1. 2+8 = 10, 10+8 = 18, 18+8 = 26

similarly

5+9 = 14, 14+9 = 23, 23+9 = 32

So the next number is 26+8 = 34

2. When the number is divided by 18, it leaves a remainder of 7.

When the number is divided by 21, it leaves a remainder of 10.

When the number is divided by 24, it leaves a remainder of 13.

Observing the constant difference of 11 between divisor and remainder:

We can conclude that if 11 is subtracted from the LCM of 18, 21 and 24, then it will leave a remainder of 7 when divided by 18, a remainder of 10 when divided by 21 and a remainder of 13 when divided by 24.

When 11 is subtracted from any multiple of LCM, it will give the same result.

Now,

18 = 2 × 3 × 3

21 = 3 × 7

24 = 2 × 2 × 2 × 3

∴ LCM of 18, 21 and 24 = 504

So, any number of the form (504n – 11) will give the required remainders, when n is an integer.

But, this number also has to be divisible by 23.

Using trial and error method, for n = 6:

We find that: 504n – 11 = 504 × 6 – 11 = 3024 – 11 = 3013

congrats excellent