# Converting between base 7 to 3

1. Oct 7, 2009

### tnutty

1. The problem statement, all variables and given/known data

convert (352416)_7 to (...)_3

the _7 means base 7 and similarly, _3 means base 3.

Idea : convert base 7 to base 10 then base 10 to base 3.

I know I can use expansion, 3 * 7^6 + 5 * 7^6 ... to convert it to base 10.

but in my exam, we are not allowed to use calculators, and I really not want to spend time
on figuring out the power of X. So do you know a faster way to convert this to base 10?
From then on I could convert it to base 3.

2. Oct 7, 2009

### Staff: Mentor

I can't think of any way other than brute force, which is what you're doing. BTW, it would be 3*75 + 5*74 + ... + 1*71 + 6

You could make a couple of shortcuts, though. 74 = 492 = (50 - 1)2 = 502 - 2*50 + 1 = 2500 - 100 + 1 = 2401.

3. Oct 7, 2009

### tnutty

Thanks for the tip.

4. Oct 7, 2009

### tnutty

Also when say the number turned out to be something like 443267. Then to change this
number of base 10 to base 3, I could divide by 3 and get the remainder. But is there another way that you can think of.

5. Oct 7, 2009

### tnutty

And could I have just divided the base 7 number by 3 to turn it into base 3?

6. Oct 7, 2009

### Staff: Mentor

Once you have converted to a base-10 number, subtract the highest power of 3 that is <= your number. It's probably easier to just show an example than to explain the process.

Suppose you want to convert 4325 to its base-4 equivalent. 4325 = 11710.

The powers of 4 are 1, 4, 16, 64, 256, and so on.

117 = 64 + 53, so my base-4 representation is going to be 1xxx.
53 = 3*16 + 5, so now I have 13xx
5 = 4 + 1, so I have 131x.
My final remainder is 1, so the number we want is 13114.

It might be possible to convert directly from base 7 to base 3, but that requires you to know arithmetic in both bases; e.g. to be able to know that 6*4 = 21 in base 7, that sort of thing.

7. Oct 7, 2009

### Donaldos

Yes, it's pretty simple in this case but be sure to express everything in base 7:

$$(352416)_7=3 \times (115352)_7+0$$

$$(115352)_7=3 \times (26340)_7 +2$$

$$(26340)_7=3 \times (6560)_7 +0$$

$$(6560)_7=3 \times (2164)_7 +2$$

$$(2164)_7=3 \times (521)_7 +1$$

$$(521)_7=3 \times (152)_7 +2$$

$$(152)_7=3 \times (40)_7 +2$$

$$(40)_7=3 \times (12)_7 +1$$

$$(12)_7=3 \times (3)_7 +0$$

$$(3)_7=3 \times (1)_7 +0$$

$$\Rightarrow (352416)_7=(10012212020)_3$$

8. Oct 7, 2009

### Donaldos

I think that $$6\times 4=33$$ in that base.

9. Oct 7, 2009

### tnutty

Yea, that what I do with power of 2's.

Thanks everyone.

10. Oct 7, 2009

My error.