Magazine article Mathematics Teaching

Letters: Reflections on MT254

Magazine article Mathematics Teaching

Letters: Reflections on MT254

Article excerpt

Dear MT

I was interested in the "First four digits puzzle". I recently asked three able year 3 pupils to do this. By introducing them to brackets and "!" we got up to 33, without using 2-digit numbers such as 23. I then went on myself to see if I could get to 100, but am stuck on 86, 89, 91, 93, 95 and 97. Can anyone help, please?

Jenny Maxwell

Responses to "Choose two points" (MT254)

1. Looked for a visual solution but didn't spot one.

2. Blow, this has to be a number problem. Worked out equation of the perpendicular bisector of (0, 0) and (4, 1).

3. No solution because integer [double dagger] integer + 1.

4. Try another pair of points: (0, 0) and (2, 1). No solution for the same reason.

5. Try equation for (0, 0) and (p,q), p and q are integers.

6. Realise from geometry that one only needs to check for the points in half of one quadrant. 1 < p < q. All other half quadrants follow by symmetry, the symmetries fixing (0,0) and the lattice are the 8 symmetries of a square.

7. Get equation qy + px = (p2 + q2)/2 for line of possible vertices.

8. For a solution, must have p2 + q2 even, that is, either both p,q even, or both p,q odd. So the original figure has no solutions.

9. If p,q are both even, then (p/2, q/2) is a lattice point and so lots of solutions.

10. If p and q are both odd in qy + px = (p2 + q2)/2, look at 2 cases:

Case (i) hcf (p, q) = 1. Then there exist integers x, y such that qy + px = 1, and a multiple of this equation gets a solution.

Case (ii) hcf (p, q) = d [double dagger] 1, then d divides (p2 + q2)/2, so the whole equation may be divided through by d, and we have a solution as in case (i). …

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