Puzzles
XYZ
Which digits \(X\), \(Y\) and \(Z\) fill this sum?
$$
\begin{array}{cccc}
&X&Z&Y\\
+&X&Y&Z\\
\hline
&Y&Z&X
\end{array}
$$
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Both the units and tens columns contain \(Y+Z\). The results are different (\(X\) and \(Z\)), so \(Y+Z=X+10\) and \(Z=X+1\) (because the 1 carries into the next column).
Therefore, \(Y+X+1 = X+10\), so \(Y=9\).
From the hundreds column, we see that \(X+X+1=Y\), so \(X=4\) and \(Z=5\).
Extension
Which digits \(X\), \(Y\) and \(Z\) fill this sum?
$$
\begin{array}{cccc}
&X&Z&Y\\
+&X&Y&Z\\
\hline
&Z&Y&X
\end{array}
$$
Where is Evariste?
Evariste is standing in a rectangular formation, in which everyone is lined up in rows and columns. There are 175 people in all the rows in front of Evariste and 400 in the rows behind him. There are 312 in the columns to his left and 264 in the columns to his right.
In which row and column is Evariste standing?
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The common factors of 175 and 400 are 1, 5, 25. Hence in front and behind him there are either:
| In front | Behind |
| 175 rows of 1 | 400 rows of 1 |
| 35 rows of 5 | 80 rows of 5 |
| 7 rows of 25 | 16 rows of 25 |
Therefore the shape of the formation is either 716×1, 126×5, or 24×25.
The length of the columns (716, 126, or 24) must be a factor of 312 and 264. Therefore it is 24.
Evariste is standing in the 8th row from the front and the 14th column from the left.
Extension
Evariste is standing in a rectangular formation, in which everyone is lined up in rows and columns. There are \(a\) people in all the rows in front of Evariste and \(b\) in the rows behind him. There are \(c\) in the columns to his left and \(d\) in the columns to his right.
For which numbers \(a\), \(b\), \(c\) and \(d\) is the row and column in which Evariste standing uniquely determined?
Bending a straw
Two points along a drinking straw are picked at random. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?
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A triangle will be made if none of the segments of straw is longer than the other two added together. This is the same as requiring that each segment must be less than half the straw.
Let the length of the straw be 1 unit. Call the points \(x\) and \(y\). A triangle is made if either:
- \(x\lt y\), \(x\lt\tfrac12\), \(y-x\lt\tfrac12\), \(1-y\lt\tfrac12\); or
- \(y\lt x\), \(y\lt\tfrac12\), \(x-y\lt\tfrac12\), \(1-x\lt\tfrac12\).
For the second condition, the allowable region is shown below.
This region covers \(\tfrac18\) of the whole square. By switching \(x\) and \(y\) it can be seen that the first condition's region is the same size as the second's, plus they don't overlap. Therefore the probability of making a triangle is \(\tfrac18+\tfrac18=\tfrac14\).
Extension
One point along a drinking straw is picked, then a coin is flipped. If the coin shows heads, a second point above the first is chosen; If tails, a second point below the first is chosen. The straw is then bent at these points. What is the probability that the two ends meet up to make a triangle?
Turning squares
Each square on a chessboard contains an arrow point up, down, left or right. You start in the bottom left square. Every second you move one square in the direction shown by the arrow in your square. Just after you move, the arrow on the square you moved from rotates 90° clockwise. If an arrow would take you off the edge of the board, you stay in that square (the arrow will still rotate).
You win the game if you reach the top right square of the chessboard. Can I design a starting arrangement of arrows that will prevent you from winning?
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No, I can't.
If I could, then my arrangement would cause you to follow an infinitely long pattern without visiting this pattern. As there are only a finite number of squares, within this pattern there must be a square that you visit infinitely often. The arrow on this square will point in each direction an infinite number of times, so you must also visit the squares next to this one infinitely often.
For the same reason, you must visit the squares next to them infinitely often, and the squares next to them, and so on. In this way, we see that you visit every square, including the all-important winning square, infinitely often.
Elastic numbers
Throughout this puzzle, expressions like \(AB\) will represent the digits of a number, not \(A\) multiplied by \(B\).
A two-digit number \(AB\) is called elastic if:
- \(A\) and \(B\) are both non-zero.
- The numbers \(A0B\), \(A00B\), \(A000B\), ... are all divisible by \(AB\).
There are three elastic numbers. Can you find them?
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15, 18 and 45 are elastic.
15's factors are 5 and 3. 105, 1005, 10005, etc will all be multiples of 5 (because they end in 5) and multiples of 3 (as their digits add to 6). Hence they are all multiples of 15.
Similarly, 108, 1008, 10008, etc are all multiples of 9 (adding digits) and 2 (they are even), so they are multiples of 18; and 405, 4005, 40005, etc are all multiples of 9 (adding digits) and 5 (last digits are 5), so they are multiples of 45.
Extension
How many elastic numbers are there in other bases?
Square pairs
Source: Maths Jam
Can you order the integers 1 to 16 so that every pair of adjacent numbers adds to a square number?
For which other numbers \(n\) is it possible to order the integers 1 to \(n\) in such a way?
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Yes: 8, 1, 15, 10, 6, 3, 13, 12, 4, 5, 11, 14, 2, 7, 9, 16.
It is clearly possible for the numbers 1 to 15 (remove the 16 from the end of the sequence above) and 1 to 17 (add 17 to the start of the sequence above).
The OEIS sequence
A090461 gives other numbers for which this is possible. It starts 15, 16, 17, 23, then includes every number from 25 onwards. It is conjectured, but not proven, that it is possible for every number above 25.
Factorial pattern
$$1\times1!=2!-1$$
$$1\times1!+2\times2!=3!-1$$
$$1\times1!+2\times2!+3\times3!=4!-1$$
Does this pattern continue?
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Yes. It can be shown by induction:
First it's easy to check that \(1\times1!=2!-1\). Next assume that \(1\times1!+2\times2!+...+k\times k!=(k+1)!-1\). Now we try to show the pattern holds for \(k+1\).
$$1\times1!+2\times2!+...+k\times k!+(k+1)\times(k+1)!$$$$=(k+1)!-1+(k+1)\times(k+1)!$$
$$=(1+k+1)(k+1)!-1$$
$$=(k+2)!-1$$
Hence, by induction, the pattern holds for all \(k\geq1\).
Placing plates
Two players take turns placing identical plates on a square table. The player who is first to be unable to place a plate loses. Which player wins?
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The first player can always win by first placing a plate on the exact centre of the table. Then the first player can copy what the second player does, but rotated 180°, and hence can always place a plate if the second player could.
Extension
What if the two players play on a regular hexagonal table? Or a regular octagonal table? Or a regular pentagonal table? Or a regular \(n\)-gonal table?
![](javascript:showlimage("straws_1.png"))