Problems for children from 5 to 15 by V.I. Arnold

My solutions to a select number of problems .

By Vamshi Jandhyala in mathematics

May 14, 2020

Problem 1

Masha was seven kopecks short to buy a first reading book, and Misha lacked one kopeck. They combined their money to buy one book to share, but even then they did not have enough. How much did the book cost?

Solution

Let the cost of the book be $b$ kopecks. Let $Masha$ have $M$ kopecks and $Misha$ have $m$ kopecks.We have

$$ \begin{align*} M &= b-7 \
m &= b-1 \end{align*} $$

We have $b>7$ and $M+m=2b-8 < b$. Therefore, the cost of the book is between $7$ and $8$ kopecks.

Problem 2

A bottle with a cork costs 10 kopecks, while the bottle itself is 9 kopecks more expensive than the cork. How much does the bottle without the cork cost?

Solution

Let the cost of the bottle with cork be $b + c = 10$ kopecks. As the bottle is $9$ kopecks more than the cork, we have $b = c + 9$. Therefore $2b-9 = 10$ which mean the bottle without cork costs $9.5$ kopecks.

Problem 3

A brick weighs one pound and half the brick. How many pounds does the brick weigh?

Solution

We have

$$ \begin{align*} b &= 1 + \frac{b}{2} \
\implies b &= 2 \end{align*} $$

Therefore the brick weighs $2$ pounds.

Problem 4

A spoon of wine is poured from a barrel of wine into a (not full) glass of tea. After that, the same spoon of the (in homogeneous) mixture from the glass is taken back into the barrel. Now both in the barrel and in the glass there is a certain volume of the foreign liquid (wine in the glass and tea in the barrel). In which is the volume of the foreign liquid greater: in the glass or in the barrel?

Solution

Let $W$ be the volume of wine in the barrel, $s$ be the volume of the spoon and $T$ be the volume to tea. The amount of wine added to the glass of tea the first time is $s$. The amount of wine removed from the glass of tea using the spoon is $ss/(T+s)$. The amount of wine left in the glass is given by $s-ss/(T+s) = sT/(T+s)$. The volume of tea in the spoon after a spoon of wine has been added to the glass of tea is $sT/(T+s)$. Therefore, the volume of the foreign liquid in the glass is same as that in the barrel.

Problem 5

Two old ladies left from $A$ to $B$ and from $B$ to $A$ at dawn heading towards one another (along the same road). They met at noon, but did not stop,and each of them carried on walking with the same speed. The first lady came (to $B$) at $4$p.m., and the second (to $A$) at $9$p.m. What time was the dawn that day?

Solution

Let $u$ be the speed of the first lady and $v$ be the speed of the second lady and $D$a.m be the time of dawn. We have $(16-D)u = (21-D)v = (12-D)(u+v)$. Hence,

$$ \begin{align*} \frac{12-D}{16-D} + \frac{12-D}{21-D} &= 1\
\implies (D-6)(D-18) = 0 \end{align*} $$

Therefore, the time of dawn that day was $6$a.m.

Problem 6

The hypotenuse of a right-angled triangle(in a standard American examination) is $10$ inches, the altitude dropped onto it is $6$ inches. Find the area of the triangle. American school students had been coping successfully with this problem over a decade. But then Russian school students arrived from Moscow, and none of them was able to solve it as had their American peers (giving $30$ square inches as the answer). Why?

Solution

The circumradius of the triangle is $5$. The altitude of the triangle cannot be greater than $5$. The triangle mentioned in the above problem cannot exist.

Problem 7

Vasya has $2$ sisters more than he has brothers. How many daughters more than sons do Vasya’s parents have?

Solution

Let $b$ be the number of brothers Vasya has. The number of sisters Vasya has is $2+b$. The total number of sons is $b+1$ and the total number of daughters is $b+2$. Therefore Vasya’s parents have $1$ more daughter compared to the number of sons.

Problem 8

There is a round lake in South America. Every year, on June $1$, a Victoria Regia flower appears at its centre (its stem rises from the bottom,and its petals lie on the water like those of a water lily). Every day the area of the flower doubles, and on July $1$, it finally covers the entire lake, drops its petals, and its seeds sink to the bottom. On which date is the flower’s area half the area of the lake?

Solution

If the flower covered the entire lake on July $1$, on June $30$, the flower must have covered half the lake as the area of the flower doubles each day.

Problem 9

A peasant must take a wolf, a goat and a cabbage across a river in a boat. However the boat is so small that he is able to take only one of the three on board with him.How should he transport all three across the river? (The wolf cannot be left alone with the goat, and the goat cannot be left alone with the cabbage.)

Solution

Here are the steps that need to be carried out by the peasant:

  1. Peasant leaves the wolf with the cabbage and carries the goat across to the other side of the river.
  2. Peasant leaves the goat on the other side and comes back to take the wolf.
  3. Peasant leaves the wolf on the other side and comes back with the goat.
  4. Peasant leaves the goat and takes the cabbage to the other side.
  5. Peasant comes back to take the goat to the other side.

Problem 10

During the daytime a snail climbs $3$cm up a post, and during the night, falling asleep, accidentally goes down by $2$cm. The post is $10$m high, and a delicious (for the snail) sweet is on its top. In how many days will the snail get the sweet?

Solution

Each day the snail covers $1$cm in total and reaches a maximum of $d+2$ where $d$ is the number of days the snail has climbed so far.After $997$ days, the snail climbs $997$ cm. On day $998$, the snail climbs the post and gets the sweet.

Problem 11

A ranger walked from his tent 10km southwards, turned east, walked straight eastwards 10km more, met his bear friend, turned north and after another 10km found himself by his tent. What colour was the bear and where did all this happen?

Solution

The colour of the bear is white as this happened near the north pole.

Problem 13

Two volumes of Pushkin, the first and the second, are side-by-side on a bookshelf. The pages of each volume are $2$cm thick, and the cover - front and back each - is $2$mm. A bookworm has gnawed through (perpendicular to the pages) from the first page of volume $1$ to the last page of volume $2$. How long is the bookworm’s track? [This topological problem with an incredible answer $4$mm is absolutely impossible for academicians, but some preschoolers handle it with ease.]

Solution

If the second volume is to the right of the first volume and the spine of the book is facing outward from the shelf, then the worm only needs to gnaw through the front cover of the first volume and back cover of the second volume to go from the first page of volume one to the last page of volume two.

Problem 17

The distance between cities $A$ and $B$ is $40$km. Two cyclists leave respectively from $A$ and $B$ simultaneously towards one another, one with speed $10$km/h and the other with speed $15$km/h. A fly flies out with the first cyclist from A with the speed of $100$km/h, reaches the second, touches his forehead and flies back to the first, touches his forehead,returns to the second,and so on until the cyclists’ foreheads collide and squash the fly. How many kilometres altogether has the fly flown?

Solution

The time for the cyclists to collide is $40/(10+15)$hrs. The fly continues flies during this time with a speed of $100$km/h. Therefore, the total distance travelled by the fly is $\frac{40}{25}\cdot 100 = 160 km$.

Problem 18

One domino piece covers two squares of a chessboard. Cover all the squares except for two opposite ones (on the same diagonal) with 31 pieces.

Solution

A domino piece always covers one white square and one black square. If opposite squares of a chessboard are removed, you either have $30$ white squares or $30$ black squares remaining. Therefore, the $31$ dominoes cannot fully cover the remaining chessboard.

Problem 19

A caterpillar wants to slither from a corner of a cubic room(the left on the floor) to the opposite one (the right on the ceiling). Find the shortest route for such a journey along the walls of the room.

Solution

Flattening the cube will give a straight line distance between the two opposite vertices as $\sqrt{2^2+1^2}=\sqrt{5}$.

Problem 20

You have two vessels of volumes $5$ litres and $3$ litres. Measure out one litre (obtaining it in one of the vessels).

Solution

Here is the sequence of steps:

  1. Fill the $3$ litre vessel.
  2. Empty it into the $5$ litre vessel.
  3. Fill the $3$ litre vessel again.
  4. Empty $2$ litres into the $5$ litre vessel.
  5. $1$ litre remains in the $3$ litre vessel.

Problem 21

There are five heads and fourteen legs in a family. How many people and how many dogs are in the family?

Solution

We have

$$ \begin{align*} p + d &= 5 \
2p + 4d &= 14 \end{align*} $$

Solving the above, we see that $d=2$ and $p=3$.

Problem 26

The Earth’s surface is projected onto the cylinder formed by the lines tangent to the meridians at their equatorial points along the rays parallel to the equator and passing through the Earth’s pole axis. Will the area of the projection of France be greater or smaller than the area of France itself?

Solution

The area of the cylinder on which the Earth’s surface is projected is given by $2\pi R \cdot 2R = 4\pi R^2$ where $R$ is the radius of Earth. This is same as the surface area of the Earth(sphere of radius $R$) which is $4\pi R^2$. Therefore the area of the projection of France will be the same as the area of France itself.

Problem 27

Prove that the remainder of division of the number $2^{p-1}$ by an odd prime $p$ is $1$.

Solution

We have

$$ \begin{align*} (1+1)^p &= 1 + {p \choose 1} + {p \choose 2} + \dots + 1 \
2^p - 2 &= {p \choose 1} + {p \choose 2} + \dots \end{align*} $$

Every term on the right hand side above is divisible by $p$ and none of the numbers $1,2,\dots,p-1$ divide $p$ as $p$ is prime.We have $2^p\equiv2(mod \text{ p})$.Therefore, $2^{p-1} \equiv 1 (mod \text{ p})$.

Problem 29

Polyhedra with triangular faces are,for example,Platonic solids: tetrahedron ($4$ faces), octahedron ($8$ of them), icosahedron ($20$ – and all the faces are the same;it is interesting to drawit,it has $12$ vertices and $30$ edges). Is it true that for any such (bounded convex polyhedron with triangular faces) the number of faces is equal to twice the number of vertices minus four?

Solution

From Euler’s Formula we have $V-E+F=2$. For a convex polyhedron with $F$ triangular faces, we have $3F/2$ edges. Every face has $3$ edges and each edge is counted twice.We now have $V-3F/2+F=2$ which gives us $F=2V-4$. Therefore, it is true that for any such (bounded convex polyhedron with triangular faces) the number of faces is equal to twice the number of vertices minus four.

Problem 34

How many different ways are there to permute $n$ objects?

Solution

For the first position in the permutation, we have $n$ options, for the second position, we have $n-1$ options and so on. Therefore the number of ways to permute $n$ objects is given by $n(n-1)\dots1=n!$.

Problem 36

The sum of the cubes of three integers is subtracted from the cube of the sum of these numbers. Is the difference always divisible by $3$?

Solution

We have

$$ \begin{align*} (a+b+c)^3-a^3-b^3-c^3 = 3(a+b)(b+c)(c+a) \end{align*} $$

Therefore, the difference is always divisible by $3$.

Problem 38

Calculate the sum

$$ \begin{align*} \frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{99\cdot100} \end{align*} $$

Solution

We have

$$ \begin{align*} &\frac{1}{1\cdot2} + \frac{1}{2\cdot3} + \frac{1}{3\cdot4} + \cdots + \frac{1}{99\cdot100} \
&= \frac{1}{1} - \frac{1}{2} +
\frac{1}{2} - \frac{1}{3} + \cdots +\frac{1}{99} - \frac{1}{100} \
&= 1 - \frac{1}{100} = \frac{99}{100} \end{align*} $$

Problem 43

The rabbit (or Fibonacci) numbers form a sequence $1,2,3,5,8,13,21,34,\dots$ in which $a_{n+2} = a_{n+1} + a_n$ for any $n = 1,2,\dots…$ ($a_n$ is the n-th number in the sequence). Find the greatest common divisor of the numbers $a_{100}$ and $a_{99}$.

Solution

From the Euclidean Algorithm for GCD, we have

$$ \begin{align*} GCD(a_{100}, a_{99}) = GCD(a_{99} + a_{98}, a_{99}) = GCD(a_{99}, a_{98}) \
= GCD(a_2, a_1) = GCD(2,1) = 1 \end{align*} $$

Problem 53

For the sequence of Fibonacci numbers $a_n$ from problem $43$,find the limit of the ratio $a_{n+1}/a_n$ when $n$ tends to infinity.

Solution

Let $\tau$ be the limit of the ratio $a_{n+1}/a_n$ when $n$ tends to infinity. We have,

$$ \begin{align*} &\lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n} = \tau = \lim_{n \rightarrow \infty} 1 + \frac{a_{n-1}}{a_n} = 1 + \frac{1}{\tau} \
&\implies \tau^2-\tau-1=0. \end{align*} $$

Therefore, $\tau = \frac{1+\sqrt{5}}{2}$ which is the golden ratio.

Problem 54

Calculate the infinite continued fraction

$$ \begin{align*} 1 + \cfrac{1}{2 + \cfrac{1}{1 + \cfrac{1}{2 + \cfrac{1}{1+\cfrac{1}{2 + \dots}}}}} = a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \cfrac{1}{a_3 + \dots}}} \end{align*} $$

with $a_{2k} = 1$ and $a_{2k+1} = 2$ (that is, find the limit of the fractions

$$ \begin{align*} a_0 + \cfrac{1}{a_1 + \cfrac{1}{a_2 + \ddots + \cfrac{1}{a_n}}} \end{align*} $$

for $n \rightarrow \infty$).

Solution

If the limit of the above continued fraction as $n \rightarrow \infty$ is $x$, we have

$$ \begin{align*} x &= 1 + \frac{1}{2+\cfrac{1}{x}} \
\implies 2x^2-2x-1 &= 0 \end{align*} $$

Therefore $x = \frac{1 + \sqrt{3}}{2}$.

References

Arnold’s Problems