To write 100 from a bunch of threes

Question

My 6th grader got an exercise to write an expression that equals 100 using

• only the digit 3
• no parentheses
• all four operations

We have worked on this for over 30 minutes. How should he even tackle this? We have tried many ways that come close but I may be overthinking. We have tried exponents as well.

Answer 1

Well, clearly $(333-33)\div 3 = 100$.

Now, you just have to write something equivalent that does not use parenthesis. Then figure a way to use addition.

Answer 2

You can note that $\frac 33=1$, so you can add or subtract any number you want by adding or subtracting $\frac 33$. We probably want to make things bigger, so choose addition. Now we just need to use up multiplication and subtraction. So $3 \times 3 -3=6$ and add $94$ terms of $\frac 33$ to be done.

Answer 3

Is there a limit on the number of times you can use the operation?

Otherwise it seems you could do

$$3\times 33 + 3\div 3 +3\div 3 - 3\div 3$$

Answer 4

For another variation not using "adjoined" 3s:

$$3 \times 3 \times 3 \times 3 + 3 \times 3 + 3 \times 3 + 3 - 3 \div 3 - 3 \div 3 = 81 + 9 + 9 + 3 - 1 - 1 = 100$$

Answer 5

If we can use addition and division more than one times then it can be solved easily.

By using division three times and addition two times we can write expression,

$$3÷3+3÷3+33×3-3÷3$$ Now using order of operation(PEMDAS) solve expression.

$$1+1+99-1$$ $$=100$$ It can also solved by various types, like

$$33×3+\dfrac{3+3}{3}-3÷3$$ $$99+\dfrac{3+3}{3}-1$$ $$99+2-1$$ $$=100$$ I hope it will help you.

Answer 6

$$ 3 3 \cdot 3 + 3 - 3 + 3 / 3 = 100. $$

$$3\cdot3\cdot3\cdot3 + 3\cdot3\cdot3 - 3\cdot3 + 3/3 = 100.$$

Answer 7

$$3333/33-33/3+3\cdot 3+3/3=\\101-11+9+1=100$$

If one is allowed to skip any of the 4 operations we can do

$$33\cdot 3 + 3/3=99+1=100$$