Tag Archives: Combinatorics

Answer to STEP I 2007 Q.1

I/2007/1 – A positive integer with 2n digits (the first of which must not be 0) is called a balanced number if the sum of the first n digits equals the sum of the last n digits. For example, 1634 is a 4-digit balanced number, but 123401 is not a balanced number. (i) Show that seventy 4-digit balanced numbers can be made using the digits 0, 1, 2, 3 and 4. (ii) Show that \frac{1}{6}k(k+1)(4k+5) 4-digit balanced numbers can be made using the digits 0 to k. You may use the identity \sum_1^n{r^2} \equiv \frac{1}{6}n(n+1)(2n+1) Continue reading

Maths prize for 13 October 2015: a Ramsey puzzle

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If you have a group of four people, it’s possible to have no subgroup of three all of whom know each other, and simultaneously no subgroup of three none of whom know each other. If there are five people at a party, is it possible that the party includes no subgroup of three who all already know each other, and no subgroup of three all of whom didn’t know each other before the party? If there are six, is that possible?

The prize was won by Joan Onokhua, who produced the best partial solution. Continue reading