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Neighbours

Neighbours

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wolfgang59
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6 people talking at a party discover they all live on the same street.
One says but my address is rather unique - my house number has 3 digits and the sum of those digits is equal to the product!

In unison the other 5 say "So do mine!"

How close are the nearest neighbours?

Z

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The 6 people live at the following numbers: 123, 132, 213, 231, 312, 321. The closest pairs are (123,132) and (312,321), both differ by 9. I guess you could say the closest would be (123,132) assuming 1 is opposite 2, 3 is opposite 4, etc...

wolfgang59
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Originally posted by Zuggy
The 6 people live at the following numbers: 123, 132, 213, 231, 312, 321. The closest pairs are (123,132) and (312,321), both differ by 9. I guess you could say the closest would be (123,132) assuming 1 is opposite 2, 3 is opposite 4, etc...
The pairs you quote are on opposite sides of the street so
no way of determining closeness and in British English usage
not "neighbours". The answer I was looking for was 213 & 231.

R
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Originally posted by wolfgang59
The pairs you quote are on opposite sides of the street so
no way of determining closeness and in British English usage
not "neighbours". The answer I was looking for was 213 & 231.
Is there a way to solve

a + b + c = a*b*c

To arrive definitively at those results?

Ponderable
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Originally posted by joe shmo
Is there a way to solve

a + b + c = a*b*c

To arrive definitively at those results?
one equation with three unknowns is not definitely defined.

what you can do is to apply some logic on the digits:

* no "0" involved since the product involving zero is zero, but 000 is not a valid three digit number.
* not more than one digit may be 1 since 1*1*n=n but 1+1+n=n+2

from the first two: no digit bigger than 3 since the smallest including 4 is 124 with 1+2+4=7 and 1*2*4=8, the products increase faster than the sums.

so 123 and its permutations are left over 1+2+306 1+2*3=6, all valid solutions are:

123 132 213 231 312 321 as given above.

wolfgang59
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Originally posted by Ponderable
one equation with three unknowns is not definitely defined.

Not true if we are dealing just with positive integers.
eg What is the solution to A+B+C=3

f
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Originally posted by wolfgang59
Not true if we are dealing just with positive integers.
eg What is the solution to A+B+C=3
This may be true, but the problem at hand is dealing with non-negative integers, not positive ones

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Originally posted by wolfgang59
How close are the nearest neighbours?
Not very close.

I mean, they live on the same street but apparently don't know this.

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