Question

What is the smallest multiple of 18 of the form 2a945b, where a and b are digits?

Answer 1

Answer:

for a Multi ple of 18 a number should be multiple of 9 and 2 both.

For a multiple of 2 it needs to be even

For a multiple of 9 its sum of digits should be a multiple of 9

2+A+9+4+5+B = 18 + 2 + A +B

this implies that for the sum of digts to be a multiple of 9

A+B+2 = 9

A+B+2 = 18 and so on .

Here B can be 2 , 4 , 6 etc … as it needs to be even

B = 2 , A = 5

B = 4 , A = 3

B = 6 , A = 1

The lower the value of A the smaller will be the number so A = 1 B = 6

219456

Answer 2

Before we go further lets break 18. i.e. 2*3*3
Now, 18*2 = 36,
We get 6 at unit place. Hence the number becomes 2a9456, which is also a multiple of 3.

Add all digits of the number i.e. 2+9+4+5+6 = 26 that is 1 less than 27 which os multiple of 3 hence a =1.

The number is 219456