# Using only the numbers 3, 3, 3 and 3 once and + - * / once find 7?

Hi,

At the end of our lecture today, the lecturer gave us this simple yet impossible puzzle.

My friend and I have tried to find the answer but in vain...

Using only the numbers 3, 3, 3 and 3 once and using only the four arithmetic + - * / once can you make the number 7.

The closest I have got is 6 or 8 but not 7.

3*3 all / 3 and then + 3 = 6
or
3*3 then - 3/3 = 8

How to find 7?? Is this actually possible?

## Answers and Replies

phyzguy
Science Advisor
3 + 3 + 3/3 = 7

jedishrfu
Mentor
think of it in terms of combinatorics with the operators:

3 op1 3 op2 3 op3 3

so there's 4 choices for the first, 3 for the third, 2 for the last one = 24 choices

Phyzguy's solution is almost correct except that he repeated the + operator and the problem says to use each operation once.

3+3-3*3=-3
3+3-3/3=5
3+3*3-3=9
3+3*3/3=6
3+3/3-3=1
3+3/3*3=3.33333

3-3+3*3=9
3-3+3/3=1
3-3*3-3=-9
3-3*3/3=0
3-3/3+3=5
3-3/3*3

...

Can't do 3 + 3 + 3/3 = 7 as you are using + twice

Are you saying it's therefore impossible?

jedishrfu
Mentor
Are you saying it's therefore impossible?

I can't tell you the answer only how to think about the problem as it was assigned by your prof.

Pity.

jedishrfu
Mentor
Pity.

You can't finish the other 12 choices to complete the proof?

are you sure the prof or you have the correct problem?
a classic is using 5 "3" and all the operators

(3*3 + 3)/3 + 3 = 7

But then you are using + twice too

rcgldr
Homework Helper
3 and 7 are prime numbers. Any combination of operations you try (except for 3/3) will be a multiple of 3. If you use 3/3 = 1, then you'd have to add or subtract the 3/3 to something else, since multiplying would result in a multiple of 3 again, and dividing would result in a fraction.

The problem doesn't state if you're allowed to use parenthesis to group operations.

If you're suppose to used + - * / excactly one, that's four operators, so you'd need five 3's.