Math Problem

[johnq86]

Find five different positive numbers whose sum and product are equal to one another.

[ooo]

misread

[h1]

0 and 2

[ooo]

............

[Infested Cats]

I don't understand your question, but is this what you're looking for? (This is probably too late):

123, 2141, 31113, 51121

[ooo]

i dont get what he means either

i was thinking it meant

a+b+c+d+e = abcde

but then i gave the problem to my friend and they got

2x2=4, 2+2=4
0x2=0, 2+0=2

im like huh?

so im confused

[Evil Gemini]

Originally posted by johnq86@13 October 2003 - 22:14
Find five different positive numbers whose sum and product are equal to one another.
Cant find any :blink:

[DarthInsinuate]

Originally posted by johnq86@13 October 2003 - 21:14
Find five different positive numbers whose sum and product are equal to one another.
is this one of them rhetorical questions?

[ooo]

wait is it

a+b+c+d+e=abcde

is this what your asking? what is a,b,c,d,e?

[Alucard1475]

1*0=0
2*0=0
3*0=0
4*0=0
5*0=0

woila :unsure:

[ZaZu]

fiddlin' wit numbah's make mah teeth itch

[I.am]

Originally posted by Soul814@13 October 2003 - 18:40
wait is it

a+b+c+d+e=abcde

is this what your asking? what is a,b,c,d,e?
I think that is what he is asking.

ummm

2+2=2*2

5+1.25 = 5*1.25

[HeavyMetalParkingLot]

nm, i'm getting a beer and adding it to my belly, then i'm gonna subtract it from my bladder

[Billy_Dean]

The sum of the three consecutive integers 1, 2, 3 is equal to their product: 1 + 2 + 3 = 1 * 2 * 3 = 6. Find all other sets of three consecutive integers with the same property (i.e., that their sum is equal to their product).
- Are there any sets of four consecutive integers with the same property? How many sets of five consecutive integers have that property? Be sure to support your answers with explanations.

(a) Consider three consecutive integers x-1, x, and x+1 whose sum and product are equal. Then

3x = (x-1) + x + (x+1) = (x-1)(x)(x+1) = x3 - x .
If x = 0 then the three integers are -1, 0, and 1. Otherwise, we can divide by x and obtain

x2 - 1 = 3 .
Solving for x then gives

x2 = 4
x = + 2
Therefore, there are three triples of consecutive integers with equal sum and product. These are: -1, 0, 1 (corresponding to x = 0); 1, 2, 3 (corrsponding to x = 2); and -3, -2, -1 (corresponding to x = -2).

- There are no 4-tuples of consecutive integers whose sum an product are equal, and the only 5-tuple whose sum equals its product is -2, -1, 0, 1, 2.

To see this, first consider a 4-tuple x, x+1, x+2, x+3. Note that one of these four integers has to be a multiple of 4, so that their product will also be a multiple of 4. However, their sum is

x + (x+1) + (x+2) + (x+3) = 4x + 6
which will always have remainder 2 when divided by 4. Therefore no 4-tuple of consecutive integers can have its sum equal to its product.

Next, consider a 5-tuple of consecutive integers x, x+1, x+2, x+3, x+4. The sum of this 5-tuple is

x + (x+1) + (x+2) + (x+3) + (x+4) = 5x + 10 = 5 (x+2) .
If this sum equals the product of the 5-tuple, then we must have

(x)(x+1)(x+2)(x+3)(x+4) = 5 (x+2) .
If x = -2 then both sides of the equation equal zero and we obtain the 5-tuple -2, -1, 0, 1, 2 given above. Otherwise, we can divide both sides of this equation by x+2 and get

(x)(x+1)(x+3)(x+4) = 5 .
Now either x or x+1 is even, so the left hand side of this equation is an even number. Since 5 is odd, no value of x will satisfy this equation and we get no more 5-tuples with equal sum and product.


:blink:

[Barbarossa]

Originally posted by I.am+14 October 2003 - 06:21--></div><table border='0' align='center' width='95%' cellpadding='3' cellspacing='1'><tr><td>QUOTE (I.am @ 14 October 2003 - 06:21)</td></tr><tr><td id='QUOTE'> <!--QuoteBegin-Soul814@13 October 2003 - 18:40
wait is it

a+b+c+d+e=abcde

is this what your asking? what is a,b,c,d,e?
I think that is what he is asking.

ummm

2+2=2*2

5+1.25 = 5*1.25 [/b][/quote]
also:

3 + (1 and a half) = 3 * (1 and a half)
4 + (1 and a third) = 4 * (1 and a third)
5 + (1 and a quarter) = 5 * (1 and a quarter)
6 + (1 and a fifth) = 6 * (1 and a fifth)
7 + (1 and a sixth) = 7 * (1 and a sixth)
8 + (1 and a seventh) = 8 * (1 and a seventh)
9 + (1 and a eighth) = 9 * (1 and a eighth)

etc...

[VB]

x + y = x * y

y = x / (-1+x)


x = 0, y = 0
x = 2, y = 2
x = 3, y = 3/2
x = 4, y = 4/3
x = 5, y = 5/4
x = 6, y = 6/5

[Barbarossa]

OK, my brain hurts, but....

for ( a + b + c + d + e ) = a*b*c*d*e

then a possible solution is:-

a = 1

b = 2

c = 3

d = 6/5

e = 36/31