Zeros and Nines

The problem below has been posted by H. W. Richmond (King's College, Cambridge) in The Mathematical Gazette, Vol. 10, No. 154. (Oct., 1921), p. 328, under the caption An Old Result in Novel Form.

Any number S is written down, and beneath it are written its multiples 2S, 3S,...up to 10S.

Consider the ten figures in any column. There does not appear to be any general law; yet every complete column must contain either a 9 or a 0. Prove this. (In the first column the gaps may be filled by 0's, and the same result holds for all columns.)

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Zeros and Nines

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The applet helps verify the assertion which actually holds for the numbers S, ..., 9S. The blue digits are clickable. When the "Autonomous digits" box is checked, the digits do not affect their neighbors. Otherwise, the blue number changes as a whole. Click a little off the center line of each digit.

The problem extends to positional systems with radix (base) different from 10. What would be the correct analogue?


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Copyright © 1996-2017 Alexander Bogomolny


There is a rather simple solution that makes use of the Pigeonhole principle. In every column there are 10 digits. If we assume that none of them equals either 0 or 9, at least two of them must be equal (this is if course true if we only consider multiples up to 9), say aS and bS, a > b. The corresponding digit of (a - b)S = aS - bS must be either 0, or 9 if any borrowing is involved from the previous column. But (a - b)S is one of the multiples of S in the table with a digit in the column under consideration. It follows that our assumption of the absence of 0's and 9's led to a contradiction.

Warning: there is a mistake in this argument; for S from 1050 through 1089, and also from 2050 through 2089, etc., the hundredth column contains neither 0 nor 9. You may want to ponder why and how to salvage the theorem.

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  • |Contact| |Front page| |Contents| |Algebra|

    Copyright © 1996-2017 Alexander Bogomolny


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