Concept is ok, thank you Ian!

Since 51 is evenly divisble by 3, we don't even have to worry about where in the cycle we started (which we would have had to do if the number of integers weren't divisible by 3).[/quote]

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

You need to think in that way , k=3a+1 so try few numbers to find how the sequance behave ..

K= 3*0+1 -------> 1
K=3*1+1 --------> 4
K=3*2+1 --------> 7
K=3*3+1 --------> 10
K=3*4+1 --------> 13

so you can see that between 0-10 there is 3 differents numbers that are divisible by 3 with a reminder of 1 ( never count for the sequence the first and the last item )

So because the sequance continue to behave in the same way between 0-50 it is like we had 5 sequance between 0-10 so there will be 3*5=15 items . So the next to last a of the sequance is a=15
and to get the final answer we need to add the first a when a=0 et the last one when a=16 you get to 17 items

Cheers

Hi ,

i am sowree might sound like a stupid question but how is the nearest multiple of 5 smaller than -7 -10? why not -5? In that case wouldnt the remainder be 2?

Ian Stewart wrote:

pepeprepa wrote:"What is the remainder when -7 is divided by 5?"
3
Because -7=5x(-2)+3
It would mean you go on the other way but anyway I do not master the "concept".
If you have examples to master it would be cool.
Thank you for the help to understand.

Yes, 3 is the answer. When we find the remainder when we divide, say, 23 by 5, we find the nearest multiple of 5 which is lower than 23, and work out how far away we are. The remainder is 3 because 23 is 3 more than 20, the nearest (smaller) multiple of 5. When we divide -7 by 5, the remainder is also 3, because the nearest multiple of 5 that is smaller than -7 is -10, and -7 is 3 more than -10.

bhumika.k.shah wrote:Hi ,

i am sowree might sound like a stupid question but how is the nearest multiple of 5 smaller than -7 -10? why not -5? In that case wouldnt the remainder be 2?

Ian Stewart wrote:

pepeprepa wrote:"What is the remainder when -7 is divided by 5?"
3
Because -7=5x(-2)+3
It would mean you go on the other way but anyway I do not master the "concept".
If you have examples to master it would be cool.
Thank you for the help to understand.

Yes, 3 is the answer. When we find the remainder when we divide, say, 23 by 5, we find the nearest multiple of 5 which is lower than 23, and work out how far away we are. The remainder is 3 because 23 is 3 more than 20, the nearest (smaller) multiple of 5. When we divide -7 by 5, the remainder is also 3, because the nearest multiple of 5 that is smaller than -7 is -10, and -7 is 3 more than -10.

nearest multiple of 5 on negative side lower than -7 is -10 and not -5 (-5>-7>-10)
logic is to find out the lesser value hence rem is 3

Ian Stewart wrote:

pepeprepa wrote:"What is the remainder when -7 is divided by 5?"
3
Because -7=5x(-2)+3
It would mean you go on the other way but anyway I do not master the "concept".
If you have examples to master it would be cool.
Thank you for the help to understand.

Yes, 3 is the answer. When we find the remainder when we divide, say, 23 by 5, we find the nearest multiple of 5 which is lower than 23, and work out how far away we are. The remainder is 3 because 23 is 3 more than 20, the nearest (smaller) multiple of 5. When we divide -7 by 5, the remainder is also 3, because the nearest multiple of 5 that is smaller than -7 is -10, and -7 is 3 more than -10.

brilliant question Ian!