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 Possible or not? (Posted on 2015-01-17)
Prove or disprove the following:
For any integer number N there exists at least one integer number M, such that the decimal presentation of M*N needs only two distinct digits.

 No Solution Yet Submitted by Ady TZIDON Rating: 4.0000 (1 votes)

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 possible solution | Comment 1 of 6

With the exception of N=0, the proposition is true.

Not only that, but the 'two distinct digits' can be sharpened to 0 and 9:

1. Assume that N divides 10^n M times, then N*M=10^n, for some M. Multiplying this by 9 produces 9 followed by a number of zeroes.

2. Assume that N is odd, is not divisible by 5, and does not divide 10^n. Then M divides 10^n-1 (this is, after all, the basis of the decimal system) so that 10*N*M= 9999...90. In this case M is the periodic part of the decimal expansion of 1/N, see Sloane A036275.

3. Assume that N is even, or divisible by 5, and does not divide 10^n:

(a) Extract all powers of 2 and 5 to produce an odd number, and derive its periodic part, M1.

(b) Since matched powers of 2 and 5  just produce terminal zeroes, they can be ignored. For each unmatched power of 2, multiply M1 by 5; for each unmatched power of 5, multiply M1 by 2, to produce M2. In this case M is M1*M2. The result comprises a series of 9's followed by a series of zeroes, 999...000.

4. Since 0*M = 0, only 1 digit is needed to represent it.

A nice problem, but not D4 as we know it, Jim.

Edited on January 17, 2015, 9:05 pm
 Posted by broll on 2015-01-17 20:58:17

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