• Mar 20th 2009, 12:51 AM
jzellt
Here is an example:
108 + 17 =

((1x
102 + 0x101) + 8x100) + (1x10 + 7x100)

= (10
2 + 8) + (10 + 7) = (102 + 10) + (8 + 7)
= (10
2 + 10) + 15 = (102 + 1x10) + (1x10 + 5)
= (10
2 + (1x10 + 1x10)) + 5 = (102 + (1 + 1)x101) + 5

= (1x
102 + 2x101) + 5x100 = 125

Now, I'm asked to show 1094 + 329 in a similiar manner, but I can't figure it out...Any thoughts?

I'm also asked to show 17 x 15 the same way...Any advice?

THanks!
• Mar 20th 2009, 01:07 AM
Prove It
Quote:

Originally Posted by jzellt
Here is an example:

108 + 17 =

((1x
102 + 0x101) + 8x100) + (1x10 + 7x100)

= (10

2 + 8) + (10 + 7) = (102 + 10) + (8 + 7)
= (10

2 + 10) + 15 = (102 + 1x10) + (1x10 + 5)
= (10

[LEFT]2 + (1x10 + 1x10)) + 5 = (102 + (1 + 1)x101) + 5
= (1x
102 + 2x101) + 5x100 = 125

Now, I'm asked to show 1094 + 329 in a similiar manner, but I can't figure it out...Any thoughts?

I'm also asked to show 17 x 15 the same way...Any advice?

THanks!

Why don't you just add the numbers? Surely it's easier than doing it in binary...

But if you MUST use binary...

I'm going to number the digits of the binary number going right to left.

The biggest power of 2 closest to 108 is \$\displaystyle 2^6 = 64\$. So there'll be 6 + 1 = 7 digits and the 7th column (closest to the left) will contain a 1.

\$\displaystyle 108 - 64 = 42\$.

The biggest power of 2 closest to 42 is \$\displaystyle 2^5 = 32\$.

So there'll be a 1 in the 5 + 1 = 6th column.

\$\displaystyle 42 - 32 = 10\$.

The biggest power of 2 closest to 10 is \$\displaystyle 2^3 = 8\$. So there'll be a 1 in the 3 + 1 = 4rd column (the 5th will have a 0).

\$\displaystyle 10 - 8 = 2\$

The biggest power of 2 closest to 2 is \$\displaystyle 2^1 = 2\$. So there'll be a 1 in the 1 + 1 = 2nd column (the 3rd will have a 0).

\$\displaystyle 2 - 2 = 0\$. We don't need to go any further.

So \$\displaystyle 108_{10} = 1101010_2\$.

Following the same procedure

The biggest power of 2 closest to 17 is \$\displaystyle 2^4 = 16\$. So there'll be 4 + 1 = 5 columns and a 1 in the 5th column.

\$\displaystyle 17 - 16 = 1\$.

The biggest power of 2 closest to 1 is \$\displaystyle 2^0 = 1\$. So there'll be a 1 in the 0 + 1 = 1st column. (Columns 2 - 4 will have 0's).

So \$\displaystyle 17_{10} = 10001_2\$.

Therefore

\$\displaystyle 108_{10} + 17_{10} = 1101010_2 + 10001_2\$

\$\displaystyle = 1111011_2\$

\$\displaystyle = (2^6 + 2^5 + 2^4 + 2^3 + 2^1 + 2^0)_{10}\$

\$\displaystyle = (64 + 32 + 16 + 8 + 2 + 1)_{10}\$

\$\displaystyle = 125_{10}\$.

Easy huh? Not...
• Mar 20th 2009, 01:29 AM
jzellt
It's not suposed to be binary.

Sorry...I just noticed now, but 108 = (1 x 10^2) + (0 x 10^1) + (8 x 10^0)

For some reason, in my first post, it looks like 10 sub 2 and thats not correct.

I am in a number theory class and supposed to pretend I don't know how to add large numbers the normal way. I am supposed to expand them and add in base 10 like my original example, but I can't figure it out.

Thanks.