Hint

Since it is impossible to fulfill all of the instructions in the will, you must only fulfill some of them. The method you use will depend on which part of the instructions you pick.

Solution

Since it is impossible to fulfill all of the instructions in the will, we must only fulfill some of them. The method we use will depend on which part of the instructions we pick. There are many possibilities. For example:
$\\$ 1) From the first condition of the will it follows that the son must receive 2/3 of the inheritance and from the second condition it follows that the daughter must receive half the amount the mother receives.
$\\$ 2) From the first condition it follows that the mother receives half the amount the son receives and from the second condition it follows that this portion is twice the amount the daughter receives.
$\\$ 3)In each condition the mother does not receive less than a third while the son’s inheritance is 4 times that of the daughter.
$\\$ Other options are also possible. This problem came from reality. Such a situation actually happened in Ancient Rome. The inheritance was divided by a court as in case 2): 4/7 of the property went to the son (i.e. 120 talents), 2/7 to the mother (i.e. 60 talents) and 1/7 to the daughter (i.e. 30 talents).

Hint

Since it is impossible to fulfill all of the instructions in the will, you must only fulfill some of them. The method you use will depend on which part of the instructions you pick.

Solution

Nobody wrote the word “chat” because no one in the class can write both the letter “c” and the letter “h”. So “cat” or “hat” must have been written by $10+18=28$ people. We notice that the only words written were “cat”, “hat” or “at”. The first two words were each written 15 times and so “at” was written by $50-15-15=20$ out of 28 pupils. This means that only 8 people wrote their word correctly.

Answer

8 pupils.

Hint

Solution

The first thing we might want to notice here is that $SHE$ is a 3-digit number that is a square number. Since $10^2 =100$ is the smallest 3-digit square, and $32^2 = 1024$ already has 4 digits, we only have to check $22$ possibilities. And we might just do that. $\\$
But let us be just a bit smarter (and lazier). A digit $E$ multiplied by itself has to give a number whose last digit is also $E$. The $E$ in the final product has nowhere else to come from. There are only $4$ digits that have this property: $0^2=0$, $1^2=1$, $5^2=25$ and $6^2=36$. $E$ cannot be $0$ as that would make $SHE$ divisible by $100$, so her last two digits would be equal to $0$, and they have to be distinct. $\\$
Remembering about the initial condition we found, this only leaves a couple of candidates: $15,16,21,25,26$ and $31$. Squaring each results in $225,256,441,625,676$ and $961$. The only number satisfying the conditions of the problem is therefore $HE=25$. Finally $S=6$, $H=2$, $E=5$.

Hint

Solution

Let us write the number as $1 \times 100000 + a$, where $a$ is simply the rest of the number. After relocating, all the digits of $a$ move one position to the left, what results in this part increasing $10$ times. Additionally, the $1$ is removed from the left and added to the right, simply as $1$. The number becomes $10a +1$. We can write an equation:
$$10a+1 = 3(100000+a) $$
$$10a-3a = 300000-1$$
$$7a = 299999$$
After a long division, we get $a=42857$, so the original number was $142857$.

Hint

Solution

The hint to solve this problem is the title of today’s session. Shmerlin must choose numbers $x,y,z$ and $w$ in such a way that somehow the sum given to him by Drago contains all the information about $A,B,C$ and $D$. If we just write $ABCD$ as a four-digit number, like in other problems, it can be broken down to $1000 \times A + 100 \times B + 10 \times C + D$, which is exactly what Shmerlin has to do! That is, set $x=1000,y=100,z=10$ and $w=1$. He will then get the number $ABCD$ from the dragon, which will tell him all he wants to do.

Hint

Solution

The first thing to notice is that the condition and the product whose last digit we have to find do not share any common digits! That is interesting. However, there are $4$ different digits in the equation and $6$ different digits in the product. In total, these are $10$ different digits. These are all the digits there are! We are going to make use of that fact. $\\$
Let us first solve the equation. $RR$ has to be such a number, that adding to it a single-digit number pushes the number through to a 3-digit number. Therefore, it has to be larger than $90$. The only suitable number is $99$, as both digits have to be equal. Thus $R=9$. $9+99= 108$ and $B=1$, $O=0$, $W=8$.$\\$
In the product, all the digits except for $1,0,8$ and $9$ are multiplied together. In particular, that means $2$ and $5$ go into that product. It must be divisible by $2 \times 5 =10$. And any number divisible by $10$ has the last digit equal to $0$.

Hint

Solution

We repeat the procedure from the first Example exactly:
$$SEND = 1000 \times S + 100 \times E + 10 \times N + D$$
$$MORE = 1000 \times M + 100 \times O + 10 \times R + E$$
$$MONEY = 10000 \times M + 1000 \times O + 100 \times N + 10 \times E + Y $$
Putting it all together:
$$SEND + MORE + MONEY = 11000 \times M + 1100 \times O + 1000 \times S + 111 \times E + 110 \times N + 10 \times R + D +Y$$
Thus we assign digits in the following way: $M = 9$, $O=8$, $S=7$, $E=6$, $N=5$, $R=4$ and $D$ and $Y$ equal to $2$ and $3$ in any order. The largest possible sum becomes:
$$11000 \times 9 + 1100 \times 8 + 1000 \times 7 + 111 \times 6 + 110 \times 5 + 10 \times 4 + 3 +2 = 116061$$

Hint

Solution

Let us call Jane’s original number $a$. By writing a $5$ at the end, she increases the original number $10$ times, by moving all digits one position to the left and also adds $5$ to it. The resulting number is supposed to be larger by $1661$. This can be written in a form of an equation:
$$10a+5 = 1661+ a$$
Solving:
$$9a = 1656$$
We can now perform long division to find $a = 184$. It does not include $5$, indeed.

Hint

Solution

Let us break down the equation and write it all in the decimal representation. $\\$
$$NO = 10 \times N + O$$
$$MORE = 1000 \times M + 100 \times O + 10 \times R + E$$
$$MATH = 1000 \times M + 100 \times A + 10 \times T + H$$
Thus $$NO + MORE + MATH = 10 \times N + O + 1000 \times M + 100 \times O + 10 \times R + E + 1000 \times M + 100 \times A + 10 \times T + H$$ We can already see that the largest coefficient by far stands before the digit represented by $M$. Let us factor this a bit more.
$$NO + MORE + MATH = 2000 \times M + 101 \times O + 100 \times A + 10 \times (N+R+T) + E + H$$
It would be a good strategy to assign digits to letters based on the coefficients standing before them in this sum. The larger the coefficient, the larger the digit, because we want to maximize the sum. This should be intuitively true; for those needing more rigorous explanation why, it is because of the rearrangement inequality. $\\$
Thus, we assign $9$ to $M$, $8$ to $O$, $7$ to $A$, and then $6,5$ and $4$ in any order to $N, R$ and $T$ and finally $2$ and $3$ in any order to $E$ and $H$. $\\$
Thus, the maximum value of the expression is
$$2000 \times 9 + 101 \times 8 + 100 \times 7 + 10 \times (6 + 5 + 4) + 1 \times (3 + 2) = 19663$$.

Hint

Solution

The equation can be rewritten in a form:
$$I + 8 \times HE = US$$
If $H \ge 2$, then $I + 8 \times HE \ge I + 8 \times 20 >160$. Therefore, $H=1$, otherwise we would have a three digit number on the other side. $\\$
Let’s think about $E$ now. If $E \ge 3$, we have $8 \times HE \ge 8 \times 13 = 104$, so our number is a three digit number again. $1$ is taken already, so $E$ can be either $0$ or $2$. The problem with $0$, however, is that if $E=0$ then the last digit of the number on the left hand side is $I$, and on the right hand side it is $S$. These were supposed to be different, however. Therefore $E$ must be exactly $2$. On the left hand side we get $I + 8 \times 12 = I + 96$. Therefore $U=9$. $\\$
Let’s think what $I$ can be. It can be $0$, because then $S=6$ and all the letters correspond to different digits. This is therefore one solution: $I =0, H=1, E=2, U=9, S=6$. $\\$
What else can $I$ be equal to? Well, $1$ and $2$ are already taken. Can $I$ be equal to $3$ or larger digits? If $I=3$, then $US=99$, but this number has both digits equal. If $I>3$, the number on the RHS becomes a three digit number already. $\\$
Thus different values of $I$ are not allowed and the above solution is the only one.

Hint

Solution

In this problem we need to use $6$ numbers instead of $5$. We also know that we can’t use all $10$ digit cards, because then the sum of all the digits is $45$, which is divisible by $9$. We can use at most $9$ digits, then without $4$. The other possibility is to use $8$ digits, without $1$ and $3$, but that limits us even more. $\\$
If we decide to use at least two three digit numbers, we have already used $6$ of the available digits, and we have only three digits left to make four other numbers. This is not possible. $\\$
Therefore we need to use at least one four digit number. That way we will use 4 digits, and we will have 5 digits left to make the remaining 5 numbers. Thus each of these 5 numbers will have to be a single digit number. $\\$
The largest four digit number smaller than $2000$ with all different digits is $1987$. The smallest four digit number larger than $2000$ with all different digits is $2013$. The latter is already larger than $2012$, so we have to use a number smaller than $2000$. If we use $1987$, then we have to just add the other $5$ digits to it: $1987+6+5+3+2+0= 2003$. This is too small, but unfortunately there is no way of making it larger, we can only make it smaller (assuming we keep $1$ as the first digit of the four digit number). Therefore the task is impossible in this case.

Hint

Solution

The task is basically to add five numbers together, such that each of the digits appears at most once across all these five numbers, and get $2012$ as the result. $\\$
First thing we can notice is that we will get some quite large numbers. $2012$ is a four digit number, so to construct it, we need either at least one four digit number or $3$ three digit ones. Why three? Remember that our numbers all have to have different digits. The largest number you can get by adding two three digit numbers together is $975+864=1839$. This is $173$ off $2012$, and the largest two digit numbers that can take us there are $31+20$ which go nowhere as close. Under these assumptions the numbers are just too small. $\\$
Unfortunately, if we use $3$ three digit numbers, we will have used $9$ digits already, so there will be only $1$ digit left for the remaining two numbers we have to add. This is not possible to do. $\\$
We can try using one four digit number and several two digit numbers instead. If we try to use all the $10$ digits that we have, we will find out we can make $2007$ or $2016$, but getting $2012$ is not trivial. We can remind ourselves of a useful fact about sums of digits now. The remainder of a sum of digits of a number in division by $9$ is the same as the remainder of this number in division by $9$. That means the remainder in division by $9$ of our sum of five numbers depends only on the digits we use. However, if we use all the digits, their sum is $0+1+2+ \dots + 9 = 45$, which is divisible by $9$. That means that if we use all the digits, our sum will be divisible by $9$. $\\$
$2012$ is not divisible by $9$. By adding its digits together we find out that its remainder in this division is $2+0+1+2=5$. That means the total sum of all digits used has to give remainder $5$, (or be equal to $5$ modulo $9$). The simplest way of achieving that is to just not use digit $4$. $\\$
We can try starting from $1987$ as our four digit number. It is quite close to $2012$ already. If we try adding single digit numbers to it, we might get:
$$1987+6+5+3+2=2003$$
We are quite close. Now, if we write $0$ on the right side of $2$, getting $20$, our number is $2021$, which is $9$ larger that $2012$. Now we can exchange two last digits of $1987$, making $1978$, which decreases the number by $9$. Finally we have:
$$1978+6+5+3+20=2012$$
Thus there is a way of doing that.

Hint

Solution

First, $O$ looks a bit like a digit $0$, so let’s try that. Then the right hand side is $B \times 1000$, in particular it is divisible by $1000 = 2^3 \times 5^3$, which means the left hand side also has to be divisible by the same prime factors. $5$ goes into $1000$ three times. Each number that is divisible by $5$ ends at $5$ or at $0$. None of the numbers on the left can end in $0$ because $0$ is $O$, we have assigned it already. That means at least one has to end in $5$. If that is $G$, we can only fit in one $5$, and we need three $5$-s to be somewhere on the left, and if we try to put other two in $BAN$, $N$ will have to also be $5$ or $0$. Not possible, because both are already taken. $\\$
We need to put all $5$-s into $BAN$, then. That means that $N=5$ but also, $BAN$ is odd. But there the right hand side is divisible by $2$ to at least power $3$, or in other words, byb $8$. All these $2$-s have to be in $G$, which is possible only if we make $G=8$. $\\$
Now, we know that $8$ is handled. The left hand side needs to be divisible by $125 = 5^3$, and this number has to in particular divide $BAN$. First, we can check if $BAN=125$ works, that means $B=1$ and we get $125 \times 8 =1000$ which checks out! We have one solution. But then, is it the only one? $\\$
$BAN$ is odd, so we can try odd multiples of $125$. If $BAN = 3 \times 125 = 375$, $B=3$, $A=7$, and on the right hand side we have $3000$, which also checks out because $375 \times 8 = 3000$. We have found another solution! $\\$
If we try taking larger odd multiples of $125$, the first digit is going to be larger, for example $5 \times 125 = 625$, and $625 \times 8 \neq 6000$.$\\$
For this particular case there are two solutions, $125 \times 8 = 1000$ and $375 \times 8 = 3000$.

Hint

Solution

Each of the letters represents a single digit, but we shouldn’t assume they are all different. First, M and B cannot be equal to 0, because they are the first digits of 4-digit numbers. $\\$
There are two possibilities here. Either M=B or B$>$M, because it can’t decrease after we add something to it. If M=B than we can disregard it write the equation again as M+EEE=OOO. We can see that EEE is a three digit number and OOO is a three digit number of the same form. There are only 9 such numbers, from $111$ to $999$. The problem is, the difference between any two of such numbers is at least $111$. We need at least that much to get from, say $666$ to $777$. But we can only add a single digit number to EEE. Zero could be a possibility, and then E=O, but we have already found that M cannot be zero, because it starts a four digit number. It means this case leads to a contradiction and we should consider the other one. $\\$
Namely, B$>$M. If it was bigger from M by more than 1, it would mean that there is at least a 1000 difference between two numbers MEEE and BOOO. But it is not possible, because we are adding a single digit to the first one. Therefore it must be that M +1 =B. $\\$
We can therefore subtract M000 from both sides and write our equation in the following form: $$M + EEE=1OOO$$
We need to be able to switch to the next thousand by adding a single digit to a number of a form of EEE. If E is 8 or less, we can’t do that, because we would need to add at least 112 to it to get to a thousand and flip it. That means EEE must be equal to 999. $\\$
If EEE is 999, then adding a single digit to it can give us a number ranging from 1000 to 1008. Only one of these numbers, 1000, is of the requested form. We therefore find that M=1 and we can reconstruct the original equality remembering that B = M+1 =2.$\\$
$$1 + 1999=2000$$

Hint

Solution

To answer this question, we need to think about what it means for a number to divide another number. Each natural number can be thought of as made from a certain set of prime numbers. Every number has a prime decomposition, which is unique and is basically a list of primes that makes that number with the respective powers, which indicate how many times a prime goes into that number. If we multiply out all these primes, each respective number of times, we get our number back. For example, $60$ has a decomposition $2^2 \times 3 \times 5$, which means $2$ goes in $60$ twice, and $3$ and $5$ once. No other prime divides it. $\\$
Primes cannot divide other primes and prime powers can divide other prime powers only if the dividing power is lower, in other words, $2^5$ divides $2^7$, $3^8$ divides $3^{11}$ and so on. $\\$
That means that number $a$ can divide number $b$ only if every single prime included in $a$-s prime decomposition is also included in $b$-s prime decomposition, with a higher power. $\\$
Think about this this way: $a$ can divide $b$, if $b$ is better than $a$ in every aspect, aspects being primes that divide it. We can divide $b$ by $a$ by taking each prime that makes $a$ and dividing $b$ by it, one by one. We will only get an integer, if we try dividing by every prime fewer number of times than it appears in $b$-s decomposition. $\\$
How can we use it here? We want to make three numbers in such a way that each of them contain some prime numbers, and any two of them contain as much of any primes as the third one alone. That way the problem’s conditions will be met.$\\$
Let’s imagine we have three primes, $p,q$ and $r$. Obviously they do not divide each other. If we then take three products of each pair of these numbers, we get $pq$, $pr$ and $qr$. Note that we still can’t divide $pq$ by any other, nor can we do it any other way around. But if we multiply all the three primes together, getting $pqr$, it obviously can be divided by each of the pair products, $pq, pr$ and $qr$. Now it is enough to see that products of pairs of these products (so for example $pq \times pr = p^2qr$) are all divisible by $pqr$. That means $qr$ divides $pqr$ which divides $p^2qr$. So $qr$ divides the product of the other two numbers, even though it doesn’t divide any of them. The same can be said about both $pr$ and $pq$, for the same reason. $\\$
For example, let’s take the smallest three primes, $2,3$ and $5$. Our numbers are $6,10$ and $15$. We can see none of these divides any other, but $6$ divides $10 \times 15 =150$, $10$ divides $6 \times 15 = 90$, and $15$ divides $6 \times 10 = 60$, so it works.

Hint

Solution

The XXI century began in 2001 and will end in 2100. If we look on the date format used in this problem, two digits for days, two for months and four for years, we can see that the month number has to consist of two first digits of a year, but in a reverse order. The first two digits of a year can only be $20$ or $21$, so the month can only be the second or twelfth month, February or December. In case of December, however, it needs to be December 2100, which means that the whole date looks like this: 00.12.2100, which is not valid because there is no zeroth day in a month. $\\$
So we should fix a month to be February, and so a palindrome date becomes: AB.02.20BA, where B and A are some digits. A year doesn’t introduce any further constraints for A and B, they can be any digits from 0 to 9. A day, however, does. That is because February in particular, has only days from 1st to 28th or 29th, but only if a year happens to be a leap year. Putting AB equal to any number from 01 to 28 certainly works, and we get dates like 05.02.2050 or 17.02.2071. $\\$
AB can only be equal to 29 if the corresponding year is a leap year. If we write out the full date, it is 29.02.2092. Is 2092 a leap year? It turns out that yes, it is, because 2092 is divisible by 4. Therefore 29.02.2092 is a valid palindrome date.$\\$
In conclusion there are exactly 29 palidrome dates in XXI century.

Hint

Solution

Jane’s original number is a two-digit number, so it can be written as 10a+b, where a and b are digits. After tting a 5 in the centre, it becomes 100a+50+b. We know that the latter is 11 times the former.

100a + 50 + b = 11(10a + b)
50 = 10a + 10b

Dividing both sides by 10 we get a + b = 5, which is the sum of digits of the original number. The sum of digits of the new number is 5 + 5 = 10.

Hint

Solution

a) Let us write the number in the form abcdef. Again, like in the Example, we notice that the entire number is dened by only two digits – this time, they are a and b.

c = a + b

d = b + c = 2b + a

e = c + d = 3b + 2a

f = e + d = 5b + 3a

This time, we want to make a the largest possible. We are limited by the fact that f must be a digit, in other words, it has to be smaller or equal to 9. Therefore, a cannot be larger than 3. When a = 3, we must have b = 0 and that results in the number 303369 which is exactly the number from the Example in reverse.
b) The limitation on the number of digits is now lifted. Increasing the number of digits is a better strategy for increasing the number than increasing the largest digit. Let us assume for now the number has 9 digits, we might decide to cut it to fewer if that is too many. Let us write it as abcdefghi. Again, the whole number is determined by a and b. The equations determining the next 4 digits are the same as in (a), let us write the rest:

g = e + f = 8b + 5a

h = g + f = 13b + 8a

i = h + g = 21b + 13a

An Observant Reader might notice that the pattern is exactly the Fibonacci sequence – and that is not a coincidence! Anyway, we are now limited by the fact that since a is the rst digit, it cannot be equal to 0 – and all the digits have to be, well, digits – smaller or equal to 9. That means we cannot really have the ninth and any following digit – 13 times any positive integer is larger than 9. For an 8-digit number, we can only set a = 1, but that is ne, as increasing the number of digits is a better strategy for increasing the number. Then b = 0 again, because 13b has to be smaller than 9.
The resulting number is 10112358 (Fibonacci strikes back!).

Hint

Solution

We repeat the procedure from the firrst Example exactly:

SEND = 1000 * S + 100 * E + 10 * N + D
MORE = 1000 * M + 100 * O + 10 * R + E
MONEY = 10000 * M + 1000 * O + 100 * N + 10 * E + Y

Putting it all together:
SEND+MORE+MONEY = 11000 * M+1100 * O+1000 * S+111 * E+110 * N+10 * R+D+Y
Thus we assign digits in the following way: M = 9, O = 8, S = 7, E = 6, N = 5, R = 4 and D and
Y equal to 2 and 3 in any order. The largest possible sum becomes:

11000 * 9 + 1100 * 8 + 1000 * 7 + 111 * 6 + 110 * 5 + 10 * 4 + 3 + 2 = 116061

Hint

Solution

It is not difficult to show that in the rest of the year this coincidence will not happen again until 25/12/2008.

Answer

25th December 2008.

Hint