# MOP Numbers

Discussion in 'Physics & Math' started by gerg, Feb 6, 2004.

1. Hello

I'm having a little trouble solving this problem, and would like to ask if someone can help me.

A mop expression is a way of producing a number using only multiplications, ones and pluses. The length of a mop is the number of ones it contains.

For example, 22 = 1+1+((1+1+1+1)x(1+1+1+1+1)), which has a length of 11.

Another mop of 22 is 1+((1+1+1)x(1+((1+1)x(1+1+1)))) which has a length of 10. There are no shorter mops for 22.

DERIVE A METHOD that inputs a single integer n (1<n<10000) and outputs the length of the shortest mop of n.

Thankyou for anyone who has a look at this

Gerg

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3. ### NanoTecRegistered Senior Member

Messages:
30
(1+1)x(1+((1+1)x(1+1+1+1+1)))=22 [length=10]
(1+1)x(1+1+((1+1+1)x(1+1+1)))=22 [length=10]

Any “mop” should satisfy n=C+(XxY)
where C,X,Y are recursive “mop”s
the sum of the resulting “mop” values C+X+Y is the length.

Prime case P: initial C must be >0
(XxY)= lower number “mop”
Factorable case: initial C can be 0
Find factors “mop”s {X,Y}>=2
Take the factors distribute them between X and Y
For every distribution find the lowest length “mop”s

A table of lower order “mop”s would be helpful.

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5. ooh, thanks for that - this will help a lot with the programing of a computer program to solve the problem.

Gerg

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7. ### NanoTecRegistered Senior Member

Messages:
30
Attached is example output data for the recursive process.
No instances of C > 4, except n=5, exist.

Last edited: Feb 8, 2004