HP 49G: Minimum Multiplier M of Integer N such that M*N Consists Only of 1's & 0's

01292020, 01:58 PM
Post: #14




RE: HP 49G: Minimum Multiplier M of Integer N such that M*N Consists Only of 1's &...
(01292020 11:18 AM)Gerald H Wrote: For input Since N is odd, not divisible by 5, we know P last digit is 1 P = 123456789 M ≡ M ≡ 1 (mod 10) M ≡ 1 ≡ 9 (mod 10) 123456789 * 9 = 1111111101 Another example, P(17) is not as lucky, but still not too bad: P = 17 M ≡ 3 M ≡ 1 (mod 10) M ≡ 1/3 ≡ 9/3 ≡ 3 (mod 10) 17 * 3 = 51 17 * 13 = 221, 1001/17 ≈ 58.88 17 * 63 = 1071 17 * 73 = 1241, 10001/17 ≈ 588.29 17 * 593 = 10081 17 * 603 = 10251, 11001/17 ≈ 647.12 17 * 653 = 11101 We could also try P(17) that ends in 01 vs 11, and pick the smaller P: P = 17*M ≡ 1 or 11 (mod 100) Using my InverseMod trick: 1 2 floor(1/2*15) = 7, 15*17=255 > 100 15 17 floor(7/15*100) = 47 = 17^{1} (mod 100) 100 M ≡ (1 or 11)/17 ≡ (1 or 11)*(47) ≡ 53 or 83 (mod 100) 17 * 53 = 901 17 * 83 = 1411, 10001/17 ≈ 588.29 17 * 653 = 11101 

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