David from Mount Carmel School sent us his working to one of the problems:

The numbers which can be divided by five but leave a remainder of four are:4, 9, 14, 19, 24, 29, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 89, 94 and 99.

The numbers which can be divided by three but leave a
remainder of two are:

2, 5, 8, 11, 14...

2, 5, 8, 11, 14...

...wait,

there it is!

14!

To complete REMAINDERS, I took all the numbers which when
divided by five left a remainder of four, and took all the numbers
which when divided by three left a remainder of two.

With these numbers, I then tried to match a number in one
category with a number in the other category which was the same,
and I got the answer!

The final question asked you to find the
smallest number with the property that when it is divided by each
of the numbers 2 to 10, the remainder is always one less than the
number it is has been divided by. Anurag from Queen Elizabeth's
Grammar School in Horncastle explained how to solve it:

When 59 is divided by 5, the remainder is 4

When 59 is divided by 4, the remainder is 3

When 59 is divided by 3, the remainder is 2

When 59 is divided by 2, the remainder is 1

The Lowest Common
Multiple, or LCM, of 5, 4, 3 and 2 is 60.

60 - 1 = 59.

60 / 5 = 12.

59 / 5 = 11 remainder 4.

There is a remainder of 4 because we have subtracted the 1
that would be necessary to get a remainder of 0.

The same applies to the question we have been asked.

We find the LCM of the numbers 2 to 10, and subtract 1.

The LCM of 2, 3, 4, 5, 6, 7, 8, 9 and 10 is 2520

(2 x 3 x 3 x 4 x 5 x 7 = 2520)

So the number we are looking for is 2520 - 1 = 2519

Kien from Warminster School also sent
us the solution to this question:

We have:

x / 2 remainder 1

x / 3 remainder 2

x / 4 remainder 3

....

x/ 10 remainder 9

So, (x + 1) must divisible by 2, 3, 4, 5, 6, 7, 8, 9,
10.

So the smallest value for (x + 1) = 10 * 9 * 7 * 4 =
2520

( the smallest number that has 2, 3, 4, 5, 6, 7, 8, 9 and 10
as factors)

So the smallest number that has the property is 2519.

Kamal from Holy Angel School
approached it like this:

Let N be the required number.

We can write,

N = 2a + 1 = 2(a + 1) - 1

N = 3b + 2 = 3(b + 1) - 1

N = 4c + 3 = 4(c + 1) - 1

N = 5d + 4 = 5(d + 1) - 1

N = 6e + 5 = 6(e + 1) - 1

N = 7f + 6 = 7(f + 1) - 1

N = 8g + 7 = 8(g + 1) - 1

N = 9h + 8 = 9(h + 1) - 1

N =10i + 9 =10(i + 1) - 1

=> (N + 1) is divisible by all of 2, 3, 4, 5, 6, 7, 8, 9
and 10.

The smallest such number will be N = LCM of (2, 3, 4, 5, 6, 7,
8, 9, 10) - 1 = 2520 - 1 = 2519

Well done to you all.