# Four 4's

1. Apr 27, 2004

### Joseph

I need to make 35,37,39,and 41 using four 4's.

2. Apr 27, 2004

If the only operations you can perform on the fours is addition, subtraction, multiplication, and division, then you can't get any of them.

You'll always get an even number, regardless of the operations you use or the order you use them in.

3. Apr 27, 2004

### Bob3141592

No, you can use two of the fours to produce one, as in 4 + 4 + 4/4 = 9. Or 4*4 + 4/4 = 17. Subtracting the quotient gives 7 and 15 respectively.

Offhand, given the four function constraint, I don't see a way to get a higher odd number than 17.

Can we use the four fours as strings instead of individual digits? 444/4 = 111, an odd number. 44/4 + 4 = 15. (44/4)^4 = 14,641, a very big odd number.

On the other hand, can we use factorials? 4! is 24, so 4! + 4! - 4/4 = 47. Drats, it's not on the target list. (4 * 4! - 4) / 4 is 23. Still no go.

Sorry, but I don't see a trick to get the desired numbers.

Last edited: Apr 27, 2004
4. Apr 28, 2004

### arildno

This is to cheat, but: ((4*4)4)/4=164/4=41

5. Apr 28, 2004

### JasonRox

Aren't all numbers primes or products of primes?

6. Apr 28, 2004

### JasonRox

I came pretty close for one of them.

$$(\frac{4^2}{ \sqrt{1/4}}) - 4^0 = 31$$

Can you see what I did?

Here's another one.

$$(\frac{4}{ \sqrt{1/4}}) 4 - 4^0 = 31$$

$$(\frac{4^2}{ \sqrt{1/4}}) + 4^0 = 33$$

Last edited: Apr 28, 2004
7. Apr 28, 2004

### JasonRox

Another one.

$$44 - 4 - 4^0 = 39$$

8. Apr 28, 2004

### JasonRox

I got to go, so I'll do the rest when I get back.

Use my tricks and you are good to go.

9. Apr 28, 2004

### JasonRox

$$44 + 4^0 - 4 = 41$$

Got back from lunch. Time to study boring accounting.

10. Apr 28, 2004

Aha! I forgot to consider 4/4 = 1.

Good point.

11. Apr 28, 2004

### Bob3141592

I'd presumed that using other numerals like 2 and 0 for exponents wouldn't be allowed. At least that's the case in problems of this sort I've seen before. Non-numeric mathematic symbols like square root are OK, but a cube root isn't be, since you have to use a numeral to specify it. If you can use a 4^2 to mean squared, why cant you say 4*5 or anything else?

Last edited: Apr 28, 2004
12. Apr 28, 2004

### JasonRox

I put squared to make it easier. If you count again, you will see there is four.

Remember all numbers have exponent 1's.

If I'm allowed square roots, I did nothing wrong. Square root is the same as exponent 1/2.

I didn't break any rules if you allow 1/2.

I believe that you are restricted to the number one. If you add 1/2 and 1/2 you get 1, and since you allowed 1/2, I can manipulate it to do other things. Here is how it works:

-1/2-1/2=-1 (I used 1/4 instead of 4^-1, because it didn't work for some reason)
1/2-1/2=0
1/2+1/2=1 (The regular exponent.)

See I broke no rules. :)

13. Apr 28, 2004

### JasonRox

More...

$$4! + 4^2 + 4^0 = 41$$

There is 4 Four's.

14. Apr 28, 2004

### JasonRox

Let's here the opposite of Fermat's Last Theorem!

$$x^n + y^n = z^n$$, is possible in infinite amounts for n<2.

15. Apr 28, 2004

### Bob3141592

Was this a class assignment? If so, let me know if changing a four to a one by raising it to the zeroth power is acceptable. I'd be surprised, but I'd like to know.

Also, earlier you wrote:

All natural numbers are, that is, positive integers. Obviously negative numbers and fractional numbers aren't.

Further, each natural number is either prime or the product of primes in a unique way. There is one and only one such representation for each. That's the fundamental theorem of arithmetic.