Laura0806
Without knowing what the question is I'm sort of stumped to offer advice.
First off definition of factor is here: www.mathsisfun.com/definitions/factor.html - so it's all the ways to make a number.
I'm not clear whether your son is being asked to only find the factor of a number using the 11 times table or all factors.
In case the issue is multiples of 11 beyond 9 - I thought this might help.
Let's start with multiples of 11 between 0 - 9.
0 - well anything x 0 = 0 so 11 x 0 = 0. That's just straightforward.
1 - 9 x 11 - is simply writing to multiple of 11 down twice.
1 x 11 = 11
2 x 11 = 22
3 x 11 = 33
4 x 11 = 44
5 x 11 = 55
6 x 11 = 66
7 x 11 = 77
8 x 11 = 88
9 x 11 = 99
now the trick when you have multiples >9. Just works for two digit numbers so 10 - 99.
take the first and second digit - separate them and then put their sum in the middle.
so 10 x 11
1 - (1+0) - 0 that gives you
1-1-0
110
so 10 x 11 = 110 (and knowing your x 10 trick (just add zero at end of multiple) you would know that anyway.
but 11 x 11 = 1 - (1+1) - 1 = 121
12 x 11 = 1 - (1+2) - 2 = 132
(you can have some where the middle number requires carrying:
19 x 11 = 1 - (1+9) - 9 = 1 - (10) - 9 (you'll have to carry the ten from the middle number) so that gives you (1+1) - 0 - 9 = 209
let's try another one
78 x 11 = 7 - (7+8) - 8
that gives you 7 - (15) - 8
(so you'll have to carry the ten from the middle number)
(7 + 1) - 5 - 8 = 858.
Have him play around with it and check his answers with a calculator - once he's used to the method than he can pretty well work out any factor of 11 times table.
So if your DS was asked what are the 11 times table factors of 154 - and you know the first and last number together make five (so no carrying) - then factors of 154 are 11 and 14.
Factors of 528 -- well adding 5 + 8 (first and last number) gives you 13 - so something is wrong - so there must have been carrying. So try one less than 5 (one less on the first number because you would carry from tens to hundreds). So what if you had 4 (one less than 5) and 8. Middle number would be 4 + 8 = 12 - you would leave the 2 and carry the ten to the hundreds column to get 5 - so yes that works.
Factors of 528 are 48 and 11.
Now just thinking about it you can have something like 528 which has all sorts of factors - not just 48 x 11. Just working it through you need to think of a line of numbers between 1 - 528 and work to the point you are only repeating numbers:
1 x 528
2 x 264
3 x 176 (by the way to work out if a number is divisible by 3 just add all the digits - if that number is divisible by 3 the original number is).
4 x 132
6 x 88
8 x 66
11 x 48
12 x 44
16 x 33
22 x 24
(now you can work that out with 11 x 48 (knowing 48 = 3 x 2 x 2 x 2 x 2) - and then 1 and itself (the prime version - a number is prime if its factors are only 1 and itself)
so you can see 2 and 264 (remaining factors 11 x 24 [3 x 2 x 2 x 2])
3 and 176 (remaining factors 11 x 8 [2 x 2 x 2 x 2])
4 and 132 ( 2 x 2 and remaining factors (11 x 3 x 2 x 2)
6 and 88 (2 x 3 and remaining factors (11 x 2 x 2 x 2)]
8 x 66 (2 x 2 x 2 and remaining factors (11 x 3 x 2)]
11 x 48 (11 and remaining factors 3 x 2 x 2 x 2 x 2)
16 x 33 (2 x 2 x 2 x 2 and remaining factors 11 x 3)
22 x 24 (2 x 11 and remaining factors 3 x 2 x 2 x 2)
so the only number left is 23 which doesn't work - 528 divided by 23 doesn't produce a whole number - so that brings us either down to 22 or up to 24 (we've crossed over or we're repeating numbers - however you want to think about it) so that's all factors.
HTH