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Home/ Questions/Q 845905
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T06:31:46+00:00 2026-05-15T06:31:46+00:00

I’m trying to solve Problem #5 in Project Euler. The code works for the

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I’m trying to solve Problem #5 in Project Euler. The code works for the example, when I check the numbers from 1 to 10 I get 2520 as a result, which is right. But when I check for the numbers from 1 to 20, the code doesn’t stop running.

Here it is:

num = 0

while true

    num += 1
    check = true

    for i in 1..20

        break unless check

        check = num%i==0

    end

    break if check

end

File.open("__RESULT__.txt", "w+").write num
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  1. Editorial Team
    Editorial Team
    2026-05-15T06:31:47+00:00Added an answer on May 15, 2026 at 6:31 am

    The solution for that problem can not be found by just calculating every possible solution. The solution is so big, that it will take days (maybe years) to calculate.

    There is a smarter solution using prime numbers to write down the numbers.

    The example that is given (2520 is the smallest number that is divisable by the numbers 1 through 10) can be written down like this:

    1 = 1 (can be skipped)  = 2^0 * 3^0 * 5^0 * 7^0
    2 = 2 (prime)           = 2^1 * 3^0 * 5^0 * 7^0
    3 = 3 (prime)           = 2^0 * 3^1 * 5^0 * 7^0
    4 = 2^2                 = 2^2 * 3^0 * 5^0 * 7^0
    5 = 5 (prime)           = 2^0 * 3^0 * 5^1 * 7^0
    6 = 2 * 3               = 2^1 * 3^1 * 5^0 * 7^0
    7 = 7 (prime)           = 2^0 * 3^0 * 5^0 * 7^1
    8 = 2^3                 = 2^3 * 3^0 * 5^0 * 7^0
    9 = 3^2                 = 2^0 * 3^2 * 5^0 * 7^0
    10= 2 * 5               = 2^1 * 3^0 * 5^1 * 7^0
    

    Now the smallest number that can be divided by these, can be calculated by using the maximum power that is used on each prime:

    2^3 * 3^2 * 5^1 * 7^1 = 2520
    

    The same can be performed (even by hand) on the numbers 1 through 20

    Last hint: the answer is larger than 100.000.000 but less that a billion, so it can be calculated in minutes if done efficiently

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