### More Mods

What is the units digit for the number 123^(456) ?

### N000ughty Thoughts

How many noughts are at the end of these giant numbers?

### Mod 3

Prove that if a^2+b^2 is a multiple of 3 then both a and b are multiples of 3.

# Squaresearch

##### Stage: 4 Challenge Level:

Pete Inglesby from Wolverhampton; Patrick Snow, age 16, Dame Alice Owens School, Potters Bar; Yatir Halevi, age 18, Maccabim-Reut High-School, Israel; Andrei Lazanu, age 12, School 205, Bucharest, Romania all solved this problem. Here is Pete's solution:

Consider numbers of the form

$$u(n) = 1! + 2! + 3! +...+ n!.$$

How many such numbers are perfect squares?

In fact only when $n=1$ or $n=3$ does $u(n)$ give a perfect square ($1$ and $9$ respectively).

The solution lies in the fact that all factorials above $5!$ are multiples of $10$. This is because they all have $2$ and $5$ as prime factors. For $n> 5$

\eqalign { n!&=1 \times 2\times 3\times 4\times 5\times \cdots \times n \cr &= 2\times 5(1\times 3\times 4\times \cdots \times n) \cr &= 10(1\times 3\times 4\times \cdots n).}

When adding a multiple of $10$, to another integer, $k$, the last digit of $k$ will remain the same. Suppose $k = 10j + l$ and $m =10n$, with $l$ and $n$ integers, then $k + m = 10(j + n) + l$.

So when adding all factorials equal to or higher than $5!$ then you are adding a multiple of ten, so the last digit of the sum will be the same. This is important because the last digit of $u(4)$ is a $3$. We have $u(4) = 1! + 2! + 3! + 4! = 1 + 2 + 6 + 26 = 33$. Therefore for any $u(n)$ with $n> 5$ the last digit will be $3$. For any integer $p = (10a + b)$ with $a$ and $b$ integers and $b< 10$, $p^2 = (100a^2 + 20ab + b^2)$. Therefore the last digit is of $p^2$ is only dependent on $b^2$.
 b b2 Last digit of b2 0 0 0 1 1 1 2 4 4 3 9 9 4 16 6 5 25 5 6 36 6 7 49 9 8 64 4 9 81 1

This table shows that the squares of all integers between $0$ and $9$, and therefore all integers, end in either $1$, $4$, $5$, $6$ or $9$.

No perfect squares end in $3$. Therefore, $u(n)$ with $n > 4$ can never be a perfect square as it has been shown to always end in $3$.

All that remains is to find $u(n)$ for all integers of $4$ or less:

$u(1) = 1! = 1$;
$u(2) = 1! + 2! = 3$;
$u(3) = 1! + 2! + 3! = 9$;
$u(4) = 1! + 2! + 3! + 4! = 33.$

Only $u(1)$ and $u(3)$ are perfect squares, and so they are the only sums of factorials to be perfect squares.