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Using fractions and fractional numbers of the digits $1$ to $10$ and only the mathematical operations of addition, subtraction, multiplication, and division, including parenthesis (e.g. $1 + 2 - (3 * 4)$), can you write $11$ equations to get the numbers $0$ to $10$ (one number per equation)?

Rules
  • Each equation must use ALL the digits from $1$ to $10$ but ONLY ONCE.
  • Fractions and fractional numbers cannot result in integers when evaluated individually. For example, in the case of this puzzle, $\frac{3}{9}$ is considered a valid fraction but $\frac{9}{3}$ is not since it evaluates to $3$. A valid fractional number in the context of this puzzle contains a fraction like $\frac{8}{3}$ or $\frac{9}{7}$ etc.
  • Each equation as written must show the fractions or fractional numbers.
  • No concatenation.

Thus the LHS of each equation will show five fractions or fractional numbers. RHS would be from $0$ to $10$ ― so eleven equations. All $10$ digits must be used but only once.


No partial answers please and no programming.

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  • 1
    $\begingroup$ "Digits 1 to 10" means "numbers 1 to 10", right? $\endgroup$
    – Gareth McCaughan
    Oct 26 '21 at 13:07
  • $\begingroup$ That is correct. $\endgroup$
    – DrD
    Oct 26 '21 at 13:07
  • $\begingroup$ You say the LHS will contain 5 fractions, so does that mean we cannot use for example 1/(2+3) or (4+5)/6 ? $\endgroup$ Oct 26 '21 at 13:10
  • $\begingroup$ Correct @JaapScherphuis. Five seperate fractions using all 1 to 10 only once. $\endgroup$
    – DrD
    Oct 26 '21 at 13:12
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    $\begingroup$ Interesting; I wonder how high one still can get all (much higher than 10). $\endgroup$
    – Retudin
    Oct 26 '21 at 21:25
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The technique I found helpful was

to get groups of 3 and 2 fractions yielding small integers, and then combine those. For 2, we need things broadly like odd/2 + even/4. For 3, we need things broadly like .../2 + .../3 + .../6.

First of all, notice that

9/2 + 4/3 + 7/6 = 7 ... (1)
(9/2 + 4/3) / (7/6) = 5 ... (1')
1/5 + 8/10 = 1 ... (2)
8/10 / (1/5) = 4 ... (2')

From these

we can get 5-4=1, 7-4=3, 5-1=4, 7-1=5, 5+1=6, 7/1=7, 7+1=8, 5+4=9. So we still need 0, 2, 10.

Now tweak (1) and (2) by swapping two numerators and fixing up:

9/2 + 1/3 + 7/6 = 6 ... (3)
4/5 - 8/10 = 0 ... (4)

which

by multiplying gets us 0. Now we need 2, 10.

A similar switcheroo gets us

1/2 + 4/3 + 7/6 = 3 ... (5)
9/5 - 8/10 = 1 ... (6)

from which

we get 2 by subtraction. So now we just need 10.

This one we get in a slightly different way. (Maybe there's an easier way, but this is the first thing I found.)

(9/2 + 7/3) x (6/4) = 41/4 = 10+1/4 ... (7)
(10/8) x (1/5) = 1/4 ... (8)
so subtracting these yields 10

and we're done.

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  • $\begingroup$ nice approach in little pieces. $\endgroup$
    – ThomasL
    Oct 28 '21 at 18:24

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