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Say you can use the numbers 1,2,3,4,5 once, and in the format of _ _ X _ _ _ where the underscores are numbers. Is it possible to find out the lowest possible product without guessing and checking? I believe it is 13 X 245, but i just want to be able to apply a rule to other numbers as well.

2007-09-02 16:26:43 · 5 answers · asked by frogster99 2 in Science & Mathematics Mathematics

5 answers

the question was "Is there a general minimizing formula to apply to questions such as this?"

I believe the answer to the question is "yes."

That's not to say I know it, but...at the very least, if the pattern is as follows:

_a_ _b_ X _c_ _d_ _e_

then, because a & c are the most significant, they should be the least significant digits.

{a,c} = {1,2}
{b,d} = {3,4}
{e} = {5}

so we can say a = 3-c, b = 7-d

_a_ _b_ X _3-a_ _7-b_ _5_

or

(10a+b)*(100(3-a)+10(7-b)+5)=P

So you want to minimize P with respect to a and b, where a in {1,2} and b in {3,4}

you can take partial derivitives, set them to zero, then have two equations...

ultimately, though...the calculation gets REALLY messy and finding a quick way to just check trial and error is best. A computer program would be easy:

for each permutation Pi of {1,2,3,4,5}
PROD(i) = _a_ _b_ X _c_ _d_ _e_

find i of min PROD(i)

if you want to see it in java, let me know.

2007-09-02 16:49:08 · answer #1 · answered by Anonymous · 0 0

Well this doesn't really quality as a elegant method but this is how I would do this:

If you have ab x cde, then the product is equal to:

(10a+b)(100c+10d+e)
= 1000ac + 100ad + 10ae + 100bc + 10bd + be
= 1000ac + 100(ad+bc) + 10(ae+bd) + be

now I would "prioritize" the letters, to see which one should be the lowest.
obviously a or c should be 1. But a also appears in the 10 term, so it is more important. so a = 1, c = 2.

next you look at b and d. they both appear in the 100 and the 10 term, but b also appears in the last term. so b might be more important. so b = 3, d = 4.

and finally e=5. So 13 x 245 is most likely the minimum product.

2007-09-02 23:53:15 · answer #2 · answered by Derek C 3 · 0 0

Don't worry about the number of digits. Just look at it as to the digit space. Such as you have one number in the hundreds space, two in the tens and one in the ones.

The 1 should go in the hundred space, then the 2 and 3 in the 10s then the 4 and 5 in the ones. It doesn't matter which number they are in as long as they are in the correct position.

So:
124 x 35
or 134 x 25
or 125x34
or 135x24

They will all be the lowest...and they all equal the same thing.

2007-09-02 23:45:36 · answer #3 · answered by Patty C 3 · 0 2

From the looks of things I would try putting the lowest value in the left-most position of the 2 digit number then the next lowest value in the left-most position of the 3 digit number. Continue altenating likewise till all positions are filled.

2007-09-02 23:36:16 · answer #4 · answered by GeekCreole 4 · 0 0

I don't think there is any particular rule. Just common sense and trial and error.

2007-09-02 23:38:25 · answer #5 · answered by ironduke8159 7 · 0 0

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