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Difference between revisions of "1971 AHSME Problems/Problem 11"

(Created page with "== Problem 11 == 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 integers, ...")
 
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== Problem 11 ==
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== 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.  
  
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<math>15+9=24</math>, which is <math>\mathrm{XXIV}</math> as a roman numeral, and thus the answer is <math>(\textbf{D})</math>
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<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>
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== See Also ==
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{{AHSME 35p box|year=1971|num-b=10|num-a=12}}
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{{MAA Notice}}

Latest revision as of 11:03, 1 August 2024

Problem

The numeral $47$ in base a represents the same number as $74$ in base $b$. Assuming that both bases are positive integers, the least possible value of $a+b$ written as a Roman numeral, is

$\textbf{(A) }\mathrm{XIII}\qquad \textbf{(B) }\mathrm{XV}\qquad \textbf{(C) }\mathrm{XXI}\qquad \textbf{(D) }\mathrm{XXIV}\qquad \textbf{(E) }\mathrm{XVI}$

Solution

$4a+7=7b+4\implies 7b=4a+3\implies b=\frac{4a+3}{7}$. $4a+3\equiv 0\implies 4a\equiv 4\implies a\equiv1 \pmod{7}$. The smallest possible value of $a$ is $8$. Then, $b=\frac{4\cdot8+3}{7}=\frac{35}{7}=5$. However, the digit $7$ is not valid in base $5$, so we have to try a larger value. $a=8+7=15$, gives a value of $\frac{15*4+3}{7}=9$, for $b$, which is valid.

$15+9=24$, which is $\mathrm{XXIV}$ as a roman numeral, and thus the answer is $\boxed{\textbf{(D) }\mathrm{XXIV}}.$

See Also

1971 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
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 • 31 • 32 • 33 • 34 • 35
All AHSME Problems and Solutions

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