Puzzles

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.

The product is 252.

By adding a point inside a triangle, you can turn one triangle into three triangles. By adding a point outside all the current triangles, you can add either one or two more triangles. As we are after the maximum number of triangles, we will choose to add two triangles for each point we add.

Four points make 3 triangles. Adding 286 more points will add 572 more triangles, giving a total of 575.

Every hour, the hour hand moves 360÷12 = 30°. Every minute, the hour hand moves 30÷60 = ½° and the minute hand moves 360÷60 = 6°.

At 8:22, the hour hand will be 8×30+22×½ = 251° from the top of the clock and the minute hand will be 22×6 = 132° from the top of the clock. The angle between the hands will be 251-132 = 119°.

We can look for a pattern as we increate the number of 9s that make up n:

If n has k digits, then n³ is k-1 9s, followed by a 7, followed by k-1 0s, followed by a 2, followed by k 9s, The sum of all these digits will by 18k.

Hence, the answer is 18×55 = 990.