JOHN
Lv 7
JOHN asked in Science & MathematicsMathematics · 5 years ago

For a given integer n let D(n) be the number of divisors of n, including 1 and n itself. For instance 6 is divisible bt 1, 2, 3 and 6, so?

D(6) = 4. Determine whether D(1) + D(2) + D(3) +....+ D(1998) is odd or even.

3 Answers

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  • 5 years ago
    Favourite answer

    In most cases, the number of divisors in a number is even. That's because each divisor will pair up with another divisor to multiply to be the original number.

    For example:

    The number 6 has four divisors {1, 2, 3, 6}. They pair up as follows:

    1 x 6 = 6

    2 x 3 = 6

    Or the number 12 has six divisors {1, 2, 3, 4, 6, 12}. They pair up as follows:

    1 x 12 = 12

    2 x 6 = 12

    3 x 4 = 12

    The exceptions are the perfect squares. Those will always have an *odd* number of divisors. That's because the square root of the number "pairs up" with itself.

    For example:

    The number 9 has three divisors {1, 3, 9}

    1 x 9 = 9

    3 x 3 = 9 <-- 3 pairs up with itself, so that results in an *odd* number of divisors.

    So the problem reduces to finding the number of perfect squares between 1 and 1998.

    √1998 ≈ 44.7

    So you'll have the following perfect squares:

    1², 2², 3², ..., 43², 44²

    That's an *even* number of perfect squares. So if there are an even number of perfect squares with an odd number of divisors, their sum will be even.

    even x odd = even

    Answer:

    The sum of all those divisors will be even.

  • Nick
    Lv 6
    5 years ago

    1998^(1/2) = 44 (rounded down)

    There are 44 square numbers between 1 and 1998:

    1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225 ....

    Square numbers all have an odd number of divisors. All other number have even divisors (since they come in pairs). There are 44 odd divisors which sum to an even number and 1998-44=1954 even divisors which sum to an even number hence:

    even + even = even

    The divisors sum to an even number.

  • Anonymous
    5 years ago

    Good answers so far, but they don't fully explain why only perfect squares have an odd number of divisors.

    If n = the product from i=1 to k of p_i^(a_i), where the p_i are distinct primes, then the number of divisors is:

    D(n) = the product from i=1 to k of (a_i + 1)

    This is because any divisor can be found by choosing one prime power from each of the following rows:

    1, p_1, p_1^2, ..., p_1^a_1

    1, p_2, p_2^2, ..., p_2^a_2

    ...

    1, p_k, p_k^2, ..., p_k^a_k

    The product of (a_i + 1) is odd if and only if a_i is even.

    That's equivalent to saying n is a perfect squares, since all prime factors of n have an even power.

    √1998 = 44.7 approx., so there are 44 perfect squares in the list from 1 to 1998.

    44 is even. The sum of an even number of odd numbers is even.

    All the other 1968-44 numbers have a even number of divisors.

    So the sum of all the numbers is even.

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