The 20th International Mathematics and Science Olympiad (IMSO) – Mathematics Short Answer Problems challenged young mathematicians with a variety of thought-provoking and conceptually rich problems. In this post, we provide detailed solutions to these problems, helping students understand the key strategies, logical reasoning, and problem-solving techniques required to tackle them effectively. Whether you’re a participant, a coach, or an enthusiast, these solutions will offer valuable insights into the mathematical concepts tested in this prestigious competition.
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Problem Number 1
The diagram below is the expanded net of a cube. We label each of the faces of the cube by using the numbers $1$, $2$, $3$, $4$, $5$, and $6$ exactly once, so that the sum of any two opposite faces is always $7$. What is the largest possible value of $A \times (B^2 + C)$?
We are given the net of a cube, where each face is labeled with the numbers $1, 2, 3, 4, 5,$ and $6,$ ensuring that the sum of any two opposite faces is always $7.$ Our goal is to determine the maximum possible value of the expression $A \times (B^2 + C).$
Since the sum of opposite faces is always $7,$ we can establish the following face pairs: $(1, 6),$ $(2, 5),$ and $(3, 4).$ To maximize the expression $A \times (B^2 + C),$ we should first maximize $B$ first, since it is squared in the expression. The largest number available is $B = 6,$ so its opposite face must be $1.$ Next, we want to maximize $A.$ The next highest number is $5,$ so we set $A = 5$ with its opposite face as $2.$ Since $4$ is already given in the net, its opposite face must be $3,$ meaning $C = 3.$
Thus, we assign $A = 5, B =6,$ and $C = 3.$ Now, we compute the given expression.
$$A \times (B^2 + C) = 5 \times (6^2 \times 3) = 195.$$Thus, the maximum value of the expression is $\boxed{195}.$
Problem Number 2
The average of $20$ numbers is $18$. If the first number is increased by $2$, the second number is increased by $4$, the third number is increased by $6$, and so on, until the twentieth number is increased by $40$. What is the average of the $20$ numbers now?
Given that the average of $20$ numbers is $18.$ This means that the sum of these $20$ numbers can be calculated as $20 \times 18 = 360.$ Now, each number in the sequence is increased according to a pattern: the first number is increased by $2,$ the second by $4,$ the third by $6,$ and so on, until the twentieth number is increased by $40.$ The total increase in the sum can be found by summing the sequence
$$2 + 4 + 6 + \cdots + 40.$$This sequence is an arithmetic series with the first term $a = 2,$ the common difference $d = 2,$ and the number of terms $n = 20.$ The sum of an arithmetic series is given by
$$S_n = \dfrac{n}{2} \times (a + \text{U}_n)$$where $\text{U}_n$ is the last term. Substituting the given values:
$$S_{20} = \dfrac{20}{2} \times (2 + 40) = 10 \times 42 = 420.$$Thus, the new total sum of the numbers after the increases is $360 + 420 = 780.$ The new average is then $\dfrac{780}{20} = 39.$
Thus, the new average of the $20$ numbers is $39.$
Problem Number 3
Twelve consecutive positive integers are written on a board. Maria erases one of the numbers. If the sum of the remaining numbers is $2023$, what number did Maria erase?
We are given twelve consecutive positive integers written on a board. Let these integers be
$$n, n+1, n+2, \cdots, n+11.$$The sum of these twelve consecutive integers can be computed using the formula for the sum of an arithmetic sequence as follows.
$$S = \dfrac{12}{2}(n+(n+11) = 12n+66.$$Maria erases one of the numbers, say $n+y,$ where $y$ is an integer between $0$ and $11$ (inclusively). The sum of the remaining numbers is given as $2023.$ This leads to the equation
$$12n+66-(n+y) = 2023$$which simplifies to
$$11n-y = 1957.$$We need to find an integer $n$ such that the equation holds. Noting that $11 \times 178 = 1958,$ we set
$$11n = 1958 \Rightarrow n = 178.$$Consequently, we have $y = 1.$ Since the erased number is $n + y,$ we find $178 + 1 = 179$ as the erased number.
Verify the solution:
The original sequence is $178, 179, \cdots, 189.$ If we remove $179,$ the sum of the remaining numbers should be $2202-179 = 2023,$ which matches the given condition.
Thus, the number that Maria erased is $\boxed{179}.$
Problem Number 4
The left-most digit of a six-digit number is $5$. If you remove the left-most digit, the original number is $26$ times the new number. What is the sum of the digits of the new number?
We are given that the left-most digit of a six-digit number is $5$. Let the six-digit number be represented as $\overline{5abcde}$ where $\overline{abcde}$ represents the five-digit number obtained by removing the left-most digit. From the problem, the original six-digit number is 26 times the five-digit number obtained after removing the leading digit, so we have
$$\begin{aligned} \overline{5abcde} & = 26 \times \overline{abcde} \\ 500,\!000 + \overline{abcde} & = 26 \times \overline{abcde} \\ 25 \times \overline{abcde} & = 500,\!000 \\ \overline{abcde} & = 20,\!000. \end{aligned}$$The number $\overline{abcde} = 20,\!000$ consists of the digits $2, 0, 0, 0, 0.$ Thus, the sum of these digits is $\boxed{2+0+0+0+0=2}.$
Problem Number 5
Mrs. Anne has two kinds of fruits: apples and oranges. The number of apples is $4$ times the number of oranges. If some of the fruits were eaten by the children, then the number of remaining apples is $7$ times the number of remaining oranges. What is the smallest possible total number of fruits that Mrs. Anne had at the beginning?
We aim to find the smallest possible total number of fruits that satisfies the condition that the remaining apples is $7$ times the number of remaining oranges.
- Suppose that Mrs. Anne has $4$ apples and $1$ oranges. However, reducing both apples and oranges in such a way is impossible while maintaining this ratio. Thus, this case does not work.
- Now, suppose that she has $8$ apples and $2$ oranges. Then, if $1$ apple and $1$ orange were taken, then Mrs. Anne now has $7$ apples and $1$ oranges, meaning the condition holds. In this case, the total number of fruits Mrs. Anne originally had is $8 + 2 = 10.$
Thus, the smallest possible total number of fruits is $\boxed{10}.$
Problem Number 6
Given that $\dfrac{a+b+c+d-29}{6} = 21,$ where $a$, $b$, $c$, and $d$ are positive integers. If $(a+c):(b+d) = 2:3,$ $a>c,$ and $b>d,$ then what is the largest possible value of $a+b$?
The equation $\dfrac{a+b+c+d-29}{6}=21$ can be simplified to $a+b+c+d=155.$ Since $(a+c) : (b+d) = 2 : 3,$ we can express $a+c=2k$ and $b+d=3k$ for some integer $k.$ Substituting these values into the equation gives
$$2k+3k=5k=155.$$Solving for $k,$ we find $k = 31.$ Consequently, we have $a+c=2(31)=62$ and $b+d=3(31) = 93.$ To maximize $a+b,$ we should maximize $a$ and $b$ while ensuring $c$ and $d$ remain positive integers. Setting $a = 61,$ $c = 1,$ $b = 92,$ and $d = 1,$ we obtain
$$a+b=61+92=153.$$Thus, the largest possible value of $a+b$ is $\boxed{153}.$
Problem Number 7
In the diagram, ABCD is a square. If all the arcs in the figure are the arcs of the circles with the same size, what is the area, in $\text{cm}^2$, of the shaded region? (Use $\pi = \dfrac{22}{7}$)
Problem Number 8
Sarah bought some number of roses. She gave half of it and another $13$ roses to Patty. Then, she gave one-third of the remaining roses and another $8$ roses to Sunny. Later on, she gave one-fourth of the remaining roses and another $11$ roses to Nicky. After all of this, Sarah has $4$ roses left for herself. What is the number of roses Sarah bought at the beginning?
Let $x$ be the number of roses Sarah initially bought. After all the transactions, Sarah has $4$ roses left.
(Reverse Nicky’s transaction) Sarah gave one-fourth of her remaining roses plus $11$ additional roses to Nicky.
Let $y$ be the number of roses she had before giving roses to Nicky. Then, we have
$$\begin{aligned} y-\left(\dfrac{1}{4}y+11\right) & = 4 \\ \dfrac34y & = 15 \\ y & = 20. \end{aligned}$$So before giving roses to Nicky, she had $20$ roses.
(Reverse Sunny’s transaction) Sarah gave one-third of her remaining roses plus $8$ additional roses to Sunny.
Let $z$ be the number of roses she had before giving roses to Sunny. Then, we have
$$\begin{aligned} z-\left(\dfrac{1}{3}z+8\right) & = 20 \\ \dfrac23z & = 28 \\ z & = 42. \end{aligned}$$So before giving roses to Sunny, she had $42$ roses.
(Reverse Patty’s transaction) Sarah gave half of her roses plus $13$ additional roses to Patty.
Let $x$ be the number of roses she had before giving roses to Patty. Then, we have
$$\begin{aligned} x-\left(\dfrac{1}{2}x+13\right) & = 42 \\ \dfrac12x & = 55 \\ x & = 110. \end{aligned}$$Thus, Sarah initially bought $110$ roses.
Problem Number 9
Arrange the positive integers $1,$ $2,$ $3,$ $4, \cdots$ in the order as shown in the diagram. The integers $2,$ $3,$ $5,$ $7,$ $10, \cdots$ are called the ‘turning integers’ as the arrow-in and arrow-out of these integers changed direction at the corner. How many ‘turning integers’ are there between $529$ and $1000$?
Consider the sequence $$2, 3, 5, 7, 10, 13, 17, 21, 26$$which follows the pattern:
$$\begin{aligned} 2 & = 1^2 + 1 \\ 3 & = 1^2 + 1 + 1 \\ 5 & = 2^2 + 1 \\ 7 & = 2^2 + 1 + 2 \\ 10 & = 3^2 + 1 \\ 13 & = 3^2 + 1 + 3 \\ 17 & = 4^2 + 1 \\ 21 & = 4^2 + 1 + 4 \\ 26 & = 5^2 + 1 \\ & \vdots \end{aligned}$$Observing the pattern, each pair consists of $n^2+1$ and $n^2+1+n$ for increasing values of $n.$ By continuing this pattern, we find
$$\begin{aligned} 530 & = 23^2 + 1 \\ 553 & = 23^2 + 1 + 23 \\ & \vdots \\ 962 & = 31^2 + 1 \\ 993 & = 31^2 + 1 + 31. \end{aligned}$$Since the integers range from $23$ to $31,$ there are $9$ values of $n,$ each contributing two numbers to the sequence. Thus, the number of “turning integers” between $529$ and $1000$ is $\boxed{9 \times 2 = 18}.$
Problem Number 10
In a badminton tournament with $14$ players, each player plays every other player exactly once. A win gets $2$ points, a draw gets $1$ point, and a loss get $0$. A prize is given to each player whose total points is at least $18$ points. What is the largest possible number of players who can get a prize?
Problem Number 11
There are $10$ positive integers. Starting from the third integer, each integer is equal to the sum of the preceding two integers. If the tenth integer is less than or equal to $2022$, what is the largest possible sum of the first two integers?
We are given a sequence of 10 positive integers where, starting from the third integer, each term is the sum of the two preceding terms. Let the first two integers be $a$ and $b.$ Then, the sequence follows the pattern: $a,$ $b,$ $a+b,$ $a+2b,$ $2a+3b,$ $3a+5b,$ $5a+8b,$ $8a+13b,$ $13a+21b,$ and $21a+34b.$ The tenth integer is given by
$$21a+34b \le 2022.$$To find the largest possible sum of the first two integers, $a+b,$ we should maximize $a,$ since its coefficient $(21)$ is smaller than that of $b$ $(34),$ meaning increasing $a$ has a smaller impact on the inequality. We start by determining the largest integer value for $a$ that keeps $21a+34b \le 2022$ while ensuring $b$ remains a positive integer.
- Testing $a = 96,$ we have $21(96) + 34b = 2016 + 34b \le 2022.$ To satisfy the inequality, we get $34b \le 6,$ which forces $b = 0,$ but $b$ must be positive. So, this is not valid.
- Testing $a = 95,$ we have $21(95) + 34b = 1995 + 34b \le 2022.$ To satisfy the inequality, we get $34b \le 27,$ so $b=0$ is again the only valid integer solution, which is not allowed.
- Testing $a = 94,$ we have $21(94) + 34b = 1974 + 34b \le 2022.$ We get $34b \le 48,$ and the largest positive integer value for $b$ is $1.$
Therefore, the maximum valid choice is $a = 94$ and $b = 1,$ giving $a+b=94+1=95.$
Thus, the largest possible sum of the first two integers is $\boxed{95}.$
Problem Number 12
There are $80$ cards in a box. Each card has a number on it from $1$ to $80$. What is the largest possible number of cards that we can take out from the box so that no two of them have the sum of $123$?
Suppose we are in the worst-case scenario. Since $123 = 61 + 62,$ we begin by selecting the first $61$ cards, consisting of the numbers $1, 2, \cdots, 61.$ At this point, there are no two selected cards that sum to $123.$ Now, if we attempt to take one more card from the remaining numbers ($62$ to $80$), we will inevitably form a pair that sums to $123.$ For example, selecting $62$ would pair with $61$ to make $123,$ selecting $63$ would pair with $60,$ and so on.
Thus, the largest possible number of cards that we can take out from the box so that no two of them have the sum of $123$ is $\boxed{61}.$
Problem Number 13
A palindromic number is a number that remains the same when its digits are reversed. It is known that $1000001$ is the smallest seven-digit palindromic number and $9999999$ is the largest seven-digit palindromic number. What is the $2023$rd seven-digit palindromic number?
A seven-digit palindrome takes the form:
$$\overline{1abcdba1}, \overline{2abcba2}, \cdots, \overline{9abcba9}$$where $a,b,c$ are any digits ($0$ to $9$). Since each of the three middle digits can be any digit from $0$ to $9,$ the number of such palindromes for a given first digit is $10 \times 10 \times 10 = 1000.$ Thus, for each leading digit $d$ (from $1$ to $9$), there are $1000$ palindromes of the form $\overline{dabcbad}.$ Now, let’s counting the palindromes sequentially.
- $\overline{1xxxxxx1}$ forms the first $1000$ palindromes. They take positions $1$ to $1000.$
- $\overline{2xxxxxx2}$ forms the first $1000$ palindromes. They take positions $1001$ to $2000.$
- $\overline{3xxxxxx3}$ begins at position $2001.$ The first $\overline{3xxxxxx3}$ palindrome, $3000003,$ is at position $2001.$ The second one, $3001003,$ is at position $2002.$ Continuing this pattern, $3009003$ is at position $2010.$ Next, the sequence continues with $3010103$ at position $2011,$ and follows the increasing order. By extending this pattern: $3020203$ is at position $2021,$ $3021203$ is at position $2022,$ and $3022203$ is at position $2023.$
Thus, the $2023$rd seven-digit palindromic number is $\boxed{3022203}.$
Problem Number 14
What is the largest possible three-digit number $\overline{abc}$ which satisfy $\overline{abc} = 10a + 10b^3 + c^2,$ where $a$, $b,$ and $c$ are different digits?
We are given that the three-digit number $\overline{abc}$ satisfies the equation
$$10a+10b^3+c^2$$where $a, b,$ and $c$ are distinct digits.
Since $c^2$ must have the same unit digit as $c$ itself, the possible values of $c$ are those digits for which $c^2$ ends in $c.$ Checking the squares of digits from $0$ to $9,$ only $c = 0, 1, 5,$ and $6$ satisfy this condition. To maximize $\overline{abc},$ we select the largest possible value of $c,$ which is $6.$ Substituting $c = 6$ into the given equation, we have
$$\begin{aligned} \overline{ab6} & =10a + 10b^3 + 6^2 \\ 100a + 10b + 6 & = 10a + 10b^3 + 36 \\ 90a + 10b & = 10b^3 + 30 \\ 9a & = (b^3-b) + 3. \end{aligned}$$Since $a$ is a digit from $0$ to $9,$ the maximum possible value of $a$ is $9.$ To satisfy the equation, we check when
$$(b^3-b) + 3 \le 81.$$Calculating for $b \ge 5,$ we find that they do not satisfy this inequality. Therefore, we set $b = 4,$ so that $9a = (4^3-4)+3,$ or $a = 7.$
Thus, the largest possible three-digit number $\overline{abc}$ is $\boxed{746}.$
Verify the solution:
We can check that $746 = 10(7) + 10(4)^3 + 36,$ satisfying the equation.
Problem Number 15
In the diagram, $ABCD$ is a parallelogram with perimeter $36$ cm, $DF = CF,$ $\angle DAB = 60^\circ,$ $\angle FCB = 30^\circ,$ $\angle DEC = 90^\circ,$ and $\angle DBC = 90^\circ.$ What is the length, in cm, of $AB$?
[</spoiler] What is the sum of all four-digit numbers that are not divisible by $3$ which consist of the digits $1,$ $2,$ $3,$ $5,$ and $7$ without repetition? We are given the digits $1, 2, 3, 5, 7,$ and need to find the sum of all four-digit numbers that are not divisible by $3,$ using these digits without repetition. Each of these sets has $4! = 24$ permutations. Therefore, there are $4 \times 24 = 96$ valid numbers. Summing all the valid cases, we have $$73326 + 86658 + 106656 + 113322 = 379962.$$Thus, the total sum of all four-digit numbers that are not divisible by $3$ is $\boxed{379962}.$Problem Number 16
The only case that results in a sum divisible by $3$ is when $3$ is removed, meaning numbers containing $1, 2, 5, 7;$ $1, 3, 5, 7;$ $2, 3, 5, 7;$ and $1, 2, 3, 7$ are not divisible by $3.$ Thus, the valid sets of four digits are:
Each digit in a four-digit number contributes equally to each place value $(1000s, 100s, 10s, 1s)$ across all permutations. Since there are $24$ permutations per set and each digit appears $6$ times in each place value, the contribution of each digit set to the total sum follows this pattern:
$$\boxed{\text{Sum of digits} \times 6 \times (1000+100+10+1) = \text{Sum of digits} \times 6 \times 1111}$$Let’s consider that four sets of four digits.
Since $1+2+3+5 = 11,$ we have total sum equals $11 \times 6 \times 1111 = 73326.$
Since $1+2+3+7 = 13,$ we have total sum equals $13 \times 6 \times 1111 = 86658.$
Since $1+3+5+7 = 16,$ we have total sum equals $16 \times 6 \times 1111 = 106656.$
Since $2+3+5+7= 17,$ we have total sum equals $17 \times 6 \times 1111 = 113322.$
Problem Number 17
How many minutes does it take for both hands of an analogue clock to form an angle of $110^\circ$ for the first time and for the second time from $05:00$ AM to $06:00$ AM?
Problem Number 18
Let $a$ and $b$ be positive integers such that
$$\dfrac{1}{1\times 9} + \dfrac{2}{9 \times 25} + \dfrac{3}{25 \times 49} + \cdots + \dfrac{15}{841 \times 961} = \dfrac{a}{b}.$$If $\text{gcd}(a, b) = 1,$ what is the sum of $a$ and $b$?
We are given the sequence where each term is of the form $\dfrac{a}{(2a-1)^2(2a+1)^2}$ for $a \in \{1, 2, \cdots, 15\}.$ Our goal is to find the sum of this sequence. We observe the identity:
$$\begin{aligned} \dfrac{1}{(2a-1)^2}-\dfrac{1}{(2a+1)^2} & = \dfrac{(2a+1)^2-(2a-1)^2}{(2a-1)^2 \cdot (2a+1)^2} \\ & = \dfrac18 \cdot \dfrac{a}{(2a-1)^2 \cdot (2a+1)^2}. \end{aligned}$$Arranging the form, we have
$$8\left(\dfrac{1}{(2a-1)^2}-\dfrac{1}{(2a+1)^2}\right) = \dfrac{a}{(2a-1)^2 \cdot (2a+1)^2}.$$By using this identity, we will execute the telescoping technique as follows.
$$\begin{aligned} & \dfrac{1}{1\times 9} + \dfrac{2}{9 \times 25} + \dfrac{3}{25 \times 49} + \cdots + \dfrac{15}{841 \times 961} \\ & = 8\left(\dfrac{1}{1}-\dfrac19\right) + 8\left(\dfrac19-\dfrac{1}{25}\right) +8\left(\dfrac{1}{25}-\dfrac{1}{49}\right)+\cdots + 8\left(\dfrac{841}-\dfrac{1}{961}\right) \\ & = 8\left(\dfrac{1}{1}-\dfrac19 + \dfrac19-\dfrac{1}{25} + \dfrac{1}{25}-\dfrac{1}{49} + \cdots + \dfrac{841}-\dfrac{1}{961}\right) \\ & = 8\left(\dfrac11-\dfrac{1}{961}\right) \\ & = \dfrac{7680}{961}. \end{aligned}$$Since $\text{gcd}(7680, 961) = 1,$ the fraction is in its simplest form. We conclude that the value of $a=7680$ and $b=961,$ so that $a+b = 7680+961 = 8641.$
Problem Number 19
In the diagram, $ABC$ is a right-angled triangle where $\angle A = 90^\circ.$ $M$ is the midpoint of $BC,$ $D$ is on $BC$ such that $AD$ is perpendicular to $BC,$ and $E$ is on the extension of $BC$ such that $AM$ is perpendicular to $AE.$ If $AM = 22$ and $DM = 10,$ what is the length of $CE$?
Problem Number 20
There are $20$ passengers in the bus, where the fare for each passenger is $5$ dollars. They have only $10$-dollar, $15$-dollar and $20$-dollar bills. By trading among themselves, they come up with the correct combined fare, and each also receives the correct change. What is the smallest possible number of bills involved?
Problem Number 21
Look at the diagram. Andry wants to go from M to S, but he doesn’t want to pass O. If Andry only can go down from one box to the adjacent box below, what is the number of ways he can do it?
To determine the number of ways Andry can go from $M$ to $S$ without passing through $O,$ we use a combinatorial approach. We fill in the grid step by step.
There is one way to reach $M$ (starting point). Each box is filled by summing the ways from the two boxes above it. This continues downward until reaching $S.$ The value at $O$ is ignored (set to $0$). Consequently, any box below $O$ doesn’t receive contributions from it. After applying the rule, the final values at the last row (just above $S$) are $220$ and $260.$ Thus, the total number of valid paths to $S$ is $220+260=480.$ In other words, the number of ways Andry can go from $M$ to $S$ without passing through $O$ is $\boxed{480}.$
Problem Number 22
Coky and Boby together can paint a room in $8$ days, Andy and Boby together can do it in $6$ days, Andy and Coky together in $4$ days. What is the number of days does it take all of them working together to paint $4$ identical rooms if Coky absent only in the second day when paints each room and Boby absent from the third day to the end when paints each room?
Problem Number 23
In the diagram, $BE : EA = 2 : 3,$ $AD : DC = 1 : 2,$ and $BF : FC = 1 : 4.$ If the difference of the areas $AED$ and $CDF$ is $75$ square units, what is the area, in square units, of $DEF$?
Suppose $[ABC] = x$ where $[ABC]$ denotes the area of the triangle $ABC.$ We need to determine the area of $\triangle DEF,$ given that the difference in areas of $\triangle AED$ and $\triangle CDF$ is $75$ square units.
Since $EA = \dfrac{3}{5} AB$ and $AD = \dfrac13 AC,$ the area of $\triangle AED$ is given by
$$\left[AED\right] = \dfrac35 \cdot \dfrac13 \cdot [ABC] = \dfrac15x.$$Similarly, since $FC = \dfrac45BC$ and $DC = \dfrac23AC,$ the area of $\triangle CDF$ is given by
$$\left[CDF\right] = \dfrac45 \cdot \dfrac23 \cdot [ABC] = \dfrac{8}{15}x.$$We are given that
$$\begin{aligned} \left[CDF\right]-\left[AED\right] & = 75 \\ \dfrac{8}{15}x-\dfrac15x & = 75 \\ \dfrac13x & = 75 \\ x & = 225. \end{aligned}$$This means that the area of triangle $ABC$ is $225$ square units. From this, we calculate $\left[AED\right] = \dfrac15(225) = 45$ and $[CDF] = \dfrac{8}{15}(225) = 120$ square units. Next, we need to find the areas of triangle $BEF.$ Since $BE = \dfrac25AB$ and $BF = \dfrac15BC,$ then we have
$$[BEF] = \dfrac25 \cdot \dfrac15 \cdot [ABC] = \dfrac{2}{5} \cdot \dfrac15 \cdot 225 = 18.$$In order to find the area of triangle $DEF,$ we just need to subtract the areas of $\triangle AED,$ $\triangle CDF,$ and $\triangle CDF$ from $[ABC].$
This will result as follows.
$$\begin{aligned} \left[DEF\right] & = \left[ABC\right]-(\left[AED\right] + \left[CDF\right] + \left[BEF\right]) \\ & = 225-(18+45+120) = 42. \end{aligned}$$Thus, the area of $DEF$ is $\boxed{42}$ square units
Problem Number 24
Alex is Betty’s elder brother, and they were born in different years. In $2023$, Alex’s age is equal to the sum of the digits of the year when he was born, and Betty’s age is also equal to the sum of the digits of the year when she was born. What is Alex’s age when Betty was born?
Suppose Betty was born in the year $\overline{20ab},$ where $a$ and $b$ are digits. In $2023,$ her age is
$$2023-\overline{20ab} = 23-(10a+b).$$Since her age equals the sum of the digits of her birth year, we have
$$23-(10a+b) = 2 + a + b.$$Rearranging the equation, we get $11a+2b = 21.$ The only solution for this equation is $a = 1$ and $b = 5.$ Thus, Betty was born in $2015.$
Now, suppose Alex was born in the year $\overline{19cd},$ where $c$ and $d$ are digits. In $2023,$ his age is
$$2023-\overline{19cd} = 123-(10c+d).$$Since his age equals the sum of the digits of his birth year, we have
$$123-(10c+d) = 1 + 9 + a + b.$$Rearranging the equation, we get $11c + 2d = 113.$ The only solution for this equation is $c = 9$ and $d = 7.$ Thus, Alex was born in $1997.$
Since Betty was born in $2015$ and Alex was born in $1997,$ his age in $2015$ was $2015-1997=18$ years old.
Problem Number 25
There were $4$ couples who came to the celebration event. Among four wives (namely A, B, C, D), A has eaten $5$ cakes, B has eaten $1$ cake, C has eaten $3$ cakes and D has eaten $4$ cakes. Among four husbands (namely I, M, S, O), I have eaten cakes $4$ times which his wife has eaten, M has eaten cakes $3$ times which his wife has eaten, S has eaten cakes $2$ times which his wife has eaten, and O has eaten as many cakes as his wife has eaten. If the total number of cakes the four couples ate was $39$, what is the number of cakes did S and his wife eat in total?
Let the number of cakes eaten by A, B, C, dan D represented as $a, b, c,$ and $d,$ respectively. According to the given conditions, the husbands’ cake consumption follows these rules:
- Husband I eats 4 times the number of cakes his wife eats.
- Husband M eats 3 times the number of cakes his wife eats.
- Husband S eats 2 times the number of cakes his wife eats.
- Husband O eats the same number of cakes as his wife.
Thus, the total number of cakes consumed by all four couples can be expressed as the Diophantine equation $4a+3b+2c+d = 39$ where $a, b, c, d$ belong to the set $\{1, 3, 4, 5\},$ as given in the problem. Note that a Diophantine equation is a polynomial equation where the solutions are required to be integers. These equations are named after the ancient Greek mathematician Diophantus of Alexandria, who studied them extensively. We will solve the equation using parity considerations.
Case 1: $d$ is an odd number
If $d$ is odd, then $b$ must be even. The only even number in the set is $4,$ so we set $b = 4.$ The equation then simplifies to $4a+2c+d=27.$ Trying different values for $d,$ we check the following.
- If $d = 1,$ then solving $4a+2c=26$ gives $a=5$ and $c=3,$ which are valid values from the given set.
- If $d = 3$ or $d = 5,$ there is no valid solution for $a$ and $c.$
Thus, the only valid solution in this case is $(a, b, c, d) = (5, 4, 3, 1).$
Case 2: $d$ is an even number
If $d$ is even, then $b$ must be odd. The possible odd values for $b$ are $1, 3,$ and $5.$ Setting $d = 4,$ we get $4a + 3b + 2c = 35.$ Trying different values for $b,$ we find that none lead to valid values of $a$ and $c$ from the given set.
Since Case 2 does not provide a valid solution, the only possibility is the one found in Case 1: $(a, b, c, d) = (5, 4, 3, 1).$ From this, we determine that wife C has eaten 3 cakes. Since S eats twice as many cakes as his wife, we conclude that S has eaten
6 cakes. Thus, the total number of cakes eaten by S and his wife C is $\boxed{3 + 6 = 9}.$