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Since, there are 17 integers between 0 and 50 that are divisible by 3 (including 0) and since the first such integer is 0 and the last is 48....hence there should be 17 such integers that have remainder as 1 when divided by 3 as 0+1 = and 48 + 1 = 49 and both these numbers are within the limits of 0 and 50.

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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To find what you seek in the road of life, the best proverb of all is that which says: "Leave no stone unturned." -Edward Bulwer Lytton