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Smallest number

Posted by Chris on June 1, 2011 – 2:44 pm

What is the smallest number that when you divide it by 2,3,4,5 and 6 will have remainders 1, 2, 3, 4 and 5 respectively?


This post is under “MathsChallenge” and has 8 respond so far.
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8 Responds so far- Add one»

  1. 1. Karys Said:

    With modulos it becomes : Find the smallest integer n such that
    n = 1 (2)
    n = 2 (3)
    n = 3 (4)
    n = 4 (5)
    n = 5 (6)

    Redundant equations :
    n = 1 (2) is already implied by n = 3 (4) and n = 5 (6) (it just means that n is odd).

    n = 1 (2) is also equivalent to :
    n = 1 OR 3 OR 5 (6)

    n = 2 (3) is :
    n = 2 OR 5 (6)

    so we end up with only 3 equations :
    n = 3 (4)
    n = 4 (5)
    n = 5 (6)

    you can also write from n = 3 (4) :
    n = 3 OR 7 OR 11 OR 15 OR 19 (20)
    and from n = 4 (5)
    n = 4 OR 9 OR 14 OR 19 (20)

    so we get n = 5 (6) and n = 19 (20)

    same thing at 60 :
    n = 5 (6) is n = 5 OR 11 OR 17 OR 23 OR 29 OR 35 OR 41 OR 47 OR 53 OR 59 (60)
    n = 19 (20) is n = 19 OR 39 OR 59 (60)

    so we end up with 59.

    I admit that may have been too quick an explanation, I mainly did not bother to explain how I “expanded” the equations to get equations with the same modulo.

  2. 2. Mody Said:

    it’s 59

  3. 3. Michelle Said:

    If you go with 59 as the number than you don’t get the remainders of 1, 2, 3, 4, and 5! Explain that!!

  4. 4. Michelle Said:

    I thought it would be 59 as well, but the remainders are not what the riddle is looking for!!

  5. 5. Chris Said:

    Hi Karys, you got it right.

    Hi Michelle, I think you must be jumbling something up.
    59/2 = 29 r 1, 59/3 = 19 r 2, 59/4 = 14 r 3, 59/5 = 11 r 4 and 59/6 = 9 r 5.

    — SOLUTION
    Let n be the number sought after. For each case we have n = m – 1 (mod m).
    Adding 1 to both sides gives: n + 1 = m = 0 (mod m), where m = 2,3,4,5,6

    So we need n + 1 = lcm(2,3,4,5,6) = 60, so n = 60 – 1 = 59.

    To get the lcm, note that the only prime factors for 2-6 are 2,3,5.
    But 4 = 2², so the lcm is 2²*3*5 = 60

  6. 6. anastacio Said:

    the answer is 11

  7. 7. Chris Said:

    Hi anastacio. Nice try, 11 nearly works. It only fails for the divide by 5 case, but then you get a remainder 1 or -4.

    I also note that if I’d extended the original problem, to an arbiitrary extent, e.g. n = 99 (mod 100), then n would have been lcm(2,3,4,5,…,100) – 1 = 69720375229712477164533808935312303556799

  8. 8. prashANT Said:

    ITS 59 AS-:
    IT LEAVES 1 AS REMAINDER WHEN DIVIDED BY 2 .SO ITS AN ODD NO.
    IT LEAVES 2 AS REMAINDER WHEN DIVIDED BY 3 . SO IT CAN BE 5,11,17 AND SO ON
    IT LEAVES 3 AS REMAINDER WHEN DIVIDED BY 4 . SO IT CAN BE 11,23,35,47,59,71 AND SO ON
    IT LEAVES 4 AS REMAINDER WHEN DIVIDED BY 5 SO IT CAN END IN 4 OR 9 BUT ITS AN ODD NO SO IT ENDS IN 9.IN ABOVE SERIES 59 IS THE SMALLEST NO AND IT ALSO GIVES 5 AS REMAINDER WHEN DIVIDED BY 6.

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