Question 1207599: How many numbers between $1000$ and $2000$ leave a remainder of $3$ when divided by $47?$ Found 2 solutions by Edwin McCravy, mccravyedwin:Answer by Edwin McCravy(20060) (Show Source):
Such an integer must be 3 more than a multiple of 47. They form an arithmetic
sequence, with common difference d=47
To find the first term, we first find the first multiple of 47 above 1000.
1000/47 = 21.27659574
So we know from that, that since the whole part is 21, that 21x47 is the last
multiple of 47 less than 1000, and that the first multiple of 47 above 1000 will
be 22x47=1034.
Add 3 to that and we get the first term,
We require the terms to be less than or equal to 2000.
Solve that inequality and get
So the answer is the largest integer less than 21.4893617, which is 21.
Edwin
It just occurred to me that we could have known the answer from that first
step:
1000/47 = 21.27659574
We could deduce the answer is 21, because there are just as many positive
multiples of 47 between 0 and 1000 as there are between 1000 and 2000, and
therefore just as many integers that are 3 more than a multiple of 47 as well.
That's because they are a long way from the endpoints of the intervals, 0, 1000,
and 2000.
Hey, don't turn in what I just wrote! I just wrote it to maybe help your
thinking about numbers. LOL.
Edwin
