# math homework?

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

### 2 Answers

- PuzzlingLv 73 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

Answer:

x has 12 possible values

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

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- 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

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