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## Homework Statement

A sequence of 3-digit numbers divisible by 7 is given. Find:

a) first three and last three numbers.

b) how many numbers are there in that sequence.

c) sum of all the numbers in the sequence.

## Homework Equations

## The Attempt at a Solution

Cause they are 3 digit numbers that means: 100-999.

In order for a number to be divisible by 7 you need to get the last digit, double it and subtract it from the number without the last digit. If that number is divisible by 7, you stop there, if not, repeat the steps. I tried this way, but it got me nowhere. So it came to my mind that 7,14,21,28 are all divisible by 7 cause the distance between those numbers is 7. So with a bit of trial and error I found that 105 is divisible by 7. So the first three numbers of the sequence are 105, 112, 119. The last two are 994, 987, 980.

Now how do I find how many numbers between 100 and 999 are divisible by 7?

And for the sum I know the formula. Sn = ((a1 + an) / 2) * n