Puzzles

Each number will appear 7 times: one time paired with the numbers six numbers 5 to 10, plus an extra appearance on the tile containing the same number twice. The total of all the numbers is therefore 7×(5+6+...+10)=7×45=315.

First, consider placing 5 tiles in a 1×9 grid. There is only one way to do this:

To get the number of ways of placing 5 tiles in a 1×10 grid, imagine adding an extra blank square to either the start or end of the grid or between two of the counters. There are 6 places this tile could be inserted leading to 6 arrangements of 5 tiles in a 1×10 grid.

For 5 tiles in a 2×10 grid, you can first pick the columns the tiles go in (as a tile being in a column means nothing can be placed the columns either side, the number of ways to pick columns is the same and the number of wats to arrange 5 tokens in a 1×10 grid). For each of these column choices, there are two locations for each tile (top or bottom). This leads to a total number of arrangements of 6×2⁵=192.

The totals that are equally likely add up to 7 times the number of dice.

This can be seen by using the fact that the opposite sides of a dice add up to 7: for each way of making a given total with \(n\) dice, there is a way of making \(7n\) minus that total by looking at the dice from below. Therefore \(T\) and \(7n-T\) are equally likely. (This also holds true (but is harder to explain) if you rearrange the faces of the dice so that the opposite faces no longer add to 7.)

Therefore today's number is \((521+200)/7\), which is 103.

If the final digit of the number is 0, then some carrying takes place when 1 is subtracted. Otherwise, no carrying happens.

If no carrying happens, call the three digits of today's number \(A\), \(B\), and \(C\). We know that \(A^3+B^3+C^3\) is one more than \(A^3 + B^3 + (C-1)^3\). This implies that \(C^3=(C-1)^3+1\), which is only possible if \(C\) is 1.

Therefore either the final digit of today's number is 0 or the final digt of one less that today's number is 0. In both cases, we need to find a number with the desired property whose final digit is 0: we are looking for digit \(A\) and \(B\) such that \(A^3+B^3\) is a multiple of 10.

Looking at all the cube numbers, there are a few combinations that add up to multiple of 10:

The only one of these that has the required property is 370. By checking 369 and finding it doesn't have the property, we see that the two numbers must be 370 and 371.

The largest number you can make with the digits in the red boxes is 321.

By drawing an appropriate diagram, it can be seen that the small square has half the area of the large square.

Therefore the area of the large square is 124.

There are 3 one-digit numbers using these digits (1, 3 and 5).

To make two-digit odd numbers, there are 3 choices for the units digit, and 4 remaining choices for the tens digit.

To make three-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, and 3 remaining choices for the hundreds digit.

To make four-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, 3 remaining choices for the hundreds digit, and 2 remaining choices for the thousands digit.

To make five-digit odd numbers, there are 3 choices for the units digit, 4 remaining choices for the tens digit, 3 remaining choices for the hundreds digit, 2 remaining choices for the thousands digit and 1 remaining choice for the ten-thousands digit.

In total, this gives 3 + 3×4× + 3×4×3 + 3×4×3×2 + 3×4×3×2×1 = 195 odd numbers.

The digits of the number must be 1, 2 and 3, so the largest number is 321.