To write 100 from a bunch of threes 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.
 A: $$ 3 3 \cdot 3 + 3 - 3 + 3 / 3 = 100. $$
$$3\cdot3\cdot3\cdot3 + 3\cdot3\cdot3 - 3\cdot3 + 3/3 = 100.$$
A: 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.
A: 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.
A: For another variation not using "adjoined" 3s:
$$3 \times 3 \times 3 \times 3 + 3 \times 3 + 3 \times 3 + 3 - 3 \div 3 - 3 \div 3 = 81 + 9 + 9 + 3 - 1 - 1 = 100$$
A: Is there a limit on the number of times you can use the operation?
Otherwise it seems you could do
$$3\times 33 + 3\div 3 +3\div 3 - 3\div 3$$
A: 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.
A: $$3333/33-33/3+3\cdot 3+3/3=\\101-11+9+1=100$$
If one is allowed to skip any of the 4 operations we can do
$$33\cdot 3 + 3/3=99+1=100$$
