Another interesting CAT question and solution

Question:
A number when divided by 5, 6, 8, 9, 12 leaves a remainder 1 in every case. But, when divided by 13 doesn't leave any remainder. Find the least such possible number.

Solution:
Looking at first line, straight away we should find L.C.M(5,6,8,9,12). And that is 360. Now any mulitple of 360 increased by 1 would satisfy the first line. Now the second line is the tricky one. 361 as such is not divisible by 13. 361 % 13 = 10 and not 0. So some other mulitple of 360 increased by 1 should be tested. Rather than brute forcing on every multiple. Assume that 'alpha'*360 when increased by 1 is perfectly divisible by 13. Then that means...
('alpha'*360) % 13 = 12 (Note this carefully)
=> [('alpha')%13] * [360%13] = 12
=> [('alpha' * 9)%13] = 12 (since 360%13 = 9)

So this looks more manageable, [('alpha') *9] should be one less than mulitple of 13 i.e. 12 or 25 or 38 or 51 or 64 or 77 or 90 <<< there you got a mulitple of 9. So least 'alpha' value is 90/9 = 10.
Hence the required answer is [(10 * 360) + 1] = 3601.

However, if in the options there is no 'none' and the other options look weird enough (as it happened in the answer choices I got), i.e. doesn't even satisfy the first requirement. Elimination of the answers and quickly arriving at the correct answer should be the preferred way of tackling a CAT question like this. CAT setters are interestingly nice, they set tough questions but keep some steppign stones in the path to arrive at solution without actually formally solving the question. Rewarded is the one, who identifies these stepping stones.

Regards
Jayachandra