Puzzles

Plotting \(y = 4x - 9\) and \(y = x^2-2x+2\) gives:

As \(p(x)\) has integer coefficients, \(p(3)\) must be an integer. If you click to view a larger version of the image, you'll see that the only option for \(p(3)\) that satisfies \(4x-9<p(x)<x^2-2x+2\) is \(p(3)=4\).

As \(p(x)>4x-9\), it must at least a polynomial of degree 1. As \(p(x)<x^2-2x+2\), it must be at most a polynomial of degree 2. Playing with coefficients, you can see that \(p(x)\) is either \(x^2-2x+1\) or \(4x-8\). \(4x-8\) is negative when \(x=0\), so \(p(x)=x^2-2x+1\) and \(p(23)\) is 484.

All the integers whose digits are all non-zero and add up to 9 can be made by taking a list of 9 ones, and between each pair deciding whether to move on a digit or not. For example, 1111|111|11 (where | means move on) would represent the number 432; and 1|1|1111111 would represent the number 117.

There are 8 gaps between 9s, so there are 2⁸ = 256 numbers that can be made.

Why are there not 512 positive integers whose digits are all non-zero and add up to 10?

If \(k\) is a multiple of 28, then let \(k=28a\) and see that:

This is an integer if either \(a=0\), \(a=1\), or \(1+a\) is a factor of 28. Hence the largest value of \(a\) is 27, leading to \(k\) = 756.

Similar working out can be done for the cases where the hcf of 28 and \(k\) is 14, 7, 2, or 1, and in each case a lower answer would be obtained.

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 three 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.