Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 months ago

# math homework?

If 2020 is divided by a positive integer x, the remainder is 20. How many possible values of x are there?

Relevance
• 3 months ago

Another way to say that, 2000 must be evenly divisible by x. But since the remainder is 20, it has to be bigger than 20.

2000 has a prime factorization of

2⁴ * 5³

The number of factors is found by incrementing the exponents and multiplying them.

4 becomes 5

3 becomes 4

Total factors would be 5 * 4 = 20

Now just remove the factors that are 20 or smaller.

1, 5

2, 10

4, 20

8

16

x has 12 possible values

Specifically {25, 40, 50, 80, 100, 125, 200, 250, 400, 500, 1000, 2000}

• 3 months ago

2020 = c * x + 20

2000 = c * x

You just need divisors of 2000 that are greater than 20

2000 = 2 * 10^3 = 2 * 2^3 * 5^3 = 2^4 * 5^3

2^0 * 5^0 , 2^1 * 5^0 , 2^2 * 5^0 , 2^3 * 5^0 , 2^4 * 5^0

2^0 * 5^1 , 2^1 * 5^1 , 2^2 * 5^1 , 2^3 * 5^1 , 2^4 * 5^1

2^0 * 5^2 , 2^1 * 5^2 , 2^2 * 5^2 , 2^3 * 5^2 , 2^4 * 5^2

2^0 * 5^3 , 2^1 * 5^3 , 2^2 * 5^3 , 2^3 * 5^3 , 2^4 * 5^3

1 , 2 , 4 , 8 , 16

5 , 10 , 20 , 40 , 80

25 , 50 , 100 , 200 , 400

125 , 250 , 500 , 1000 , 2000

25 , 40 , 50 , 80 , 100 , 125 , 200 , 250 , 400 , 500 , 1000 , 2000