Question

Which numbers between 10 and 99 have digits that differ by 4 and add to 6

Answer 1

A

Explanation:

Correct option is A)

Let a 3-digit number ABC

Case (1)

If A=7, then B and C can be any digit but not 7

so, total 9 digits (0,1,2,3,4,5,6,8,9) can replace B and C

Then total such combinations =1(9)(9)=81

Case (2)

If B=7 then A can be any digit but not 0 and 7, because if A=0 it will be a 2-digit number.

And C can be any digit but not 7

Then total combinations =(8)(1)(9)=72

Case (3)

If C=7 then A can be any digit but not 0 and 7, because if A=0 it will be a 2-digit number.

And B can be any digit but not 7

Then total combinations =(8)(9)(1)=72

So, total numbers between 99 and 1000 which have exactly one of their digits as 7 are =81+72+72=225