# Make numbers from 1-10 with 5 3s

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?

Complete answer(thanks to Ross Millikan and BAWS):

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

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

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

• Hi, on this site, for answers, you need to hide them in spoiler blocks(>!) most of the times. Thank you. Dec 13, 2020 at 5:37

Here are $$1,2$$ and $$8$$

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

We obtain the missing numbers from the following.

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

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

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

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