In remainder -LCM-HCF problems... you get 3 types of problems...
1. same remainder so, N= LCM(divisors)+the remainder
2. Common Diff between divisors and remainders (CD), so N= LCM( Divisors) x K - Common diff
3. no same remainder and no common diff. you need to take all the equations formed then solve them.
tip: but some questions sometimes comes in a bit tricky conditions...you may get to see that no common diff is there..but usually in these questions if you multiply all the equation by some common no you get the standard equations which will show you common differences...... so always suspect a que from this category if it can be arranged in some way to make it fall under the 2nd category.
eg:
If n is the smallest number which when divided by 3, 5, 7 and 9 leave a remainder of 1, 2, 3 and 4 respectively, find the product of the digits of n.
solution: n will be of the form:
n = 3a + 1 n = 5b + 2 n = 7c + 3 n = 9d + 4
Multiplying each of the equations by 2,
2n = 6a + 2 = 3A + 2 2n = 10b + 4 = 5B + 4
2n = 14c + 6 = 7C + 6 2n = 18d + 8 = 9D + 8
This means that when 2n is divided by 3, 5, 7 and 9, the remainders are 2, 4, 6 and 8 respectively. This is a standard problem with difference between divisors and remainders being 1.
So, smallest value of 2n = LCM (3, 5, 7, 9) - 1 = 315 - 1 i.e. 314 and the smallest value of n is 157. The product of the digits is 35.