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dtan5457
 one year ago
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?
dtan5457
 one year ago
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?

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welshfella
 one year ago
Best ResponseYou've already chosen the best response.1(x + y + z)^3 = 7624 y = 13 x + z = 27

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.2i have that im trying to solve x^3+z^3=7455 x+y=27

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1u sure you have the question right? 7624 is not a perfect cube

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.2i think they mean the sum of each cube

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0@welshfella I think you need to do x^3 + y^3 + z^3 = 7624

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1x^3 + y^3 + z^3 = 7624

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Isaac 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 $$(a+b+c)^3=a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc\\40^3=7624+3(13a^2+169a+a^2c+ac^2+169c+13c^2)+78ac$$

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1as a first step I would try x = 12 and z = 15 and x = 11 and z = 16

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.2is it possible to solve algebraically?

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0x^3 + 13^3 + z^3 = 7624 x^3 + 2197 + z^3 = 7624 x^3 + z^3 = 5427 x + z = 27

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1weel you can generate an equation in x by plugging in x = 27  z

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now since \(a+c=27\) we get \((a+c)^2=a^2+c^2+2ac\implies a^2+c^2=27^22ac\) 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
 one year ago
Best ResponseYou've already chosen the best response.1x^3 + (27x)^3 = 7455

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, 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
 one year ago
Best ResponseYou've already chosen the best response.0we have that \(a+c=27\implies a+c+26=53\) and further that \(a^2+c^2=7292ac\) so $$64000=7624+3\cdot(169\cdot27+13\cdot(7292ac)+ac\cdot53)$$

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.2Never mind guys, i solved this using some quadratic formula. thanks for your help though

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so $$\frac{640007624}3169\cdot2713\cdot729=27ac\\\implies ac=176$$

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.2I 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^22187z+27^3+z^3=5427 81z^22187z+14256=0 ^^quadratic formula and you get the other 2 integers

mathstudent55
 one year ago
Best ResponseYou've already chosen the best response.0x = 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
 one year ago
Best ResponseYou've already chosen the best response.1i8 tried that and got 27x^2 as first term  must have made a mistake somewhere

welshfella
 one year ago
Best ResponseYou've already chosen the best response.1yea i see where i went wrong

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now we know $$ac=176\\a+c=27$$ so$$a(27a)=176\\27aa^2=176\\a^227a+176=0\\a=\frac{27\pm\sqrt{7292\cdot176}}2\\\implies a=11\\\implies c=16$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but yeah, you can plug in for \(b\) at the start and get $$a^3+c^3=\dots\\a^3+c^3=(a+c)^33ac(a+c)$$ and get \(ac\) much faster

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0thing 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
 one year ago
Best ResponseYou've already chosen the best response.2you are right but i guess its just a little easier to go with things we learned before sometimes

dtan5457
 one year ago
Best ResponseYou've already chosen the best response.2even if it takes longer
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