Monthly Archives: October 2015

Word Play: A=1, B=2, C=3, D=4, E=5, ..

    A = 1              B = 2            C = 3            D = 4           E = 5           F = … Continue reading

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Palindromes P divisible by 19 & DigitSum(P) = 2,3,…,100

    Find other palindromes divisible by 19.   Paul found:      4   :     11000000011   =   7*11*11*13*19*52579    6   :     1200021   =   3*19*37*569    7   :     200111002   = … Continue reading

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PPT & x^2 – 2*y^2 = 7

    Using the parametrization of primitive Pythagorean triples as   where   the sum p of its two legs can be expressed as the Pell-like equation,          Eq #1 When p is a prime If   Eq #1   … Continue reading

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Palindromes P divisible by 17 & DigitSum(P) = 2,3,…,100

    Find more palindromes divisible by 17   David & Paul found: 35 :     35 :   35 :   35 :   35 :   35 :   35 :   35 :   35 :   … Continue reading

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PPT (a,b,c) | 2(b*c)^2 = b^4 + c^4 – a^4

  The integers form a Pythagorean triple. And, that is,   All the Primitive Pythagorean Triples (PPT) with     :     …………     ……….     ……….     ……….     ………     ………   … Continue reading

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n-digit integers which are also an n-th power

    Here are some examples:   1-digit:       2-digit:       3-digit:     4-digit:     5-digit:     6-digit:     7-digit:     8-digit:     9-digit:               10-digit:               11-digit:     12-digit:     13-digit:     14-digit:     15-digit: … Continue reading

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Palindromes P divisible by 13 & DigitSum(P) = 2,3,…,100

    Can you find palindromes divisible by 13 and whose sum of digits is: 32,   38,   39,   …,   100?   Pipo and Paul found:   32:     88088 = 2*2*2*7*11*11*13   38:     1974791 = … Continue reading

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Palindromes P divisible by 7 & DigitSum(P) = 2,3,…,40

    Can you find palindromes divisible by 7 and whose sum of digits is: 5, 27, 29, 31, 33, 35, 37, 38, 39, 40?   Pipo found : 4 :   1002001 5 :   1011101 6 :   … Continue reading

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Equation : x^2 ± k*y^2 are square numbers, x a prime number less than 1000

  Here are the prime numbers, less than 1000, that are solutions to the two equations :                                    

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Cubes with same digits

    The digits of can be permuted to form   cube whose digits can be permuted to produce two other cubes :               and three other cubes :     Paul found:   … Continue reading

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