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s3017789 wrote:

Can someone explain the division test for 14.

If there is one.

The test seems to be if the number is divisible by 2 and 7 since 2*7=14.

ex: 56 is divisible by 14 because 56/2=28 and 56/7=8

**Quote:**

I had a test that wanted to know the sum of numbers <500 that had a remainder of 1 when divided by 14.

For this one the only solution I could come up with is as follows:

36*14=504 which is >500, so we are interested in values under 36.

Try 35*14=490. So we are interested in values 35 to 1.

Now any number divided by 14 can have remainders 0,1,2,3,4...13.

499/14 = 35*14 + 9

..

..

492/14 = 35*14 + 2

491/14 = 35*14 + 1

490/14 = 35*14 + 0

489/14 = 34*14 + 13

..

..

1/14 = 0*14 + 1

So the thing is we are interested in values like 491, i.e (multiple of 14) + 1

The sum would look something like this;

(35*14) + 1 + (34*14) + 1 + (34*14) + 1 + ..... + (14*1) + 1 + (14*0) + 1

= 36 + 14(35 + 34 + 33 + ... +1)

= 36 + 14( (35+1)(35)/2 )

= 36 + (14*18*35)

= 36 + 8820

= 8856

I get the 36 by adding up all the 1's

And for the sum of 35+34..+1, I'm just using the formula n(n+1)/2. I think it's part of HH's notes.

However I haven't verified the answer, and I might be miscounting some where along the line, so this answer is half baked

UPDATE: Ok I plugged it into the computer and that's the answer I'm getting.

I'd hate to get this on an exam though.