Difference between revisions of "1971 AHSME Problems/Problem 11"
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− | == Problem | + | == Problem == |
The numeral <math>47</math> in base a represents the same number as <math>74</math> in base <math>b</math>. Assuming that both bases are positive | The numeral <math>47</math> in base a represents the same number as <math>74</math> in base <math>b</math>. Assuming that both bases are positive | ||
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<math>4a+7=7b+4\implies 7b=4a+3\implies b=\frac{4a+3}{7}</math>. <math>4a+3\equiv 0\implies 4a\equiv 4\implies a\equiv1 \pmod{7}</math>. The smallest possible value of <math>a</math> is <math>8</math>. Then, <math>b=\frac{4\cdot8+3}{7}=\frac{35}{7}=5</math>. However, the digit <math>7</math> is not valid in base <math>5</math>, so we have to try a larger value. <math>a=8+7=15</math>, gives a value of <math>\frac{15*4+3}{7}=9</math>, for <math>b</math>, which is valid. | <math>4a+7=7b+4\implies 7b=4a+3\implies b=\frac{4a+3}{7}</math>. <math>4a+3\equiv 0\implies 4a\equiv 4\implies a\equiv1 \pmod{7}</math>. The smallest possible value of <math>a</math> is <math>8</math>. Then, <math>b=\frac{4\cdot8+3}{7}=\frac{35}{7}=5</math>. However, the digit <math>7</math> is not valid in base <math>5</math>, so we have to try a larger value. <math>a=8+7=15</math>, gives a value of <math>\frac{15*4+3}{7}=9</math>, for <math>b</math>, which is valid. | ||
− | <math>15+9=24</math>, which is <math>\mathrm{XXIV}</math> as a roman numeral, and thus the answer is <math> | + | <math>15+9=24</math>, which is <math>\mathrm{XXIV}</math> as a roman numeral, and thus the answer is <math>\boxed{\textbf{(D) }\mathrm{XXIV}}.</math> |
+ | |||
+ | == See Also == | ||
+ | {{AHSME 35p box|year=1971|num-b=10|num-a=12}} | ||
+ | {{MAA Notice}} |
Latest revision as of 12:03, 1 August 2024
Problem
The numeral in base a represents the same number as
in base
. Assuming that both bases are positive
integers, the least possible value of
written as a Roman numeral, is
Solution
.
. The smallest possible value of
is
. Then,
. However, the digit
is not valid in base
, so we have to try a larger value.
, gives a value of
, for
, which is valid.
, which is
as a roman numeral, and thus the answer is
See Also
1971 AHSC (Problems • Answer Key • Resources) | ||
Preceded by Problem 10 |
Followed by Problem 12 | |
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All AHSME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.