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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?

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  1. welshfella
    • one year ago
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    (x + y + z)^3 = 7624 y = 13 x + z = 27

  2. dtan5457
    • one year ago
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    i have that im trying to solve x^3+z^3=7455 x+y=27

  3. dtan5457
    • one year ago
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    *x+z=27

  4. welshfella
    • one year ago
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    u sure you have the question right? 7624 is not a perfect cube

  5. dtan5457
    • one year ago
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    i think they mean the sum of each cube

  6. mathstudent55
    • one year ago
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    @welshfella I think you need to do x^3 + y^3 + z^3 = 7624

  7. welshfella
    • one year ago
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    x^3 + y^3 + z^3 = 7624

  8. welshfella
    • one year ago
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    ok

  9. dtan5457
    • one year ago
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    yea

  10. anonymous
    • one year ago
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    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 $$(a+b+c)^3=a^3+b^3+c^3+3(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)+6abc\\40^3=7624+3(13a^2+169a+a^2c+ac^2+169c+13c^2)+78ac$$

  11. welshfella
    • one year ago
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    as a first step I would try x = 12 and z = 15 and x = 11 and z = 16

  12. welshfella
    • one year ago
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    trial and error

  13. dtan5457
    • one year ago
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    is it possible to solve algebraically?

  14. mathstudent55
    • one year ago
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    x^3 + 13^3 + z^3 = 7624 x^3 + 2197 + z^3 = 7624 x^3 + z^3 = 5427 x + z = 27

  15. welshfella
    • one year ago
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    weel you can generate an equation in x by plugging in x = 27 - z

  16. anonymous
    • one year ago
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    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 $$13a^2+169a+169c+13c^2+a^2c+ac^2+78ac\\\quad =169(a+c)+13(a^2+c^2)+ac(a+c+78)$$

  17. welshfella
    • one year ago
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    sorry z = 27 - x

  18. welshfella
    • one year ago
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    x^3 + (27-x)^3 = 7455

  19. anonymous
    • one year ago
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    oops, that should be $$3(13a^2+169a+169c+13c^2+a^2c+ac^2)+78ac$$ so $$3\cdot (169(a+c)+13(a^2+c^2)+ac(a+c+26))$$

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

  21. dtan5457
    • one year ago
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    Never mind guys, i solved this using some quadratic formula. thanks for your help though

  22. anonymous
    • one year ago
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    so $$\frac{64000-7624}3-169\cdot27-13\cdot729=27ac\\\implies ac=176$$

  23. dtan5457
    • one year ago
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    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

  24. mathstudent55
    • one year ago
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    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

  25. welshfella
    • one year ago
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    i8 tried that and got -27x^2 as first term - must have made a mistake somewhere

  26. welshfella
    • one year ago
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    yea i see where i went wrong

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

  28. anonymous
    • one year ago
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    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

  29. anonymous
    • one year ago
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    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 \(176\)?

  30. dtan5457
    • one year ago
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    you are right but i guess its just a little easier to go with things we learned before sometimes

  31. dtan5457
    • one year ago
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    even if it takes longer

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