adding will cause the "fives" digit to change by if it changes at all, so , and thus since and are the only digits left (we already know which letters are assigned to , , and ), we must have and . Thus , which is answer .
See also
1987 AHSME (Problems • Answer Key • Resources) | ||
Preceded by Problem 15 |
Followed by Problem 17 | |
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 • 26 • 27 • 28 • 29 • 30 | ||
All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.