# algebra!

posted by .

How many positive integers, a, are there such that 2027 divided by a leaves a remainder of 7?

• algebra! -

There is probably an easier way to go about doing this but I can't seem to think of it at the moment so you'll have to suffice with this until someone else can answer:

I came up with this formula:
a x b + 7 = 2027 where a is the the divisor and b is the quotient and 7 is the remainder. We know that if we multiple the divisor and quotient and add the remainder we should get the dividend.

a x b = 2020

Now we must find all of the factors of 2020.

1 x 2020 = 2020 - therefore, a = 2020
2 x 1010 = 2020 - therefore, a = 1010
4 x 505 = 2020 - therefore, a = 505
5 x 404 = 2020 - therefore, a=404
10 x 202 = 2020 - therefore, a = 10 or 202
20 x 101 = 2020, therefore, a = 20 or 101

So the # of + integers for a such that 2027 divided by a leaves a remainder of 7 would be 8.

I'm not too sure if I did that correctly. Hopefully someone can verify this.

• algebra! -

IT is Absolutely right . The answer is 8