Problems & Puzzles: Puzzles

Puzzle 487. Follow up to Puzzle 485 Farideh Firoozbakht sent the following extension to Puzzle 485 I. Use first n natural numbers and allow any number of natural numbers in the right side of the equation ( A - B = C, C <= B < A and 1 in factorization of C ). n = 3 3 - 2 = 1 n = 5 2*4 - 5 = 1*3 n = 6 4*5 - 3*6 = 1*2 n = 7 4*7 - 3*6 = 1*2*5 n = 8 7*8 - 2*3*6 = 1*4*5 n = 9 4*5*9 - 3*7*8 = 1*2*6 n = 9 2*7*9 - 3*4*8 = 1*5*6 n = 9 3*5*8 - 2*6*7 = 1*4*9 n = 10 6*7*8 - 2*3*5*10 = 1*4*9 n = 10 2*3*7*8 - 5*6*10 = 1*4*9 n = 10 5*6*10 - 4*7*9 = 1*2*3*8 n = 10 2*3*5*10 - 4*7*9 = 1*6*8 Number of such representations for n = 3, 4, ... , 12 : 1, 0, 1, 1, 1, 1, 3, 4, 0, 0 Q1. Can you get more examples for other n values? II. Use first n composite numbers and allow any number of composites in the right side of the equation ( A - B = C, C <= B < A ). 8*12 - 6*10 = 4*9 ( first 6 composite numbers) 4*6*10 - 12*14 = 8*9 ( first 7 composite numbers) 4*10*15 - 6*8*9 = 12*14 ( first 8 composite numbers) 4*6*9*18 - 12*14*16 = 8*10*18 ( first 10 composite numbers) 8*9*10*14 - 16*18*20 = 4*6*12*15 ( first 11 composite numbers) Q2. Can you get more examples for other first k composite numbers?

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Contribution came only from Jacques Tramu.

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Mr. Tramu wrote:

n = 11,12,13,14 : ZERO solution

n = 15 4*5*10*11*15 - 2*3*6*7*9*14 = 1*8*12*13
n = 15 4*5*10*11*15 - 3*7*9*12*14 = 1*2*6*8*13
n = 15 : 2 solutions

n= 16,17,18 : ZERO solution

n = 19 2*3*4*5*9*10*11*15 - 6*7*12*13*14*19 = 1*8*16*17*18
n = 19 2*5*9*10*11*12*15 - 3*4*6*7*13*14*19 = 1*8*16*17*18
...
n = 19 : 15 solutions


n = 20 2*3*4*5*7*10*14*19 - 6*8*9*13*15*18 = 1*11*12*16*17*20
n = 20 2*3*7*10*14*19*20 - 4*9*12*13*15*18 = 1*5*6*8*11*16*17
...
n = 20 42 solutions
n = 21 : ZERO solution

n = 22 2*3*4*11*13*14*18*22 - 5*6*8*10*15*20*21 = 1*7*9*12*16*17*19
n = 22 2*3*4*11*12*13*21*22 - 5*7*8*10*15*18*20 = 1*6*9*14*16*17*19
n = 22 2*3*4*11*12*13*21*22 - 5*7*9*10*15*16*20 = 1*6*8*14*17*18*19
...
n = 22 113 solutions
n = 23 ZERO solution
n = 24 ??

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