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In this puzzle book, I came across a question that goes like this:

Make the numbers from 1-10 using 5 3s.

Rules:

You are only allowed to use the four basic arithmetic operators ($+, -, \times, \div$). You are allowed to use brackets.

Here is what I have so far:

$3+3-3+3-3=3$
$3+3+3+3-3=9$
$(3\times3)+(3\div3)-3=7$
$3\div3+3+3+3=10$
$3+3\div3-3+3=4$

Can someone help me figure out the rest?

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Complete answer(thanks to Ross Millikan and BAWS):

$3-3\div3-3\div3=1$
$3-3\div3\times3\div3=2$
$3+3-3+3-3=3$
$3+3\div3-3+3=4$
$((3\times3)+(3+3))\div3=5$
$(3\times3)-(3-3)-3 = 6$
$(3\times3)+(3\div3)-3=7$
$3+3+((3+3)\div3)=8$
$3+3+3+3-3=9$
$3\div3+3+3+3=10$

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$((3\times3)+(3+3))\div3 = 5$
$(3\times3)-(3-3)-3 = 6$

I'll leave the rest with this new way of looking at it.

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  • $\begingroup$ Hi, on this site, for answers, you need to hide them in spoiler blocks(>!) most of the times. Thank you. $\endgroup$
    – Pilot
    Dec 13 '20 at 5:37
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Here are $1,2$ and $8$

$$3-\frac 33 - \frac 33 =1$$
$$3-\frac 33 \cdot \frac 33=2 $$
$$3+3+\frac{3+3}3=8$$

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We obtain the missing numbers from the following.

$(3-3)\times3+\frac{3}{3}=1$

$3-3+\frac{3+3}{3}=2$

$3+\frac{3}{3}+\frac{3}{3}=5$

$3+3+3\times (3-3)=6$

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