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

[Natasha1]

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?

 

[phyzguy]

3 + 3 + 3/3 = 7

 

[jedishrfu]

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

...

 

[Natasha1]

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

 

[Natasha1]

Are you saying it's therefore impossible?

 

[jedishrfu]

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.

 

[Natasha1]

Pity.

 

[jedishrfu]

Pity.

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

 

[maltman]

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

 

[Natasha1]

But then you are using + twice too

 

[rcgldr]

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.