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There are 16 multiples of 3 and hence there are 16 numbers that leave reminder of 1 (Note checked 48+1 = 49 <50)

1 also leaves a remainder 1 when divided by 3!

Hence 16+1 = 17 _________________

"To dream anything that you want to dream, that is the beauty of the human mind. To do anything that you want to do, that is the strength of the human will. To trust yourself, to test your limits, that is the courage to succeed."

sorry, I know this post is dead but I would like to know how 1 divided by 3 has a remainder of 1? I thought the numerator had to be greater than the denom. in order to have an integer plus an integer remainder. What am I missing?

I think I may have answered my own question. (1/3) will have a remainder of 1 when the quotient is 0. (x/y)=q+r ......x=y(q) + r , for integers x and y, with the remainder r. Therefore, 1=3(0) + 1. Is this correctly stated?

I always forget about zero and the tricky roles it plays in questions....

Let me give it a shot (checking my own understanding in the process)...

Formula for Arithmetic Progression:

an = a1 + (n-1)d
where an = nth term; a1 = first term; n = number of terms; d = difference of succesive numbers

using 4 (which leaves a remainder of 1) as the first term and 49 (which leaves a remainder of 1) as the last term, we can solve for n

49 = 4 + (n-1)3 ---> n = 16

HOWEVER, n = 16 when the first term, i.e. the first number leaving a remainder of 1, is 4. But we have to remember that zero is an integer as well. Therefore, the number 1, when divided by 3 leaves a remainder of 1 as well. The expression would look like this:

Find 'x' for which above equation is true or appox true
your will get x = 17 (giving answer 51)

In other words, just remember table of three & check which position gives you number below/equal to 50. 3*16 = 48. So, 16 numbers, but table of 3 doesn't include '1'. So include '1', 16+1 = 17.

Now, why tables of 3? Because each element has remainder of 0 when divided by 3?
This is a trick, each element + 1, will give remainder of 1. This can be used for any such problem