Difference between revisions of "1967 AHSME Problems/Problem 16"
(Created page with "== Problem == Let the product <math>(12)(15)(16)</math>, each factor written in base <math>b</math>, equals <math>3146</math> in base <math>b</math>. Let <math>s=12+15+16</math>...") |
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== Solution == | == Solution == | ||
− | <math>\fbox{B}</math> | + | |
+ | Converting everything into base <math>10</math>, we have <math>(b + 2)(b+5)(b+6) = 3b^3 + b^2 + 4b + 6</math>. Looking ahead, the constant term of the polynomial will be <math>2*5*6 -6 = 54</math>. By the Rational Root Theorem, the only possible integer roots are <math>1, 2, 3, 6, 9, 18, 27, 54</math>. Bases <math>1, 2, 3, 6</math> do not have a <math>6</math> as a digit. Testing <math>b=9</math> gives a solution that works. | ||
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+ | Therefore, we are working in base <math>9</math>. Adding the units place in base <math>9</math>, <math>2 + 5 + 6 = 14_9</math>, so we carry the <math>1</math> to get a total of <math>44_9</math>, which is option <math>\fbox{B}</math>. | ||
== See also == | == See also == |
Latest revision as of 01:59, 13 July 2019
Problem
Let the product , each factor written in base , equals in base . Let , each term expressed in base . Then , in base , is
Solution
Converting everything into base , we have . Looking ahead, the constant term of the polynomial will be . By the Rational Root Theorem, the only possible integer roots are . Bases do not have a as a digit. Testing gives a solution that works.
Therefore, we are working in base . Adding the units place in base , , so we carry the to get a total of , which is option .
See also
1967 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 |
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