How many primitive Pythagorean triples have hypotenuse less than 50? A triple is primitive when the three numbers share no common factor greater than 1.

Seven primitive triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (20,21,29), (12,35,37) and (9,40,41).

Every primitive triple comes from a pair of coprime integers m > n of opposite parity, via a = m^2 - n^2, b = 2mn, c = m^2 + n^2.

Factor the number 111111 completely into primes, and explain how the factorisation follows from the identity 111111 = \frac{10^6 - 1}{9} rather than from trial division.

111111 = 3 \cdot 7 \cdot 11 \cdot 13 \cdot 37.

10^6 - 1 factors as a difference of cubes *and* as a difference of squares — exploit both.

The row sum is 2^n, so the only rows in range are n = 10 (sum 1024) and n = 11 (sum 2048).

There is a one-line answer for the row sum, and you can get it by expanding (1+1)^n.

The parity statement is forced: the number of odd-degree people is always even, so 3 is impossible.

Count the total number of handshake-ends rather than the handshakes.

\frac{3}{4}, achieved by a parity strategy.

Count configurations rather than players: you need a rule that is correct on 6 of the 8 hat assignments and silent on the other 2.

I = 3 interior lattice points; Pick's formula gives A = 3 + \frac{8}{2} - 1 = 6, matching the area \frac{1}{2} \cdot 4 \cdot 3 = 6.

Pick's theorem reads A = I + \frac{B}{2} - 1, where I is the number of interior lattice points and B the number on the boundary.

Label the card positions 0 to 51 and work out where position p goes in one out-shuffle.

The path length must be even; every closed knight's tour has an even number of moves.

The obvious invariants — row, column, distance from the centre — all change in an uncontrolled way. Try colouring.

Work modulo 3: once you can make three consecutive amounts, you can make every larger amount by adding 3-cent stamps.

\frac{1}{3} for 'at least one is a boy'; \frac{1}{2} when you meet a randomly chosen child who turns out to be a boy.

Write down all four equally likely ordered outcomes first, then delete the ones the information rules out.

\frac{\sqrt{2}}{2}, attained by putting one point at each corner and the fifth at the centre.

Five points and one number that is not five — cut the square into the right number of equal pieces first.

Two or three moves chained together. These are the ones worth writing down.

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