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 Dark Divisibility (Posted on 2013-04-27)
The natural numbers a,b,c,d are such that their least common multiple equals a+b+c+d. Prove that abcd is divisible by 3 or by 5.

 No Solution Yet Submitted by Danish Ahmed Khan Rating: 5.0000 (2 votes)

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 Is this list exhaustive? | Comment 11 of 18 |

'Primitive' examples only, (i.e. all other cases are multiples of these)

A.   Let 2n=abc, d=(a+b+c):

(1)  First kind: {a,b,c,d, (a+b+c+d), (abcd), (2(a+b+c)/b, (2(a+b+c)/c}=

{1, 1, 4, 6, 12, 24, 12, 3}
{1, 2, 2, 5, 10, 20, 5, 5}
{1, 2, 6, 9, 18, 108, 3, 9}
{1, 4, 5, 10, 20, 200, 5, 4}

(2)  Second kind {a,b,c,d, (a+b+c+d), (abcd), (2(a+b+c)/b, (2(a+b+c)/c}=

{1, 3, 8, 12, 24, 288, 8, 3}
{1, 6, 14, 21, 42, 1764, 7, 3}

B.  Let 3n=abc, 2d=(a+b+c)

{a,b,c,d, (a+b+c+d), (abcd), (a+b+c+d)/b, (a+b+c+d)/c}=

{1, 3, 4, 4, 12, 48, 4, 3}
{2, 3, 3, 4, 12, 72, 4, 4}

C. Let a=k, b=(k+1), c=2(k+1), d=3(k+1)

{a,b,c,d, (a+b+c+d), (abcd), (a+b+c+d)/b, (a+b+c+d)/c}=

{2, 3, 10, 15, 30, 900, 10, 3}

Comment: ((a+b+c+d)/c}=3 for any given {a,b} is a minimum; it's not necessary to check any larger values of c.

Edited on April 29, 2013, 11:08 am
 Posted by broll on 2013-04-29 08:13:56

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