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$pOnGeBoB : )
Lv 4
$pOnGeBoB : ) asked in Science & MathematicsMathematics · 1 decade ago

math riddle - hard (again with different answers)?

only using the numbers 1953

make a equation to get the answers of any of the following

7

9

20

30

35

37

40

choose one you like to do or try all

you can use -, +, / x and parenthesis, square root, absolute value, exponents, ect

you can also put the number together ex: 53-19=

2 Answers

Relevance
  • Omega_lx
    1 decade ago
    Favourite answer

    7 = 9^1 - 5 - 3

    9 = 1^(5-3) * 9

    20 = (9 + 1) * (5 - 3)

    30 = (9 - 3) * 5 / 1

    35 = (9 + 1 - 3) * 5

    37 = (9 - 1) * 5 - 3

    40 = (5 + 3)^(1/3) * (9 + 1)

  • Dave Castle
    1 decade ago

    So like, 1x9-(5-3)=7 ... is that what you're going for?

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