Question

For how many natural numbers n between 1 and 2020, then number (8n/9999-n) is an integer?

Answer 1

8n

9999−n

=

−8(9999−n)+8∗9999

9999−n

8n9999−n=−8(9999−n)+8∗99999999−n

=−8+

8∗9∗11∗101

9999−n

=−8+8∗9∗11∗1019999−n

Now the denominator can be max  

9998

9998

and min  

7985

7985

Now we need to find the factors of  

(8∗9∗11∗101)

(8∗9∗11∗101)

in the interval  

[7985,9998]

[7985,9998]

The only factor in the interval is  

8888

8888

wondering how?

If we remove  

11

11

, the product will be  

7272

7272

which is not in the interval,

If we remove  

9

9

, the product will be  

8888

8888

which is in the interval,

If we remove  

8

8

, the product will be  

9999

9999

which is not in the interval,

We know these three factors are consecutive,

So there is only one case when the denominator gets canceled at  

n=1111

n=1111

So there is only one value of  

n

n