Did you solve it? The maths problem for five-year-olds 'stumping' the web

A still from the Seven Samurai, [Toshiro Mifune, Japan, 1954]. Photograph: RONALD GRANT

Earlier today I set you the following three puzzles.

1. In each of the four sectors of the outer circle, there is a two-digit number which is equal to the sum of the three numbers at the corners of its sector. The numbers in the individual circles can only be 1 to 9 and each number can be used only once. One number has been provided to get you started. Find the remaining four numbers

Solution: clockwise from top 6, 1, 8, 9

Call the positions North, East, West and South. The candidates for W and S are 8 and 9, since S + W + 3 = 20, or S + W = 17. But we also know that S + E + 3 = 12, or S + E = 9. But S cannot be 9 since this would make E = 0 which is not allowed. So S = 8, W = 9, N = 6 and E = 1.

2. Turn this fake viral puzzle that stumped the internet into a meaningful one:

Many people made suggestions, but I thought this one from Tom Flannery the best because it was beautifully simple.

Populate each circle with an integer such that the sum of quadrants in each semi circle is equal to the sum of circles

Good stuff Tom! A copy of Can You Solve My Problems? on its way to you. Thanks to everyone else who took part, and sorry I cant respond to you all individually.

3. The final problem was from the 1727 Japanese puzzle book Wakoku Chie-Kurabe.

Write the numbers from 1 to 9 in the black circles such that the sum of the numbers around each blue circle (and including the centre circle) and along both horizontal and vertical lines is the same.

Solution

There are many solutions, such as this one:

The solution page in the Wakoku Chie-Kurabe

The clue to solving this one is to realize that there are only three possible numbers that can go in the centre*, which are 1 or 5 or 9. Once you have chosen the centre number, separate the remaining digits into pairs that add up to the same number, and place these pairs in opposite circles. If we exclude 1, as in my version, we get the pairs 2 and 9; 3 and 8; 4 and 7; and 5 and 6, which all add up to 11. Check the positions of the pairs, and we’re done!

CORRECTION: Sorry folks, I got it wrong. A few of you have been in touch with perfectly correct solutions with a 3 in the middle. It is indeed possible to find solutions with any odd number in the centre. Such as this one, sent in by Francisco Zampella: horizontal numbers: 9-5-3-6-1. Vertical: 7-8-3-2-4.

I set a puzzle here every two weeks on a Monday. Send me your email if you want me to alert you each time I post a new one.

I’m always on the look-out for great puzzles. If you would like to suggest one, email me.