Puzzles

The factors will all be of the form \(2^a\times5^b\times7^c\times11^d\), where \(0\leqslant a\leqslant26\), \(0\leqslant b\leqslant1\), \(0\leqslant c\leqslant5\), and \(0\leqslant d\leqslant2\). There are 27 choices for \(a\), 2 for \(b\), 6 for \(c\), and 3 for \(d\) giving a total of 27×2×6×3 = 972 factors.

The product of the numbers in the red boxes is 336.

The only three-digit number that is equal to the product of a square number and the same square number with its digits reversed is 16×61 = 976.

We can look at the first few powers and look for a pattern:

If n is even (and > 2), the last two digits of 15ⁿ are 625.

As 3D is 3 times 5A, and they both share a final digit, their final digit must be 5 or 0.

If it were 0, then 4A would also end zero, so 3D ends in 00. This would mean that 5A also ends 00. 2D therefore would end in 0, but then it's impossible for the sum of 2D's digits to be 19. Hence, the final digit of 5A cannot be 0, so must be 5:

As 4A is 2 times 5A, it ends in a 0:

3D is a multiple of 3, so must be 105, 405, or 705. These would mean 5A is 35, 135, or 235 (respectively). 5A has three digits so this rules out the first option. Both other options end in 35:

4A is 2 times 5A, so must end in 70:

The sum of the digits of 2D is 19:

If 5A were 135, then 4A would be 270. But then 1D ends with 21 and its digits cannot add to 15. Therefore 5A 235:

4A is 2 times 5A and 3D is 3 times 5A:

The sum of the digits of 1D is 15:

Therefore, today's number is 997.

Holly's sum is odd, so she must have removed the units digit and so her calculation was:

\(a\) and \(b\) cannot both be 0, so the must sum in the tens column must've caused a carry into the hundreds column. This means that \(a\) must be 2, and the calculation is:

Two single digits cannot add to 19, so there can't be a carry from the units column into the tens column. This means that \(b\) is 8:

We can see now that \(c\) was 1, so Holly's three digt number was 281.

If the set contains one number, then it must be {5}.

If the set contains three numbers, then it must contain 5, one number less than 5, and one number greater than 5. There are 4 numbers less than 5 and 6 numbers greater than 5 to choose from, giving a total of 4×6 = 24 sets.

More generally, if the set contains \(2n+1\) numbers, then it must contain 5, \(n\) numbers less than 5, and \(n\) numbers greater than 5. There are \(\left(\begin{array}{c}4\\n\end{array}\right)\) numbers less than 5 and \(\left(\begin{array}{c}6\\n\end{array}\right)\) numbers greater than 5 to choose from, giving a total of \(\left(\begin{array}{c}4\\n\end{array}\right)\times\left(\begin{array}{c}6\\n\end{array}\right)\) sets.

In total there are

sets. This is equal to 210.

The palindromes between 10 and 100 will be the numbers 11, 22, 33, ..., 99. The sum of these is 495.