From the Days of Analogue Arithmetic

In the days before the use of pocket calculators became universal, students memorized multiplication tables to 12 x 12 -- while I was memorizing, I even thought I might as well memorize up to the 16 x 16 table. And students progressed to slide rules, with names like Mannheim, for the simplest, to log-log decitrig for those with a bewildering array of scales. When my desk is cleaned out after I go, will anyone know what purpose my slip sticks served? Those columns of numbers we had to total on paper with pencils would benefit from a method for checking the result. So I learned the method of checking by dropping 9's. It is still useful today for checking bank deposit slips, say, when the pocket calculator is not at hand -- or has a dead battery! In those days the prudent physicist with a lot of calculating to do, say, for optical lens design, would do well to marry someone who was adept at operating a Marchand mechanical calculator!

Checking Arithmetic by Dropping 9's

Suppose one had an addition problem like:

$\begin{array}{r}
321\\
4567\\
\underline{+44622}\\
49510
\end{array}$

The method of checking the sum by dropping 9's proceeds as follows. One replaces each factor in the sum with the number that results from adding the numbers in each factor, dropping 9's during the process. In the present case the first factor, 321, is processed, starting from the left, by adding 3 to 2 to get 5. Then 5 is added to the remaining digit, 1, to get 6. This result is spoken of by saying 321 = 6 mod(9).

You take the next factor, 4567, and, starting from the left, add the 4 to the 5 to get 9. Now drop the 9 and continue with the result, 0. Now 0 added to the next digit, 6, gives 6, and it added to 7 gives 13. Again drop the 9 from 13 and we have the result, 4. As above we say 4567 = 4 mod(9).

Now continue with the third factor, 44622. Starting from the left the first digit, 4, is added to the second digit, giving 8. This 8 is added to the third digit, 6, giving 14. Now drop the 9 and continue with the remainder, 5. This 5 is now added to the next digit, 2, giving 7. And this 7 is added to the last digit, 2, giving 9. Now drop the 9 from this 9 and the result for the third term is written, 44622 = 0 mod(9).

Now a measure of the sum of these three terms is the sum of their mod(9) values: 321 + 4567 + 44622 = 6 mod(9) + 4 mod(9) + 0 mod(9) = 10, or 1 mod(9).

The long hand addition is then checked by comparing its mod(9) value for the terms in the sum and the result one gets by dropping the 9's from the answer, 49510. In this case the sum does check, for, starting from the left digit of the sum I have 4. The next 9 I drop -- after adding to 4 and then dropping the 9 from the sum, I have my 4 again -- and so I add the 4 to the third digit, 5, to get 9, which I drop. Continuing with 0, I add it to add to the next digit, 1, I get 1 to that point. Adding 1 to the last digit, 0, leaves the sum as 1. This says that 49510 = 1 mod(9), which checks with the result from dropping 9's from the factors that produced the sum. So the sum I did long hand is correct!

Why does this work? Let me see if I can make it plausible. If I understand what the first factor 321 means, I see I could write it as 3X100 + 2X10 + 1. Or 3X(9+1)(9+1) + 2X(9+1) + 1. The first factor can be expanded to 3X(9X9 + 2X9 + 1). So dropping 9's in the first factor leaves only the 3X1 term or just the digit 3 itself. Similarly the second term, when 9's are dropped leaves just the digit itself. Ditto for the last term. So dropping 9's has reduced the factor to the sum of its digits, in this case 6. Now these mod(9)'s, as I call them, can be added to give a number unique to the actual arithmetic sum.

That is an unnecessarily complicated process. I can see that 10 = (9 + 1) = 1 mod(9). And 100 = (99 + 1) = I mod(9), 1000 = (999 + 1) = 1 mod(9), and so on. But the mod(9) of a product equals the product of the mod(9)'s of the factors. Since 10 = 1 mod(9), then 100 = 1 x 1 mod(9). So the mod(9) of every power of 10 is simply 1 mod(9).

In the example above, 321 = 100 x 3 + 10 x 2 + 1. Substituting the mod(9) values for the powers of 10 one find 321 = 3 mod(9) + 2 mod(9) + 1 mod(9) = 6 mod(9) The other operations of difference, multiplication and division proceed in same way. For subtraction, for example:

$\begin{array}{r}
4567\\
\underline{-4462}\\
105
\end{array}$

The procedure described above reduces the minuend, 4567, to 4 + 5 = 9, drop 9, 0 +6 = 6, 6 + 7 = 13, drop 9 to have 4. The subtrahend, 4462, reduces as 4 + 4 = 8, 8 + 6 = 14, drop 9 to have 5, 5 + 2 = 7. So the problem reduces to 4 - 7 = -3. In the case where the minuend number is less than the subtrahend, one must add back a 9 to the minuend to have a positive answer. In our case this results in -3 + 9 = 6. To continue with the result of my long hand subtraction, 105, we see that it has a residue of 1 + 0 = 1, 1 + 5 = 6, which checks my result above.

For multiplication the procedure of checking the answer is similar. With the product:

$\begin{array}{r}
4567\\
\underline{x4462}\\
9134\\
27402~\:\\
18268~~~\\
\...
...~~~~~\\
28 \Longrightarrow 1\\
$and$ ~~20377954 \Longrightarrow 1
\end{array}$ when the 9's are dropped.

Division proceeds in like manner:

$\begin{array}{r}
117 \\
39 \surd \overline{4567}\\
\underline {39~~~}\\
66~\...
...rrow \underline{0}~~~~~~~~\\
117 x 39 \Longrightarrow 0~~~~~~~~\\
\end{array}$ when the 9's are dropped.

The process for division must take account of the remainder, if one exists. Obviously, this complicates the checking, but maybe one does not have to do too much long division! The way the remainder is treated here is one possible way of handling it, and probably sufficient to most problems.

Checking by 9's is valid in the usual case. It should be evident that any permutation the digits in any of the terms will leave the mod(9) value for that term unchanged, and this will give a false check on the result. Other mod(N) values could be used however to avoid this flaw if desired.

Look at mod(11). 10 = 11 - 1 = -1 mod(11). The last statement is irregular for, by definition, a mod(N) value is positive definite. It is a remainder after all the N's have been divided from the number. But never mind. Continuing, 100 = (9 x 11 + 1) = 1 mod(11). From what we have said before this is so because 100 = 10 x 10 = (-1 mod(11)) x (-1 mod(11)), the product of the mod(11) values of the factors. Thus 100 = 1 mod(11). In the same manner 1000 = -1 mod(11), or the Nth power of 10 equals the Nth power of (-1).

So for mod(11) arithmetic, 321 = 3 x 1 mod(11) + 2 x -1 mod(11) + 1 mod(11) = 2 mod(11). This result is not invariant with permutation of the order of the digits. It therefore provided a more trustworthy check of the arithmetic at a price of increased complexity.

Mod(7) shows more general properties. In this case 10 = 7 + 3 = 3 mod(7). Using the product of modular representations we find 100 = 3 mod(7) x 3 mod(7) = 9 mod(7) or 100 = 2 mod(7). In this way we find: 1000 = 10 x 100 = 3 mod(7) x 2 mod(7) = 6 mod(7), So the powers of 10 have the values 3, 2, 6, 4, 5, 1, 3, 2, . repeating cycles. And 321 = 3 x 2 mod(7) + 2 x 3 mod(7) + 1 = 6 mod(7), a result one can check by dividing 321 explicitly by 7 and noting that the remainder is 6.

The simplest check is the odd or evenness of the result. This is a mod(2) check. All powers of 10 are divisible by 2 so they are all equal to 0 mod(2). Only the unit digit must be examined. Whether it is 0 or 1 mod(2) determines if the number is even or odd.

For those skilled in algebra it should appear possible to reconstruct any N digit result -- sum, product, etc. -- given N independent mod(M) values determined from the factors of the operation.





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Malcom W. P. Strandberg
1998-05-26