Easy does it! The Smart Scholar Method

Let the typical room number be ‘ABC’. In the ‘prohibited’ room numbers, A + B = C. Let us see which are the ‘prohibited’ room numbers.

It is clear that C cannot be 0 (because the least value of A is 1.)

If C is 1, the only possible number is 101; that is, 1 room is ‘prohibited’.

If C is 2, the possible numbers are 202 and 112; that is, 2 rooms are ‘prohibited’.

If C is 3, the possible numbers are 303, 213 and 123; that is, 3 rooms are ‘prohibited’.

If C is 4, the possible numbers are 404, 314, 224 and 134; that is, 4 rooms are ‘prohibited’.

If C is 5, the possible numbers are 505, 415, 325, 235 and 145; that is, 5 rooms are ‘prohibited’.

And so on.....

Note the pattern here - If C = n, n rooms are ‘prohibited’.

Thus, total number of ‘prohibited’ rooms is 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 = 45

Altogether then, there are 45 ‘prohibited’ rooms. As the total number of rooms is 899 (Not 900, mind you), Vijay can choose from among 899 – 45 = 854 rooms.

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