Difference between revisions of "2018 UNCO Math Contest II Problems/Problem 7"
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== Problem == | == Problem == | ||
+ | Let <math>x = 2^A + 10^B</math> where <math>A</math> and <math>B</math> are randomly chosen with replacement from among the | ||
+ | positive integers less than or equal to twelve. What is the probability that <math>x</math> is a multiple of | ||
+ | <math>12</math>? | ||
+ | == Solution == | ||
+ | To have a number divisible by <math>12</math>, it must be divisible by <math>3</math> and <math>4</math>. | ||
+ | |||
+ | Consider mod 3 of x: | ||
+ | |||
+ | <math>x \: {\equiv} \: 0 \: (mod \: 3)</math> | ||
+ | |||
+ | <math>2^A + 10^B\:{\equiv}\:0\:(mod\:3)</math> | ||
+ | |||
+ | <math>2^A + 1\:{\equiv}\:0\:(mod\:3)</math> | ||
+ | |||
+ | <math>2^A\:{\equiv}\:2\:(mod\:3)</math> | ||
+ | |||
+ | Here, since <math>2^A\:{\equiv}\:1 \: or \: 2 \: (mod \: 3)</math> for A is even and odd respectively | ||
+ | |||
+ | <math>\therefore \: A</math> is odd | ||
+ | |||
+ | Consider mod 4 of x: | ||
+ | |||
+ | <math>x \: {\equiv} \: 0 \: (mod \: 4)</math> | ||
+ | |||
+ | <math>2^A + 10^B\:{\equiv}\:0\:(mod\:4)</math> | ||
+ | |||
+ | <math>2^A + 2^B\:{\equiv}\:0\:(mod\:4)</math> | ||
+ | |||
+ | <math>\because\:</math> we know A is odd, for <math>A=1</math>, taking <math>mod\:4</math> gives <math>2</math> as result; for <math>A>1</math>, taking <math>mod\:4</math> gives <math>0</math> as result, so we split the case for <math>A=1</math> and <math>A>1</math> here. | ||
+ | |||
+ | For <math>A=1</math>, | ||
+ | |||
+ | <math>2 + 2^B\:{\equiv}\:0\:(mod\:4)</math> | ||
+ | |||
+ | <math>2^B\:{\equiv}\:2\:(mod\:4)</math> | ||
+ | <math>\therefore \: B=1</math> | ||
− | = | + | For <math>A>1</math>, |
+ | |||
+ | <math>2^B\:{\equiv}\:0\:(mod\:4)</math> | ||
+ | |||
+ | <math>\therefore \: B>1</math> | ||
+ | |||
+ | <math>\therefore\:</math> Concluding our above conditions, we have <math>(A,B) = (1,1)</math> or <math>(A,B) \in (\{3,5,7,9,11\},\:\{2,3,4,5,6,7,8,9,10,11,12\})</math> | ||
+ | By counting the number of solutions, the required probability <math>=\frac{1\times1 + 5\times11}{12\times12} = \frac{56}{144} = \frac{7}{18}</math> | ||
== See also == | == See also == | ||
{{UNCO Math Contest box|year=2018|n=II|num-b=6|num-a=8}} | {{UNCO Math Contest box|year=2018|n=II|num-b=6|num-a=8}} | ||
− | [[Category:]] | + | [[Category:Intermediate Probability Problems]] |
Latest revision as of 11:57, 26 September 2022
Problem
Let where
and
are randomly chosen with replacement from among the
positive integers less than or equal to twelve. What is the probability that
is a multiple of
?
Solution
To have a number divisible by , it must be divisible by
and
.
Consider mod 3 of x:
Here, since for A is even and odd respectively
is odd
Consider mod 4 of x:
we know A is odd, for
, taking
gives
as result; for
, taking
gives
as result, so we split the case for
and
here.
For ,
For ,
Concluding our above conditions, we have
or
By counting the number of solutions, the required probability
See also
2018 UNCO Math Contest II (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 | ||
All UNCO Math Contest Problems and Solutions |