Puzzles

The across number can be 138 (and the down number 234 or 432).

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.