a)

Since question asks for smallest 4 digit number divisible by 11, we try by putting the smallest digit 3 in the thousands place.

3 _ _ _

you have 5,7 and 9 left.

We try the smallest possible number first
3 5 7 9

(putting the 2nd smallest digit in hundreds place, 2nd biggest in tens place and largest in the ones place)

3579 ÷ 11 = 325 R 4

Try the next biggest number, 3597

3597 ÷ 11 = 327

So 3597 is the answer.


(Note : we got the answer relatively fast for this question, but for other questions it might not be the case. If let's say all options with 3 being the first digit don't work, you'll need to try with 5 as the first digit. If it doesn't, try 7, and so on)



b)


Any number exactly divisible by 5 ends with a 0 or 5.

To see why it's like that, observe the following :

5 x 1 = 5
5 x 2 = 10
5 x 3 = 15
5 x 4 = 20

And so on.

Because two 5s make a 10, multiplying 5 by an odd number gives you a number ending with 5 since the last 5 is unpaired.

Multiplying 5 by an even number gives a number ending with 0 since all 5s are paired, and the total each pair is equal to 10. Any number of pairs x 10 per pair will still give you a last digit of 0.

So biggest number divisivible by 5 is 9735.


c)


Difference = 9735 - 3597 = 6138


At P5 and P6, divisibility rules/tests are not covered so it's not expected of you to memorise them. This question is more of a trial and error but with some logical thinking needed.