Puzzles

The square numbers must be 1, 9, 25, 36, and 784.

If we add up the first \(n\) even numbers then add on half of the next even number, we get \((n+1)^2\). This means that Holly added up the first \(\sqrt{465124}-1\) or 681 even numbers.

Can you show why adding up the first \(n\) even numbers and half of the next even number gives \((n+1)^2\)?

Today's number is 148.

By the same method as this puzzle, the answer is 108.

Only the units digit of the number will affect the last digit of the square and cube. This table shows how the last digits of the square and cube depend on the last digit of the number:

So numbers ending in 0, 1, 5 and 6 will have squares and cubes that end in the same digit. There are 4×9=36 two-digit numbers then end in one of these digits.

How many two-digit numbers are there in binary whose square and cube end in the same digit?

How many two-digit numbers are there in ternary whose square and cube end in the same digit?

How many two-digit numbers are there in base \(n\) whose square and cube end in the same digit?

570

819

The two numbers between 14 are a cube number and a triangle number: these must be 8 and 6. Next you can see that 23 is the sum of 8, two times a triangle number and a square number: the triangle and square numbers must be 3 and 9.

Next, call the square number at the bottom left \(a\), the square number at the top left \(b\), and the triangle number at the top right \(c\). Adding upwards, we find that \(a+b+45=106\) and \(a+141+c=198\); and so \(a+b=61\) and \(a+c=57\). The only two square numbers that add to 61 are 25 and 36. Therefore \(c\) must be 21 or 32, but must be 21 as 32 is not a triangle number. And so \(a\) is 36 and \(b\) is 25.

Putting all these numbers into the tree gives the top number as 433. This is fitting because 433 is in fact a star number: