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 Product equals sum (Posted on 2016-01-17)
Let's start with an easy problem : What three positive integers have a sum equal to their product?

This puzzle can easily be transformed into a D4 problem:

For what values of k will the question "What k positive integers have a sum equal to their product?" have only one unique set of integers for an answer?
Clearly for k=2 the answer is unique: (2,2) and so it is for k=4: (1,1,2,4).

List all other values of k below 1000.

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 re: A continuation | Comment 5 of 6 |
(In reply to A continuation by Steve Herman)

Let S be a set of positive integers whose sum equals its product.  Let f1, f2, f3, ..., fn be the members of that set which are greater than 1 (arranged in descending order).  Then there are (f1*f2*...*fn) - (f1+f2+...+fn) occurances of 1.  Then k = (f1*f2*...*fn) - (f1+f2+...+fn) + n

Continuing specifically with n=3:
x=m, y=z=2: k = 4m-(m+4)+3 = 3m - 1; m>=2 -- 5, 8, 11, 14, ...
x=m, y=3, z=2: k = 6m-(m+5)+3 = 5m - 2; m>=3 -- 13, 18, 23, 28, ...
x=m, y=4, z=2: k = 8m-(m+6)+3 = 7m - 3; m>=4 -- 25, 32, 39, 46, ...
x=m, y=5, z=2: k = 10m-(m+7)+3 = 9m - 4; m>=5 -- 41, 50, 59, ...
x=m, y=6, z=2: k = 12m-(m+8)+3 = 11m - 5; m>=6 -- 61, 72, ...

So after (x, y, 2, 1, ... 1) the list below 60 shortens to: 2,3,4,6,12,24,30,42,54

Lets move on:
x=m, y=3, z=3: k = 9m-(m+6)+3 = 8m - 3; m>=3 -- 21, 29, 37, ...
x=m, y=4, z=3: k = 12m-(m+7)+3 = 11m - 4; m>=4 -- 40, 51, 62, ...
x=m, y=5, z=3: k = 15m-(m+8)+3 = 14m - 5; m>=5 -- 65, 79, ...
x=m, y=4, z=4: k = 16m-(m+8)+3 = 15m - 5; m>=4 -- 55, 70, ...

Nothing else to sieve out of the below 60 list.

Trying n=4 combos:
w=m, x=y=z=2: k = 8m-(m+6)+4 = 7m-2; m>=2 -- 12, 19, 26, 33, 40, ...
w=m, x=3, y=z=2: k = 12m-(m+7)+4 = 11m-3; m>=3 -- 30, 41, 52, 63, ...
w=m, x=4, y=z=2: k = 16m-(m+8)+4 = 15m-4; m>=4 -- 56, 71, ...
w=m, x=3, y=3, z=2: k = 18m-(m+8)+4 = 17m-4; m>=3 -- 47, 64, ...
w=m, x=4, y=3, z=2: k = 24m-(m+9)+4 = 23m-5; m>=4 -- 87, 110
w=m, x=3, y=3, z=3: k = 27m-(m+9)+4 = 26m-5; m>=3 -- 73, 99, ...

This reduces the below 60 list a little more: 2,3,4,6,24,42
The only n=5 case that may work for k<60 is (m, 2, 2, 2, 2, 1, ..., 1): k = 16m-(m+8)+5 = 15m-3; m>=2 -- 27, 42, 57, ...

This gets us a little further: 2,3,4,6,24.  This is extremely likely to be all such k below 60.

Time to head over to the OEIS.  Searching 2,3,4,6,24 returns 7 results, one of which is A033179: "Numbers n such that exactly one multiset of n positive integers has equal sum and product."
Only eight terms are known for all k<10^10: 2, 3, 4, 6, 24, 114, 174, 444

 Posted by Brian Smith on 2016-01-18 10:47:17

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