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?

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  • 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

    Answer:

    x has 12 possible values 

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

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  • 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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