Mathematical Quickies

Problems with quick and elegant solutions.

By Vamshi Jandhyala in mathematics

April 21, 2021

Problem (credit: Muthu Veerappan): How many cards on average should we take (without replacement) from a shuffled pack of cards to get an Ace?

Solution: From symmetry, we see that the four aces divide the rest of the $48$ cards into $5$ equal groups of $9.6$ cards each. Therefore, we need to take $9.6$ cards on average to get the first ace.

Problem: Prove that no number in the sequence $11, 111, 1111, 11111, \dots$ is the square of an integer.

Solution: All terms in the above sequence are of the form $4k+3$ and no square integer can have that form.

Problem: Let $n$ be a positive integer and $x>0$, prove that $(1+x)^{n+1} \geq \frac{(n+1)^{n+1}}{n^n}x$.

Solution: By $AM \geq GM$, we have $\frac{\underbrace{1 + \cdots + 1}_{\text{n times}} + nx}{n + 1} \geq (nx)^{1/(n+1)}$.

Problem: The positive real numbers $p, q, r$ are such that $q \neq r$ and $2p = q + r$. Show that

$$ \frac{p^{q+r}}{q^qr^r} \leq 1 $$

Solution: Using $GM \geq HM$, we have

$$ \begin{align*} &\frac{q + r}{\underbrace{\frac{1}{q} + \cdots + \frac{1}{q}}_\text{q times} + \underbrace{\frac{1}{r} + \cdots + \frac{1}{r}}_\text{r times}} \leq (q^qr^r)^{\frac{1}{q+r}} \
&\implies \frac{2p}{2} \leq (q^qr^r)^{\frac{1}{q+r}} \implies \frac{p^{q+r}}{q^qr^r} \leq 1 \end{align*} $$

Problem: Prove that $i^i$ is real and find its value.

Solution: $i = e^{i\frac{\pi}{2}}$. Therefore, $i^i = e^{i^2\frac{\pi}{2}} = e^{-\frac{\pi}{2}}$ which is real.

Problem: What is the remainder when $23^{23}$ is divided by $53$?

Solution: $23^2 = 529 \equiv -1\mod(53)$. $23^{22} \equiv (-1)^{11} \mod(53)$. Therefore, $23^{23} \equiv -23 \mod(53) \equiv 30 \mod(53)$.

Problem: Let $f_n$ be a sequence for which $f_n \neq 0, \forall n$ and $f_{n+1} = -\frac{f_n^2+1}{f_{n-1}}$. Prove that $\exists k \in \mathbb{R}$ such that $f_{n+1} = kf_n - f_{n-1}$.

Solution: Since $f_n \neq 0$, we have

$$ \begin{align*} \frac{f_{n+1}+f_{n-1}}{f_n} = \frac{\frac{f_n^2-1}{f_{n-1}}+f_{n-1}}{f_n} = \frac{f_n^2 + f_{n-1}^2 - 1}{f_n f_{n-1}} \
= \frac{f_n}{f_{n-1}} + \frac{f_{n-2}}{f_{n-1}} = \frac{f_2}{f_1} + \frac{f_0}{f_1} = k \end{align*} $$

Problem: Each of the numbers $a_1, a_2, …, a_n$ is $1$ or $−1$, and we have $S = a_1 a_2 a_3 a_4 + a_2 a_3 a_4 a_5 + \dots + a_n a_1 a_2 a_3 = 0$. Prove that $n$ is divisible by $4$.

Solution: It is easy to see that $n$ cannot be odd as to get a sum of zeros you need $1$ and $-1$ to occur in pairs, so $n$ cannot be of the form $4k+1$ and $4k+3$. So $n$ can only be of the form $4k$ or $4k+2$. If $n=4k+2$, you would have $2k+1$ terms in the sum $S$ which are $-1$ and $2k+1$ which are $1$. The product of all the $4k+2$ terms will be $-1$ but the product of all the terms in $S$ which is $(a_1\dots a_n)^4$ is $1$ which is a contradiction. Therefore, $n$ has to be of the form $4k$.

Problem: Some tribesman belonging to $2$ hostile clans are sitting on a circular table such the number of tribesman with friends (same clan) on left is equal to number of tribesman with foes (opposing clan) on the left. Prove that total number of tribesman is divisible by $4$.

Solution: Let $a_1, a_2. \dots a_n$ represent the $n$ tribesmen. Let $a_i = 1$ for the first tribe and $a_i = -1$ for the second. This would mean $a_i a_{i+1} = 1$ when $i$ and $i+1$ places have members of the same tribe and $-1$ otherwise. The statement of the problem is equivalent to $a_1a_2 + a_2a_3 + \dots + a_n a_1 = 0$. Using the same logic as above, we can see show that the number of tribesmen is divisible by $4$.