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 == | ||
| − | {{AHSME box|year=1967|num-b=15|num-a=17}} | + | {{AHSME 40p box|year=1967|num-b=15|num-a=17}} |
[[Category:Introductory Algebra Problems]] | [[Category:Introductory Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:39, 16 August 2023
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 AHSC (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 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
