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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
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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