When you reverse the digits of a two digit number the new number is 18 more than the earlier number. How many such numbers are possible?

(a) 5

(b) 6

(c) 7

(d) 8

Let 10x + y be a two digit number,
where x and y are positive single digit integers and x>0.

Its reverse = 10y + x

Now, 10y + x - 10x - y = 18

∴ 9(y - x) = 18 ∴ y - x = 2

Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

∴ Other than 13, there are 6 such numbers.

Its reverse = 10y + x

Now, 10y + x - 10x - y = 18

∴ 9(y - x) = 18 ∴ y - x = 2

Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

∴ Other than 13, there are 6 such numbers.

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