Puzzles

123

The two numbers between 14 are a cube number and a triangle number: these must be 8 and 6. Next you can see that 23 is the sum of 8, two times a triangle number and a square number: the triangle and square numbers must be 3 and 9.

Next, call the square number at the bottom left \(a\), the square number at the top left \(b\), and the triangle number at the top right \(c\). Adding upwards, we find that \(a+b+45=106\) and \(a+141+c=198\); and so \(a+b=61\) and \(a+c=57\). The only two square numbers that add to 61 are 25 and 36. Therefore \(c\) must be 21 or 32, but must be 21 as 32 is not a triangle number. And so \(a\) is 36 and \(b\) is 25.

Putting all these numbers into the tree gives the top number as 433. This is fitting because 433 is in fact a star number:

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 \(Z=5\).

Which digits \(X\), \(Y\) and \(Z\) fill this sum?

The common factors of 175 and 400 are 1, 5, 25. Hence in front and behind him there are either:

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.

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?

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:

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 \(\tfrac18+\tfrac18=\tfrac14\).

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?

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.

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.

How many elastic numbers are there in other bases?

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.