Puzzles

For any digit \(V\), \(VVV\) is a multiple of 111, and \(111=37\times3\). 37 is prime, so \(VI\) must be a multiple of 37 (as \(X\) is less than 10 so cannot be a multiple of 37). Therefore \(VI\) is either 37 or 74.

If \(VI\) was 74, then \(VVV\) is 777. But then \(VI\times X\) is even and \(VVV\) is odd. This is impossible, so \(VI\) must the 37.

Now that we know that \(VI\) is 37, we see that

and so \(X=9\) and the final solution is

If 1/10890 has a finite number of digits in base \(b\), then \(b^n/10890\) is an integer for some integer \(n\). In order for this to be possible \(b\) must be a mutiple of each prime factor of 10890: the smallest such \(b\) will be when the exponent of each prime factor is 1.

The prime factorisation of 10890 is 2×3³×5×11². Therefore the smallest base in which 1/10890 has a finite number of digits is 2×3×5×11=330.

The number can contain no 1 (as all numbers are multiples of 1). It's first and last digits cannot be 2 (as it would then be even and so a multiple of 2).

The lowest three-digit palindrome containing no 0 or 1 and not ending in 2 is 323. This is not a multiple of 2 or 3 and so is the solution.

There are 6 available choices for each digit, so there are 6² and 6³ such two- and three-digit numbers. 6²+6³=252.

By the same method used to solve this puzzle, the answer if 150.

Only the units digit of the number will affect the last digit of the square and cube. This table shows how the last digits of the square and cube depend on the last digit of the number:

So numbers ending in 0, 1, 5 and 6 will have squares and cubes that end in the same digit. There are 4×9=36 two-digit numbers then end in one of these digits.

How many two-digit numbers are there in binary whose square and cube end in the same digit?

How many two-digit numbers are there in ternary whose square and cube end in the same digit?

How many two-digit numbers are there in base \(n\) whose square and cube end in the same digit?

Ted's number was 925: \(925\div25=37\).

If Ted had removed the final digit of his number, then he would be looking for a solution of \(ABC = 37\times AB\). But \(ABC\) is between 10 and 11 times \(AB\) (it is \(10\times AB + C\)) and so cannot be 37 times \(AB\). So Ted cannot have removed the final digit.

Therefore, Ted must have removed one of the first two digits: so two- and three- digit numbers have the same final digit (\(C\)). The final digit of the three-digit number (\(C\)) will be the final digit of \(7\times C\) (7 times the final digit of the two digit number). This is only possible if the final digit is \(C\) is 0 or 5.

This only leaves four possible solutions—10, 15, 20 and 25—as \(30\times37>1000\). Of these only \(925=37\times25\) works.

How many three-digit numbers are there that are a multiple of one of the two-digit numbers you can make by removing a digit?

EDCCBA is a multiple of four, so A is even. A cannot be more than 2, as otherwise EDCCBA would have more digits. So A is 2.

E must therefore be 8 or 9 (as 4 times B is less than E) and 3 or 8 (as 4 times E ends in 2). Therefore E is 8.

Carrying on like this, we find: