N000ughty Thoughts

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

Factorial

How many zeros are there at the end of the number which is the product of first hundred positive integers?

Fac-finding

Lyndon chose this as one of his favourite problems. It is accessible but needs some careful analysis of what is included and what is not. A systematic approach is really helpful.

Squaresearch

Age 14 to 16
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.

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Age 14 to 16Challenge Level

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Challenge Level