### 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.

### Sixational

The nth term of a sequence is given by the formula n^3 + 11n . Find the first four terms of the sequence given by this formula and the first term of the sequence which is bigger than one million. Prove that all terms of the sequence are divisible by 6.

# N000ughty Thoughts

##### Stage: 4 Challenge Level:

Thank you Vassil Vassilev, Yr 11, Lawnswood High School, Leeds for the solution below, well done! Danny Ng, 16, from Milliken Mills High School, Canada sent a very similar solution and so did Koopa Koo of Boston College.

First I tried to convince myself that 100! has 24 noughts. I did that by counting the number of 5s in the numbers from 1 to 100 which are all multiplied together. I did that because a zero at the end can only be produced by multiplying an even number with a 5 and there are more even numbers than multiples of 5 in the product.

Range of numbers Number of 5
1 - 10 2
11 - 20 2
21 - 30 3
31 - 40 2
41 - 50 3
51 - 60 2
61 - 70 2
71 - 80 3
81 - 90 2
91 - 100 3
total 24

Also there is another way to find the number of zeros. This is by:

 100 / 5 = 20 this is the number of multiples of 5 20 / 5 = 4 this is the number of multiples of 5 2

When we add this two together we get 24 which is exactly the number of noughts in 100!

So to see if my rule works I will find how many noughts are there in 1000!:

 1000 / 5 = 200 this is the number of multiples of 5 200 / 5 = 40 this is the number of multiples of 5 2 40 / 5 = 8 this is the number of multiples of 5 3 8 / 5 = 1.6 this is the number of multiples of 5 4 .

The number of zeros has to be a whole number so the number of multiples of 5 4 is 1 which is the integer part of 1.6 (written [1.6] ). Note that the process stops when division by 5 gives a number less than 5. If we add those answers together we will get the number of noughts. 200 + 40 + 8 + 1 = 249. From here we see that my rule works.

So to get the number of noughts in 10 000! we just divide by 5 to get the number of 5s:

 10000 / 5 = 2000 this is the number of multiples of 5 2000 / 5 = 400 this is the number of multiples of 5 2 400 / 5 = 80 this is the number of multiples of 5 3 80 / 5 = 16 this is the number of multiples of 5 4 16 / 5 = 3.2 so [3.2] = 3 is the number of multiples of 5 5

2000 + 400 + 80 + 16 + 3 = 2499

To get the number of noughts in 100 000!:

 100000 / 5 = 20000 this is the number of multiples of 5 20000 / 5 = 4000 this is the number of multiples of 5 2 4000 / 5 = 800 this is the number of multiples of 5 3 800 / 5 = 160 this is the number of multiples of 5 4 160 / 5 = 32 this is the number of multiples of 5 5 32 / 5 = 6.4 so [6.4]= 6 is the number of multiples of 5 6 6.4 / 5 = 1.28 so [1.28] = 1 is the number of multiples of 5 7

20000 + 4000 + 800 + 160 + 32 + 6 + 1= 24999

Here is how Koopa Koo gave the solution for 1 000 000!.

Let [x] denotes the greatest integer that does not exceed x.

The number of right most zeros of 1 000 000! = [1000000/5] +[1000000/5 2] +[1000000/5 3] +[1000000/5 4] +[1000000/5 5] +[1000000/5 6] +[1000000/5 7] + [1000000/5 8] = 200000 + 40000 + 8000 + 1600 + 320 + 64 + 12 + 2 = 249998.