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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 1"
 
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==Problem==
 
==Problem==
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Let <math>S_1 = \{2,0,3\}</math> and <math>S_2 = \{2,20,202,2023\}.</math> Find the last digit of
 
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<cmath>\sum_{a\in S_1,b\in S_2}a^b.</cmath>
 
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==Solution==
 
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Since the power of <math>0</math> to an integer is  always <math>0</math>, it
 
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follows that we want to find the last digit of
 
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<cmath>2^2 + 2^{20} + 2^{202} + 2^{2023} + </cmath>
 
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<cmath>3^2 + 3^{20} + 3^{202} + 3^{2023}</cmath>
 
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\end{align*}
 
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Since the powers of <math>2</math> are <math>2, 4, 8, 16, 32</math>
 
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it follows that <math>2^n</math> and <math>2^{n+4}</math> have the same last
 
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digit for <math>n \ge 1</math>. Similarily, <math>3^n</math> and <math>3^{n+4}</math> have the same last digit. (This follows as <math>\varphi(10) = 4</math> too).
 
  
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The expression then has the same last digit as
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Bobby is bored one day and flips a fair coin until it lands on tails. Bobby wins if the coin lands on heads a positive even number of times in the sequence of tosses. Then the probability that Bobby wins can be expressed in the form <math>\tfrac mn</math>, where <math>m</math> and <math>n</math> are relatively prime positive integers. Find <math>m+n</math>.
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<cmath>2^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}</cmath>
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which is just <math>8</math>.
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==Solution 1==
 +
Note that the first two flips must be tails, which occurs with a probabilty of  <math>\left(\tfrac12\right)^2=\frac14</math>th. Assuming the first two are tails, let <math>x</math> be the probability Bobby wins. Note that the coin must land an even or one more than an even number of times (when first flip is tails), meaning <math>x+\tfrac12 x=1\implies x=\tfrac23.</math> So, our answer is <cmath>\left(\frac14\right)\left(\frac23\right)=\frac16\implies1+6=\boxed{7}.</cmath>
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 +
~pinkpig
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==Solution 2==
 +
 
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Consider the probability <math>P(</math> win <math>)</math> as the sum of the probabilities of all sequences where Bobby wins:
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<math>P(</math> win <math>)=P(2</math> heads and then 1 tails <math>)+P(4</math> heads and then 1 tails <math>)+</math> <math>P(6</math> heads and then 1 tails <math>)+\ldots</math>
 +
 
 +
For any sequence with <math>2 k</math> heads followed by a tail, the probability is:
 +
<cmath>
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\left(\frac{1}{2}\right)^{2 k} \times\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2 k+1}
 +
</cmath>
 +
 
 +
We sum this for <math>k=1,2,3, \ldots</math> :
 +
<cmath>
 +
P(\text { win })=\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{2 k+1}
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</cmath>
 +
 
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Factor out the constant term <math>\frac{1}{2}</math> :
 +
<cmath>
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P(\text { win })=\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^k
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</cmath>
 +
 
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This is a geometric series with the first term <math>a=\left(\frac{1}{4}\right)</math> and common ratio
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<cmath>
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r=\left(\frac{1}{4}\right)
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</cmath>
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<cmath>
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\sum_{k=0}^{\infty} r^k=\frac{a}{1-r}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}
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</cmath>
 +
 
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Thus:
 +
<cmath>
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P(\text { win })=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}
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</cmath>
 +
 
 +
The probability <math>P(</math> win) can be expressed as:
 +
<cmath>
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\frac{1}{6}
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</cmath>
 +
 
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In this case, <math>m=1</math> and <math>n=6</math>. Therefore, <math>m+n=1+6=7</math>.
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Thus, the value of <math>m+n</math> is:
 +
<cmath>
 +
\boxed{\text{7}}
 +
</cmath>

Latest revision as of 16:59, 13 September 2025

Problem

Bobby is bored one day and flips a fair coin until it lands on tails. Bobby wins if the coin lands on heads a positive even number of times in the sequence of tosses. Then the probability that Bobby wins can be expressed in the form $\tfrac mn$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

Solution 1

Note that the first two flips must be tails, which occurs with a probabilty of $\left(\tfrac12\right)^2=\frac14$th. Assuming the first two are tails, let $x$ be the probability Bobby wins. Note that the coin must land an even or one more than an even number of times (when first flip is tails), meaning $x+\tfrac12 x=1\implies x=\tfrac23.$ So, our answer is \[\left(\frac14\right)\left(\frac23\right)=\frac16\implies1+6=\boxed{7}.\]

~pinkpig

Solution 2

Consider the probability $P($ win $)$ as the sum of the probabilities of all sequences where Bobby wins: $P($ win $)=P(2$ heads and then 1 tails $)+P(4$ heads and then 1 tails $)+$ $P(6$ heads and then 1 tails $)+\ldots$

For any sequence with $2 k$ heads followed by a tail, the probability is: \[\left(\frac{1}{2}\right)^{2 k} \times\left(\frac{1}{2}\right)=\left(\frac{1}{2}\right)^{2 k+1}\]

We sum this for $k=1,2,3, \ldots$ : \[P(\text { win })=\sum_{k=1}^{\infty}\left(\frac{1}{2}\right)^{2 k+1}\]

Factor out the constant term $\frac{1}{2}$ : \[P(\text { win })=\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^k\]

This is a geometric series with the first term $a=\left(\frac{1}{4}\right)$ and common ratio \[r=\left(\frac{1}{4}\right)\] \[\sum_{k=0}^{\infty} r^k=\frac{a}{1-r}=\frac{\frac{1}{4}}{1-\frac{1}{4}}=\frac{\frac{1}{4}}{\frac{3}{4}}=\frac{1}{3}\]

Thus: \[P(\text { win })=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}\]

The probability $P($ win) can be expressed as: \[\frac{1}{6}\]

In this case, $m=1$ and $n=6$. Therefore, $m+n=1+6=7$. Thus, the value of $m+n$ is: \[\boxed{\text{7}}\]

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