Difference between revisions of "2022 SSMO Speed Round Problems/Problem 1"

Latest revision as of 15:23, 24 May 2024

solution 1

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: $\boxed{\text{7}}$

@@ Line 1: / Line 1: @@
-==Problem==
+== solution 1 ==
-Let <math>S_1 = \{2,0,3\}</math> and <math>S_2 = \{2,20,202,2023\}.</math> Find the last digit of
-<cmath>\sum_{a\in S_1,b\in S_2}a^b.</cmath>
-==Solution==
-Since the power of <math>0</math> to an integer is  always <math>0</math>, it
-follows that we want to find the last digit of
-\begin{align*}
-&2^2 + 2^{20} + 2^{202} + 2^{2023} + \\
-&3^2 + 3^{20} + 3^{202} + 3^{2023}
-\end{align*}
-Since the powers of <math>2</math> are <math>2, 4, 8, 16, 32</math>
-it follows that <math>2^n</math> and <math>2^{n+4}</math> have the same last
-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).
-The expression then has the same last digit as
+Consider the probability <math>P(</math> win <math>)</math> as the sum of the probabilities of all sequences where Bobby wins:
-\[^2 + 2^{4} + 2^{2} + 2^{3} + 3^2 + 3^{4} + 3^{2} + 3^{3}
+<math>P(</math> win <math>)=P(2</math> heads and then 1 tails <math>)+P(4</math> heads and then 1 tails <math>)+</math> <math>P(6</math> heads and then 1 tails <math>)+\ldots</math>
-\]
-which is just <math>8</math>.
+For any sequence with <math>2 k</math> heads followed by a tail, the probability is:
+<cmath>
+\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}
+</cmath>
+Factor out the constant term <math>\frac{1}{2}</math> :
+<cmath>
+P(\text { win })=\frac{1}{2} \sum_{k=1}^{\infty}\left(\frac{1}{4}\right)^k
+</cmath>
+This is a geometric series with the first term <math>a=\left(\frac{1}{4}\right)</math> and common ratio
+<cmath>
+r=\left(\frac{1}{4}\right)
+</cmath>
+<cmath>
+\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}
+</cmath>
+Thus:
+<cmath>
+P(\text { win })=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}
+</cmath>
+The probability <math>P(</math> win) can be expressed as:
+<cmath>
+\frac{1}{6}
+</cmath>
+In this case, <math>m=1</math> and <math>n=6</math>. Therefore, <math>m+n=1+6=7</math>.
+Thus, the value of <math>m+n</math> is:
+<cmath>
+\boxed{\text{7}}
+</cmath>

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Difference between revisions of "2022 SSMO Speed Round Problems/Problem 1"

Latest revision as of 15:23, 24 May 2024

solution 1