### N000ughty Thoughts

Factorial one hundred (written 100!) has 24 noughts when written in full and that 1000! has 249 noughts? Convince yourself that the above is true. Perhaps your methodology will help you find the number of noughts in 10 000! and 100 000! or even 1 000 000!

### Mod 3

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

### Novemberish

a) A four digit number (in base 10) aabb is a perfect square. Discuss ways of systematically finding this number. (b) Prove that 11^{10}-1 is divisible by 100.

##### Stage: 3 and 4 Challenge Level:
We received lots of good solutions to this problem. Thanks to everyone who submitted a solution! Unfortunately there were so many we can't mention you all by name. A special well done to the pupils of Beaconsfield High School for all their great solutions - we're glad you enjoyed the problem so much!

Here is a really nice solution submitted by Oliver from Loreto College:

Assuming n is an integer, there are no values of n so that $2^n$ is a multiple of 10 because $2^n$ doesn't contain any necessary factors of 5.

The unit digits of $2^n$ for n=1,2,3... are 2, 4, 8, 6 then repeat. For $3^n$ they go 3, 9, 7, 1 then repeat. If n is odd, the units that are being added are either 2 + 3 or 7 + 8, which both end in 5. So $2^n + 3^n$ where n is odd always ends in 5. This is a stronger conclusion than saying 'it's a multiple of 5' as a multiple of 5 can also end in 0.

If n is a multiple of 4 the units being added are 6 + 1 so in this case it will always end in 7.

$1^n + 2^n + 3^n$ is even for all values of n. This is obvious because $1^n$ and $3^n$ are always odd and $2^n$ is always even, and odd + odd + even =even.

$1^n + 2^n + 3^n + 4^n$ is a multiple of 10 for when 4 does not divide n:

The unit digits of the powers of 1, 2, 3 and 4 are:

 x^1 x^2 x^3 x^4 1 1 1 1 2 4 8 6 3 9 7 1 4 6 4 6

Summing down the columns gives 10, 20, 20 and 14. This shows that when n is not divisible by 4, the last digit of $1^n + 2^n + 3^n + 4^n$ is 0 so it's a multiple of 10. (When n is divisible by 4, the last digit is 4).

$1^n + 2^n + 3^n + 4^n + 5^n$ ends in 5 for n not divisible by 4. This is obvious if we consider the previous result because $5^n$ ends in 5 for all n, so adding this to the multiples of 10 will give a final digit of 5.

Ryan from Renaissance College Hong Kong had another good explanation for why $2^n$ cannot be a multiple of 10:

 Power of 2 Answer Units Digit 1 2 2 2 4 4 3 8 8 4 16 6 5 32 2 6 64 4 7 128 8 8 256 6

For which values of n will $2^n$ be a multiple of 10?

As can be seen from the following table, the unit digits of the powers of
two are in a repetitious pattern of 2, 4, 8, 6, 2, 4, 8, 6…

All multiples of 10 have a unit digit of 0. However, as seen from the
pattern, none of the powers of 2 have their unit digit ending in a 0.
Therefore, no power of 2 is a multiple of 10.

Alexander from University College School noticed the following about the extension questions:

If the power n in $4^n + 5^n + 6^n$ is an odd number, then the last digit of this sum is 5; if the power n is even then the last digit of the sum is 7.

$3^n + 8^n$:  For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 1, 3, 9, 7.

$2^n + 4^n + 6^n$:  For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 2, 6, 8, 8.

$3^n +5^n +7^n$:  For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 5, 3, 5, 7.

$3^n-2^n$:  For consecutive values of n (n = 1, 2, 3, …) the last digit of the sum goes in a pattern of 1, 5, 9, 5.

Can you explain why this might be true?
Ryan also had some interesting ideas about how he could extend his results:

The unit digit of $1^n$ is 1, 1, 1, 1…
The unit digit of $2^n$ is 2, 4, 8, 6, 2, 4, 8, 6,...
The unit digit of $3^n$ is 3, 9, 7, 1, 4, 9, 7, 1,...
The unit digit of $4^n$ is 4, 6, 4, 6…
The unit digit of $5^n$ is 5, 5, 5, 5…
The unit digit of $6^n$ is 6, 6, 6, 6…
The unit digit of $7^n$ is 7, 9, 3, 1, 7, 9, 3, 1,...
The unit digit of $8^n$ is 8, 4, 2, 6, 8, 4, 2, 6,...
The unit digit of $9^n$ is 9, 1, 9, 1…
The unit digit of $10^n$ is 0, 0, 0, 0…

Thus the unit digits of the powers of any number repeat in a cycle four digits long, as all numbers end with one of the above 10 digits.Let the pattern of the unit digits of the powers of x be a, b, c, d... and let The pattern of the unit digits of the powers of another number y be e, f, g, h…

As a result, the sum/difference of the unit digits of the powers of x and y are in a repeating pattern of: a + e, b + f, c + g, and d + h, or a - e, b - f, c - g and d – h.

As a result, no matter what combination of powers, adding or subtracting, the unit digits are always in a pattern of 4 repeating numbers.

Well done everyone!