Question

Isaac has written down three integers. The sum of all 3 when cubed is 7624. The sum of the range of the numbers (smallest and largest) is 27. The middle integer is 13. What are the other integers?

Answer 1

(x + y + z)^3 = 7624
y = 13
x + z = 27

Answer 2

i have that im trying to solve

x^3+z^3=7455

x+y=27

Answer 3

*x+z=27

Answer 4

u sure you have the question right? 7624 is not a perfect cube

Answer 5

i think they mean the sum of each cube

Answer 6

@welshfella
I think you need to do
x^3 + y^3 + z^3 = 7624

Answer 7

x^3 + y^3 + z^3 = 7624

Answer 8

ok

Answer 9

yea

Answer 10

Isaac has written down three integers. The sum of all 3 when cubed is 7624. The sum of the range of the numbers (smallest and largest) is 27. The middle integer is 13. What are the other integers?
WLOG assume \(a\le b <c\)
$$a^3+b^3+c^3=7624\\a+c=27\\b=13\\\implies a+b+c=40$$

now observe $$(a+b+c)^3=a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc\\40^3=7624+3(13a^2+169a+a^2c+ac^2+169c+13c^2)+78ac$$

Answer 11

as a first step I would try x = 12 and z = 15
and x = 11 and z = 16

Answer 12

trial and error

Answer 13

is it possible to solve algebraically?

Answer 14

x^3 + 13^3 + z^3 = 7624
x^3 + 2197 + z^3 = 7624
x^3 + z^3 = 5427
x + z = 27

Answer 15

weel you can generate an equation in x by plugging in x = 27 - z

Answer 16

now since \(a+c=27\) we get \((a+c)^2=a^2+c^2+2ac\implies a^2+c^2=27^2-2ac\) so consider now $$13a^2+169a+169c+13c^2+a^2c+ac^2+78ac\\\quad =169(a+c)+13(a^2+c^2)+ac(a+c+78)$$

Answer 17

sorry z = 27 - x

Answer 18

x^3 + (27-x)^3 = 7455

Answer 19

oops, that should be $$3(13a^2+169a+169c+13c^2+a^2c+ac^2)+78ac$$ so $$3\cdot (169(a+c)+13(a^2+c^2)+ac(a+c+26))$$

Answer 20

we have that \(a+c=27\implies a+c+26=53\) and further that \(a^2+c^2=729-2ac\) so $$64000=7624+3\cdot(169\cdot27+13\cdot(729-2ac)+ac\cdot53)$$

Answer 21

Never mind guys, i solved this using some quadratic formula. thanks for your help though

Answer 22

so $$\frac{64000-7624}3-169\cdot27-13\cdot729=27ac\\\implies ac=176$$

Answer 23

I just did x^3+13^3+z^3=7624

x^3+z^3=5427

x+z=27

x=(-z+27)

(-z+27)^3+z^3=5427

-z^3+81z^2-2187z+27^3+z^3=5427

81z^2-2187z+14256=0

^^quadratic formula and you get the other 2 integers

Answer 24

x = 27 - z
x^3 = (27 - z)^3
x^3 = -z^3 + 81z^2 - 2187z + 19683

-z^3 + 81z^2 - 2187z + 19683 + z^3 = 5427
81z^2 - 2187z + 19683 = 5427
81z^2 - 2187z + 14256 = 0

z^2 - 27z + 176 = 0
(z - 11)(z - 16) = 0
z = 11 or z = 16

Answer 25

i8 tried that and got -27x^2 as first term - must have made a mistake somewhere

Answer 26

yea i see where i went wrong

Answer 27

now we know $$ac=176\\a+c=27$$ so$$a(27-a)=176\\27a-a^2=176\\a^2-27a+176=0\\a=\frac{27\pm\sqrt{729-2\cdot176}}2\\\implies a=11\\\implies c=16$$

Answer 28

but yeah, you can plug in for \(b\) at the start and get $$a^3+c^3=\dots\\a^3+c^3=(a+c)^3-3ac(a+c)$$ and get \(ac\) much faster

Answer 29

thing is, once you have \(a+c,ac\), writing out the quadratic and factoring is unnecessary because you ask the same question -- what two numbers add up to \(27\) but multiply to give \(176\)?

Answer 30

you are right but i guess its just a little easier to go with things we learned before sometimes

Answer 31

even if it takes longer