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JB
Lv 7
JB asked in Science & MathematicsMathematics · 6 years ago

Find distinct prime numbers p, q, r such that 1/(p-1) + 1/(q-1) + 1/(r-1) = 1/7.?

2 Answers

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  • atsuo
    Lv 6
    6 years ago
    Favorite Answer

    p,q,r are distinct prime numbers , so assume that p < q < r .

    I found only one solution p,q,r = 13,29,43 .

    Let P = p-1 , Q = q-1 and R = r-1 . P < Q < R and 1/P + 1/Q + 1/R = 1/7 .

    1/R = 1/7 - (1/P + 1/Q)

    ..... = 1/7 - (P+Q)/(PQ)

    ..... = PQ / (7PQ) - 7(P+Q) / (7PQ)

    ..... = [PQ - 7(P+Q)] / (7PQ)

    R = 7PQ / [PQ - 7(P+Q)]

    If we decide P and Q , then we can find the value of R . But r = R+1 must be a prime number .

    P and Q can be only 1,2,4,6,10,12,16,18,22,28,30, .... (prime numbers - 1)

    If P is 6 or less , then 1/P > 1/7 . So we let P = 10 and start to check the value of R .

    Minimum value of Q is 28 . ( If Q is 22 or less , then 1/P + 1/Q > 1/7 . )

    P ..... Q ..... R

    10 ... 28 .... 140 ...... r = R+1 = 141 is not a prime number .

    10 ... 30 .... 105 ...... r = R+1 = 106 is not a prime number .

    10 ... 36 .... 66.3 ..... R is not an integer .

    10 ... 40 .... 56 ........ r = R+1 = 57 is not a prime number .

    10 ... 42 .... 52.5 ..... R is not an integer .

    10 ... 46 .... 47.35 ... R is not an integer .

    10 ... 52 .... 42.32 ... Q > R , so we can finish to check the value of R with P = 10 .

    Next , we let P = 12 and start to check the value of R .

    Minimum value of Q is 18 . ( If Q is 16 , then 1/P + 1/Q > 1/7 . )

    P ..... Q ..... R

    12 ... 18 .... 252 ...... r = R+1 = 253 is not a prime number .

    12 ... 22 .... 71.07 ... R is not an integer .

    12 ... 28 .... 42 ........ r = R+1 = 43 is a prime number . ---> p,q,r = 13,29,43 is a solution .

    12 ... 30 .... 38.18 ... R is not an integer .

    12 ... 36 .... 31.5 ..... Q > R , so we can finish to check the value of R with P = 12 .

    Next , we let P = 16 and start to check the value of R .

    P ..... Q ..... R

    16 ... 18 .... 40.32 ... R is not an integer .

    16 ... 22 .... 28.65 ... R is not an integer .

    16 ... 28 .... 22.4 ..... Q > R , so we can finish to check the value of R with P = 16 .

    Next , we let P = 18 and start to check the value of R .

    P ..... Q ..... R

    18 ... 22 .... 23.89 ... R is not an integer .

    18 ... 28 .... 19.38 ... Q > R , so we can finish to check the value of R with P = 18 .

    If we let P = 22 , then the minimum value of Q and R is 28 and 30 . But

    1/22 + 1/28 + 1/30 = 0.1145... < 1/7 , no solution exists .

    And clearly if P > 22 then no solution exists .

  • ?
    Lv 7
    6 years ago

    1/(p-1) + 1/(q-1) + 1/(r-1) = 1/7

    (q-1)(r-1) + (p-1)(r-1) + (p-1)(q-1) = (p-1)(q-1)(r-1) / 7

    (qr - q - r + 1)+(pr - p - r + 1) + (pq - p - q + 1) = (p-1)(q-1)(r-1) / 7

    qr + pr + qp - 2p - 2q - 2r + 3 = (p-1)(q-1)(r-1) / 7

    7qr + 7pr + 7qp - 14p - 14q - 14r + 21 = (p-1)(q-1)(r-1)

    Now expand the right-hand side and solve

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