Difference between revisions of "2021 AMC 10B Problems/Problem 13"
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<math>\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16</math> | <math>\textbf{(A)} ~10 \qquad\textbf{(B)} ~11 \qquad\textbf{(C)} ~13 \qquad\textbf{(D)} ~15 \qquad\textbf{(E)} ~16</math> | ||
− | ==Solution== | + | ==Solution 1== |
− | We can start by setting up an equation to convert <math>\underline{32d}</math> base <math>n</math> to base 10. To convert this to base 10, it would be | + | We can start by setting up an equation to convert <math>\underline{32d}</math> base <math>n</math> to base 10. To convert this to base 10, it would be <math>3{n}^2+2n+d.</math> Because it is equal to 263, we can set this equation to 263. Finally, subtract <math>d</math> from both sides to get <math>3{n}^2+2n = 263-d</math>. |
− | We can also set up equations to convert <math>\underline{324}</math> base <math>n</math> and <math>\underline{11d1}</math> base 6 to base 10. The equation to covert <math>\underline{324}</math> base <math>n</math> to base 10 is | + | We can also set up equations to convert <math>\underline{324}</math> base <math>n</math> and <math>\underline{11d1}</math> base 6 to base 10. The equation to covert <math>\underline{324}</math> base <math>n</math> to base 10 is <math>3{n}^2+2n+4.</math> The equation to convert <math>\underline{11d1}</math> base 6 to base 10 is <math>{6}^3+{6}^2+6d+1.</math> |
− | Simplify <math>{6}^3 | + | Simplify <math>{6}^3+{6}^2+6d+1</math> so it becomes <math>6d+253.</math> Setting the above equations equal to each other, we have |
+ | <cmath>3{n}^2+2n+4 = 6d+253.</cmath> | ||
+ | Subtracting 4 from both sides gets <math>3{n}^2+2n = 6d+249.</math> | ||
− | We can then use | + | We can then use equations |
+ | <cmath>3{n}^2+2n = 263-d</cmath> | ||
+ | <cmath>3{n}^2+2n = 6d+249</cmath> | ||
+ | to solve for <math>d</math>. Set <math>263-d</math> equal to <math>6d+249</math> and solve to find that <math>d=2</math>. | ||
− | Plug <math>d</math> | + | Plug <math>d=2</math> back into the equation <math>3{n}^2+2n = 263-d</math>. Subtract 261 from both sides to get your final equation of <math>3{n}^2+2n-261 = 0.</math> We solve using the quadratic formula to find that the solutions are <math>9</math> and <math>-29/3.</math> Because the base must be positive, <math>n=9.</math> |
− | Adding 2 to 9 gets <math>\boxed{\textbf{(B)}11}</math | + | Adding 2 to 9 gets <math>\boxed{\textbf{(B)} ~11}</math> |
− | |||
+ | -Zeusthemoose (edited for readability) | ||
+ | -solution corrected by Billowingsweater | ||
− | - | + | ==Solution 2== |
+ | <math>32d</math> is greater than <math>263</math> when both are interpreted in base 10, so <math>n</math> is less than <math>10</math>. Some trial and error gives <math>n=9</math>. <math>263</math> in base 9 is <math>322</math>, so the answer is <math>9+2=\boxed{\textbf{(B)} ~11}</math>. | ||
+ | |||
+ | -SmileKat32 | ||
+ | |||
+ | ==Solution 3== | ||
+ | We have | ||
+ | <cmath>3n^2 + 2n + d = 263</cmath> | ||
+ | <cmath>3n^2 + 2n + 4 = 6^3 + 6^2 + 6d + 1</cmath> | ||
+ | Subtracting the 2nd from the 1st equation we get | ||
+ | <cmath>\begin{align*} | ||
+ | d-4 &= 263 - (216 + 36 + 6d + 1) \\ | ||
+ | &= 263 - 253 - 6d \\ | ||
+ | &= 10 - 6d | ||
+ | \end{align*}</cmath> | ||
+ | Thus we have <math>d=2.</math> | ||
+ | Substituting into the first, we have <math>3n^2 + 2n + 2 = 263 \Rightarrow 3n^2 + 2n - 261 = 0.</math> | ||
+ | Factoring, we have <math>(n-9)(3n+29)=0.</math> | ||
+ | A digit cannot be negative, so we have <math>n=9.</math> | ||
+ | Thus, <math>d+n=2+9=\boxed{\textbf{(B)} ~11}</math> | ||
+ | |||
+ | mathboy282 signing off | ||
+ | ~<math>\LaTeX</math> fixed by Lamboreghini | ||
+ | |||
+ | ==Solution 4== | ||
+ | Find that <math>d=2</math> using one of the methods above. Then we have that <math>3n^2 + 2n = 261</math>. We know that <math>n</math> is an integer, so we can solve the equation <math>n(3n+2)=261</math> (this is guaranteed to have a solution if we did this correctly). The prime factorization of <math>261</math> is <math>3^2 \cdot 29</math>, so the corresponding factor pairs are <math>(1, 261)</math>, <math>(3,87)</math>, and <math>(9,29)</math>. If <math>n = 9</math>, then the equation is true, so <math>n + d = 9 + 2 = \boxed{\textbf{(B)} ~11}</math>. | ||
+ | |||
+ | Note: This solution provides an alternate way to complete the final part of the problem if, like me, you really dislike quadratics. | ||
+ | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] | ||
+ | |||
+ | ==Video Solution (🚀 Under 4 min 🚀)== | ||
+ | https://youtu.be/jm2VbqRpFyI | ||
+ | |||
+ | ~<i> Education, the Study of Everything </i> | ||
+ | |||
+ | == Video Solution by OmegaLearn (Bases and System of Equations) == | ||
+ | https://youtu.be/oAc3GdAm6lk | ||
+ | |||
+ | ~ pi_is_3.14 | ||
+ | |||
+ | ==Video Solution by TheBeautyofMath== | ||
+ | https://youtu.be/L1iW94Ue3eI?t=880 | ||
+ | |||
+ | ~IceMatrix | ||
+ | |||
+ | ==Video Solution by Interstigation== | ||
+ | https://youtu.be/X86a7-pSSSY | ||
+ | |||
+ | ~Interstigation | ||
+ | |||
+ | ==See Also== | ||
+ | {{AMC10 box|year=2021|ab=B|num-b=12|num-a=14}} | ||
+ | {{MAA Notice}} |
Latest revision as of 21:54, 12 May 2024
Contents
Problem
Let be a positive integer and
be a digit such that the value of the numeral
in base
equals
, and the value of the numeral
in base
equals the value of the numeral
in base six. What is
Solution 1
We can start by setting up an equation to convert base
to base 10. To convert this to base 10, it would be
Because it is equal to 263, we can set this equation to 263. Finally, subtract
from both sides to get
.
We can also set up equations to convert base
and
base 6 to base 10. The equation to covert
base
to base 10 is
The equation to convert
base 6 to base 10 is
Simplify so it becomes
Setting the above equations equal to each other, we have
Subtracting 4 from both sides gets
We can then use equations
to solve for
. Set
equal to
and solve to find that
.
Plug back into the equation
. Subtract 261 from both sides to get your final equation of
We solve using the quadratic formula to find that the solutions are
and
Because the base must be positive,
Adding 2 to 9 gets
-Zeusthemoose (edited for readability) -solution corrected by Billowingsweater
Solution 2
is greater than
when both are interpreted in base 10, so
is less than
. Some trial and error gives
.
in base 9 is
, so the answer is
.
-SmileKat32
Solution 3
We have
Subtracting the 2nd from the 1st equation we get
Thus we have
Substituting into the first, we have
Factoring, we have
A digit cannot be negative, so we have
Thus,
mathboy282 signing off
~ fixed by Lamboreghini
Solution 4
Find that using one of the methods above. Then we have that
. We know that
is an integer, so we can solve the equation
(this is guaranteed to have a solution if we did this correctly). The prime factorization of
is
, so the corresponding factor pairs are
,
, and
. If
, then the equation is true, so
.
Note: This solution provides an alternate way to complete the final part of the problem if, like me, you really dislike quadratics. ~ cxsmi
Video Solution (🚀 Under 4 min 🚀)
~ Education, the Study of Everything
Video Solution by OmegaLearn (Bases and System of Equations)
~ pi_is_3.14
Video Solution by TheBeautyofMath
https://youtu.be/L1iW94Ue3eI?t=880
~IceMatrix
Video Solution by Interstigation
~Interstigation
See Also
2021 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 12 |
Followed by Problem 14 | |
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 | ||
All AMC 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.