dtan5457
  • dtan5457
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?
Mathematics
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SOLVED
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schrodinger
  • schrodinger
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welshfella
  • welshfella
(x + y + z)^3 = 7624 y = 13 x + z = 27
dtan5457
  • dtan5457
i have that im trying to solve x^3+z^3=7455 x+y=27
dtan5457
  • dtan5457
*x+z=27

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More answers

welshfella
  • welshfella
u sure you have the question right? 7624 is not a perfect cube
dtan5457
  • dtan5457
i think they mean the sum of each cube
mathstudent55
  • mathstudent55
@welshfella I think you need to do x^3 + y^3 + z^3 = 7624
welshfella
  • welshfella
x^3 + y^3 + z^3 = 7624
welshfella
  • welshfella
ok
dtan5457
  • dtan5457
yea
anonymous
  • anonymous
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
welshfella
  • welshfella
as a first step I would try x = 12 and z = 15 and x = 11 and z = 16
welshfella
  • welshfella
trial and error
dtan5457
  • dtan5457
is it possible to solve algebraically?
mathstudent55
  • mathstudent55
x^3 + 13^3 + z^3 = 7624 x^3 + 2197 + z^3 = 7624 x^3 + z^3 = 5427 x + z = 27
welshfella
  • welshfella
weel you can generate an equation in x by plugging in x = 27 - z
anonymous
  • anonymous
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 $$13a^2+169a+169c+13c^2+a^2c+ac^2+78ac\\\quad =169(a+c)+13(a^2+c^2)+ac(a+c+78)$$
welshfella
  • welshfella
sorry z = 27 - x
welshfella
  • welshfella
x^3 + (27-x)^3 = 7455
anonymous
  • anonymous
oops, that should be $$3(13a^2+169a+169c+13c^2+a^2c+ac^2)+78ac$$ so $$3\cdot (169(a+c)+13(a^2+c^2)+ac(a+c+26))$$
anonymous
  • anonymous
we have that \(a+c=27\implies a+c+26=53\) and further that \(a^2+c^2=729-2ac\) so $$64000=7624+3\cdot(169\cdot27+13\cdot(729-2ac)+ac\cdot53)$$
dtan5457
  • dtan5457
Never mind guys, i solved this using some quadratic formula. thanks for your help though
anonymous
  • anonymous
so $$\frac{64000-7624}3-169\cdot27-13\cdot729=27ac\\\implies ac=176$$
dtan5457
  • dtan5457
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
mathstudent55
  • mathstudent55
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
welshfella
  • welshfella
i8 tried that and got -27x^2 as first term - must have made a mistake somewhere
welshfella
  • welshfella
yea i see where i went wrong
anonymous
  • anonymous
now we know $$ac=176\\a+c=27$$ so$$a(27-a)=176\\27a-a^2=176\\a^2-27a+176=0\\a=\frac{27\pm\sqrt{729-2\cdot176}}2\\\implies a=11\\\implies c=16$$
anonymous
  • anonymous
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
anonymous
  • anonymous
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 \(176\)?
dtan5457
  • dtan5457
you are right but i guess its just a little easier to go with things we learned before sometimes
dtan5457
  • dtan5457
even if it takes longer

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