Converting between base 7 to 3

[tnutty]

Homework Statement

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.

 

[Mark44]

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.

 

[tnutty]

Thanks for the tip.

 

[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.

 

[tnutty]

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

 

[Mark44]

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.

 

[Donaldos]

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

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

 

[Donaldos]

e.g. to be able to know that 6*4 = 21 in base 7, that sort of thing.

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

 

[tnutty]

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.

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

Thanks everyone.

 

[Mark44]

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

My error.