Find the house numbers?

Question

There are 3 houses on one side of the street (meaning, the house numbers differ by 2). The product of the 3 house numbers is 60 times their sum. What are the 3 house numbers?

What about this: How many factors does the product of the three house numbers have?

Answer 1

I don't get rational numbers.

basically, you can set up the equation like this:
x(x+2)(x+4) = 60(x + x+2 + x+4)
x(x+2)(x+4) = 60(3x + 6)
x(x+2)(x+4) = 180(x+2) - divide through by (x+2)
x(x+4) = 180

x^2 + 4x - 180 = 0.

Unless I did something wrong, the above equation doesn't have rational roots.

[edit]
the only positive number that works for x is 11.5647...
so, the numbers would be 11.5647..., 13.5647..., 15.5647...

[edit - in response to "additional details"]
For the irrational answer given above, none. The product is an irrational number.

For any given set of 3 integer house numbers, the number of factors would depend on what the numbers are, but there would be at least three - the three house numbers.

Answer 2

Well you kept me busy for a while .. nice problem, just a little too many solutions.

ALL SOLUTIONS
===========================================
10 14 18
10 12 22
6 14 50
6 18 30
6 20 26
2 32 510
2 34 270
2 36 190
2 38 150
2 40 126
2 42 110
2 46 90
2 50 78
2 54 70
2 60 62

DISCUSSION
==========================================
The product of the house numbers is a multiply of 60, hence even, therefore the house numbers are even. We can write

2A * 2B * 2C = 60 * (2A + 2B + 2C)
A * B * C = 15 * (A + B + C)

Can we assume that they are neighbors? If so, write A = B-1 and C = B+1, and simplify:

(B-1)B(B+1) = 45B
(B-1)(B+1) = 45

but this has no integer solutions. So let's drop the assumption that the houses are next door to each other.

CALCULATION
============================================

A * B * C = 15 * (A + B + C)

Let A < B < C. It is easy to see that

B * C = 15 * (1 + B/A + C/A) > 45 hence C > sqrt(45), i.e. C>=7
A * B = 15 * (A/C + B/C + 1) < 45 hence A < sqrt(45), i.e. A<=6

We rewrite A * B * C = 15 * (A + B + C) as

... B = 15 (C + A) / (AC - 15) ...

and try the values A = 1, 2, ..., 6.

## A = 6 ##
B = 15 (C + 6) / (6C - 15) = (5C + 30) / (2C - 5) = [5 + 55/(2C-5)]/2
We need 2C-5 to divide 55, and C >= 7; this gives C = 8, 30, but the results have B < A.

## A = 5 ##
B = 15 (C + 5) / (5C - 15) = (3C + 15) / (C - 3) = 3 + 24/(C-3)
We need C-3 to divide 24, and C >= 7; this gives C = 7, 9, 11, 15, 27, and we find the solutions

>>> (A,B,C) = (5,7,9) <<< 10*14*18 = 60*(10 + 14 + 18) <<<
>>> (A,B,C) = (5,6,11) <<< 10*12*22 = 60*(10 + 12 + 22) <<<

## A = 4 ##
B = 15 (C + 4) / (4C - 15) = [15 + 465/(4C - 15)]/4
We need 4C-15 to divide 465 and leave a multiple of 4 minus 1. This is not possible.

## A = 3 ##
B = 15 (C + 3) / (3C - 15) = (5C + 15) / (C - 5) = 5 + 40/(C - 5)
We need C - 5 to divide 40, and C >= 7. This gives C = 7, 9, 10, 13, 15, 25 and 45. We find a few solutions:

>>> (A,B,C) = (3,7,25) <<< 6*14*50 = 60*(6 + 14 + 50) <<<
>>> (A,B,C) = (3,9,15) <<< 6*18*30 = 60*(6 + 18 + 30) <<<
>>> (A,B,C) = (3,10,13) <<< 6*20*26 = 60*(6 + 20 + 26) <<<

## A = 2 ##
B = 15 (C + 2) / (2C - 15) = [15 + 60/(2C - 15)]/2
We need 2C - 15 to divide 60 and result in an odd number. This is impossible: 2C - 15 is odd, so 60/(2C - 15) will be even.

## A = 1 ##
B = 15 (C + 1) / (C - 15) = 15 + 240/(C - 15).
We need C - 15 to divide 240. Possibilities are C = 16, 17, 18, 19, 20, 21, 23, 25, 27, 30, 31, 35, 39, 45, 55, 63, 75, 95, 135, 255. The corresponding solutions:

>>> (A,B,C) = (1,16,255) <<< 2*32*510 = 60*(2 + 32 + 510) <<<
>>> (A,B,C) = (1,17,135) <<< 2*34*270 = 60*(2 + 34 + 270) <<<
>>> (A,B,C) = (1,18,95) <<< 2*36*190 = 60*(2 + 36 + 190) <<<
>>> (A,B,C) = (1,19,75) <<< 2*38*150 = 60*(2 + 38 + 150) <<<
>>> (A,B,C) = (1,20,63) <<< 2*40*126 = 60*(2 + 40 + 126) <<<
>>> (A,B,C) = (1,21,55) <<< 2*42*110 = 60*(2 + 42 + 110) <<<
>>> (A,B,C) = (1,23,45) <<< 2*46*90 = 60*(2 + 46 + 90) <<<
>>> (A,B,C) = (1,25,39) <<< 2*50*78 = 60*(2 + 50 + 78) <<<
>>> (A,B,C) = (1,27,35) <<< 2*54*70 = 60*(2 + 54 + 70) <<<
>>> (A,B,C) = (1,30,31) <<< 2*60*62 = 60*(2 + 60 + 62) <<<

Answer 3

this is an extremly interesting sum! wish i had creative questions like that in my days.

assume that the first house number is x. that would make the second house x + 2 and the third x + 4, since they differ by 2.

the sum of the numbers would then be:
x + (x + 2) + (x + 4) = 3x + 6

the product would be:
x(x + 2)(x + 4) =
from the question it is stated that the product of the 3 house numbers is 60 times their sum.

product = 60 x sum
x^3 + 6x^2 + 8x = 60 (3x + 6)
x^3 + 6x^2 + 8x = 180 x + 360
x^3 + 6x^2 - 172x - 360 = 0

the next thing to do is to solve for x.
then once x is found, you just find x + 2 and x +4.

the funny thing is though, i dont get any rational numbers. after checking with the rest of the people that answered, they dont seem to get any too. so it seems that you have a problem with your question or this is a really funky neighbourhood.

but the focus is not so much the answer, it's the steps. notice how each part of the sentence is transformed into an equation and then you just solve for x? i do hope that the explaination is clear and that you can now solve questions like these on your own.

good luck and have fun with math. it isnt your enemy!

Answer 4

Say the house number in the middle is x, so the other two houses have the numbers x-2 and x+2. Then the product of the house numbers is

x*(x-2)*(x+2) = x^3-4*x

and the sum is

x+(x-2)+(x+2)=3x

Since the product is 60 times the sum, we have x^3-4*x=60*3x, so
x^3-184*x=0. This factors as x*(x^2-184)=0. Unfortunately, the solutions to this are x=0, and x=plus/minus sqrt(184), which is not an integer. Assuming you want the house numbers to be positive integers, this question is not solvable.

Answer 5

Assume that the number of the first house is x, then the number of the 2nd is x+2, and that of the third is x+4.

the sum is x+ x+2+ x+4 = 3x +6
the product is x(x+2)(x+4) = x( x^2 +6x +8) = x^3 +6x^2 +8x

so 60 times their sum= their product
so : 60*(3x+6) = x^3 +6x^2 +8x
180x +360 = x^3 +6x^3 +8x

so: x^3 +6x^2 -172x -360 =0

the answer for x is between 11 and 12.
So the numbers are:
11< 1st<12, 13< 2nd<14, 15<3rd<16

Answer 6

I do believe your premise is incorrect.
Most house numbers differ by 4
so you have

X-4 X X+4 as the 3 house numbers

that product is X cubed -16X
the sum is 3X (keeping in mind that X is the middle house)

X cubed -16X = 180 X
X squared - 16 = 180
X squared = 196
X = 14

The house numbers are 10 14 18

10*14*18 = 2520
60*(10+14+18) = 60*42 = 2520

Answer 7

Let the house in the middle be x

product = 60*sum
(x-2)*(x)*(x+2) = 60*(3x)
x(x^2-4) =180x
x^3 -4x =180x
x^3-184x =0
x(x^2-184) =0

Possible roots are x =0, x=2*squareroot(46), and x=-2*squareroot(46)

I have never seen addresses with non-positive numbers or irrational numbers.

Answer 8

you need to have a base number, example 400 block and you need to know if it is on the even or odd number of the street

Answer 9

does not have any answer possible