Anonymous
Anonymous asked in Science & MathematicsMathematics · 9 months ago

Using the numbers 0-9 only once, anyone able to assist with this? ((4-1)*(a-b)+c)*(d+e+f)=90?

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  • 9 months ago
    Favourite answer

    You've already used 0, 1, 4 and 9. So we have 2, 3, 5, 6, 7, 8 left to use.

    UPDATE: I originally grouped things incorrectly --> I had (4 - 1) * ((3 - 7) + 6) * (2 + 5 + 8) = 90. As Goners and TomV pointed out, when grouped like the question you get -90, not 90.

    Ultimately we have two integer expressions multiplying to be 90, so let's think about factoring 90:

    1 * 90

    2 * 45

    3 * 30

    5 * 18

    6 * 15

    9 * 10

    The smallest value for d+e+f is 10 --> 2 + 3 + 5

    The largest value for d+e+f is 18 --> 5 + 6 + 7 or 3 + 7 + 8

    The other possible value for d+e+f is 15 --> 2 + 5 + 8 or 3 + 5 + 7

    With a little trial and error using the remaining 3 digits and trying to make the other half of the product, I came up with 4 main answers:

    ((4 - 1) * (2 - 3) + 8) * (5 + 6 + 7) = 90

    ((4 - 1) * (6 - 5) + 2) * (3 + 7 + 8) = 90

    ((4 - 1) * (7 - 6) + 3) * (2 + 5 + 8) = 90

    ((4 - 1) * (8 - 7) + 6) * (2 + 3 + 5) = 90

    Note: You are still able to put d+e+f in any order so technically there are 24 different solutions.

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    • Goners9 months agoReport

      @Puzzling awesome work THANKS HEAPS :) all 4 listed equations come out with 90 as the result :D much appreciate the assistance

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  • 9 months ago

    ((4 - 1)*(a - b) + c)*(d + e + f) = 90

    ((4 - 1)*(9 - 6) + 0)*(2 + 3 + 5) = 90

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  • 9 months ago

    [3.(a - b) + c].(d + e + f) = 90

    [3.(2 - 1) + 3].(4 + 5 + 6) = 90

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