In this first example we start by adding 17 to the abacus adding 1 to 0 on rod A and 7 to 0 on rod B. Next we add 6 to 17 by adding 6 to 7 on rod B. Notice this addition will require using the 5/10 pair combination. First doing the 10 pair addition we add 1 to 1 on rod A and then subtract 4, the 10 pair complement of 6, from 7 on rod B. But since there are only 2 lower beads touching the bar we will need to subtract the 4 as a 5 pair. So subtract the 4 by adding 1, the 5 pair complement of 4, and subtract the 5 bead. The interim sum is now 23. Next add 2 to 23 by adding 2 to 3 on rod B using the 5 pair addition. Add 5 and subtract 2, the 5 pair complement of 3, on rod B for an interim sum of 25. Lastly subtract 8 from 25 by subtracting 8 from 5 on rod B using the 10 pair subtraction rule. First subtract 1 from 2 on rod A and then add 2, the 10 pair complement of 8, to 5 on rod B. The final answer is 17.