Arithmetic with big numbers (Part 2)

Ready for an elementary arithmetic problem? Here it is:

bigmult1

Nothing to it… just multiply the two numbers. Of course, we’d rather not multiply them by hand, so let’s use a calculator instead:

bigmult2

Uh oh… the calculator doesn’t give the complete answer. It does return the first nine significant digits, but it doesn’t return all 16 digits. Indeed, we can’t be sure that the final 5 in the answer is correct because of rounding.

So now what we do (other than buy a more expensive calculator)?

In yesterday’s post, I posed a similar problem involving addition. Adding two big numbers by hand is no big deal. However, multiplying two big numbers, one digit at time, would be tedious!

When I pose this question to students, the knee-jerk reaction is to groan when facing the prospect of multiplying these two big numbers by hand. However, it is possible to use modern technology to make ordinary grade-school multiplication move a lot quicker. Perhaps the fastest way to do this is to split the numbers into block of five digits instead of the usual three:bigmult3

Now we proceed as if each block of five digits was a single digit. We begin with the last block of digits on the second row, which is 48974. First, we multiply 6797 and 48974 using a calculator. Because most modern scientific calculators have a 10-digit display, we can be assured that the complete answer will be shown. (This is why I chose to divide the numbers using block of five digits and not six or more.) The last five digits in the answer are written down; the more significant digits are carried.

Next, we multiply 2236 and 48974 and then add the number that was carried.

bigmult4We then repeat using 2449, the next (and final) block of digits on the second row. First, we multiply 6797 and 2449 using a calculator. The last five digits in the answer are written down; the more significant digits are carried.

Next, we multiply 2236 and 2449 and then add the number that was carried.

bigmult5

Finally, it remains to add these two partial products to obtain the final product. For this problem, this can be accomplished with only a single addition: the block of digits 76278 simply carry down to the final answer, and so we can start by adding the second and third blocks of digits. As this sum is less than 10^{10}, there is no digit to carry, and so the leading 54 also carries down to the final answer.

bigmult6The above technique is logically equivalent to using base 100,000 as opposed to the customary use of base 10 arithmetic. So while multiplying two numbers in the billions still takes some time, judiciously using a calculator makes this exercise go a lot quicker than the ordinary grade-school method of multiplying one digit at a time.

 

 

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