Multiplying Three-Digit Number by Three-Digit Number

This is an extension of the previous section and the procedure is the same but requires a little more multiplication and bookkeeping. Let $n_1 = abc$ and $n_2 = def$, where a, b, c, d, e, f are digits. You’ll want to do the groups of $n_1$ as $a$ and $bc$ and $n_2$ as $d$ and $ef$ and perform the FOIL/LOIFing.

$abc = 100a + (bc)$ and $def = 100d + (ef)$

$[100a + (bc)] \times [100d + (ef)] = 10000ad + 100[a(ef) + d(bc)] + (bc)(ef)$

This shows us that:

1. Again, the ones and tens digit of the answer is simply the last two digits when performing the multiplication of the groups of bc and ef.
2. Again, carries are common, so keep track!