rao_1857 wrote:

Tweaking the question little bit, how about this:

How many integers from 0 to 50, inclusive, have a remainder of 1 when divided by 3 and have a remainder of 3 when divided by 5 ?

Any idea Gurus?

remainder of 1 when divided by 3: 3k+1 = 1, 4, 7, 10, (13), 16, 19, 22, 25, (28), 31, 34, 37, 40, (43), 46, 49...

remainder of 3 when divided by 5: 5k+3= 3, 8, (13), 18, 23, 28, 33, 38, (43), 48...

Here, 13 and 43 are common in the two series. Other numbers will be 58, 73, so on and so forth.

The general equation is: (28-13)k+13= 15k+ 13 (where, k=0,1,2,...)

Hence, if we want to find the number of integers even upto 1000, we can use the above formula.

Going back to the problem, I see 3 numbers that satisfies the above criteria i.e 13, 28 and 43. (The next one will be 58 which is greater than 50).

Any comments?

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