If you add the first two equations, you get 2 stars = 22 (the circles will cancel out. So STAR = 11. Substituting 11 ito either equation 1 or 2, for star gives CIRCLE = 9.
Using CIRCLE = 9 in the last equation gives TRIANGLE = 14.
This puzzle appeared as a recent Alex Bellos Monday Puzzle.
Here is a printable version and here is the digital version>
Q: Two octopuses will fight using boxing gloves but they each need 8(!) gloves and the gloves MUST be the same color.
Drawing at random from a bag that has 16 red gloves and 16 green gloves, what is minimum number of gloves to guarantee that the colors of the gloves of the two octopuses are:
(a) the same.
(b) different.
(c) either the same or different.
A: (a) 31 (b) 24 (c) 23
(a) If the lone glove left behind is red, both octopuses can wear green gloves. If the lone glove left behind is green, both octopuses can wear red gloves.
(b) Ifmagine the first 16 choices are all of one colour. You will then need 8 of the other.
(c) With 23 gloves, at least 7 are one color (say red) and at least 8 are the other (say green). If there there exactly 7 red, then both octopuses can wear green. If there are more than 7 red, one can wear red and the other green. (Likewise, if at least 7 are green, and at least 8 are red, we get the same result.)
With many thanks to Alex Bello’s Monday puzzles in the Guardian:
https://www.theguardian.com/science/2025/aug/18/did-you-solve-it-are-you-a-matcha-for-these-tasty-japanese-puzzles
Q: On each of ten cards is one of the digits 0 to 9.
These numbers can be arranged in 10! ways.
How many of these 10 digit numbers are even?
How many of these 10 digit numbers are divisible by 3?
A: 1814400 even numbers, 3628800 are divisible by 3.
Half the numbers are even as each number will end with either an even or odd digit.
10! = 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 3628800
10!/2 = 1814400
All of the 10 digit numbers are divisible by 3 as the digits add up to 45 and 45 is divisible by 3 (note; all th enumbers are also divisible by 9 because of the sum = 45).
Q: At a zoo, the number of two-legged birds is twice the number of 4-legged mammals.
If there is a total of 80 legs, how many birds and how many 4-legged mammals are there?
A: There are 20 two-legged birds (40 legs) and 10 four-legged mammals (40 more legs).
Let x = the number of 4-legged mammals
then 2x = # of 2-legged birds
2x(2) + x (4) = 80
8x = 80
x = 10 2x = 20
Math Confidence by Robin the Math Lady 