160 SPECIAL TOPICS IN THEORETICAL ARITHMETIC

This error is equivalent to dividing the third partial product
by 10. Call the incorrect third partial product, in this case
25280, b. Then the correct third partial product would be
106. Call the sum of the first two partial products a. Then
the final result as given above is @ 4+ b and the result should
be a + 106.

Now (a 4+ b) — (e + 10b) = — 9b.

Hence ¢ + b = a 4+ 100 (mod 9), that is, the incorrect re-
sult is congruent to the correct result.

The error would therfor not be reveald by casting out the
nines (1).

Neither would it be reveald in this example by a congruence
to any of the moduli 2, 3, 4, 5, 6, 8, or 10, or any other factor
of 95, that is, of 9 X 4 X 632 X 10. It would however be
reveald in this example by using as the modulus either 7 or 11,
or any other number that is not a factor of 95, provided the
check was applied without error.

(1) For other errors that would not be reveald see O’Sullivan, l. c., pp.
60, 61.