Pairs of Odd IntegersDate: 09/24/97 at 04:12:40 From: Brian Lam Subject: Very Challenging Integer Problem Dr. Maths, I am a Year 9 student from Australia and I am asking you to help me with a very hard and challenging integer problem. Find all pairs of odd integers a and b that satisfy the following equation : a + 128b = 3ab Thank you! Brian Lam Date: 09/24/97 at 08:59:05 From: Doctor Rob Subject: Re: Very Challenging Integer Problem The trick for this kind of equation is to reduce it to the form (A*a + B)*(C*b + D) = E, for some integers A, B, C, D, and E. To that end, bring all terms to the righthand side: 0 = 3*a*b - a - 128*b Factor a from the first two terms: 0 = a*(3*b-1) - 128*b This tells me that C = 3 and D = -1 in the above form. Add 128/3 to both sides to make the right side divisible by 3*b-1 exactly. 128/3 = a*(3*b-1) - 128/3*(3*b-1) = (a-128/3)*(3*b-1) Mulitply both sides by 3 to clear fractions: 128 = (3*a-128)*(3*b-1) This is in the desired form, and A = 3, B = -128, and E = 128. Now factor E = 128 into its prime power factors. 2^7 = (3*a-128)*(3*b-1) Observe that both 3*b-1 and 3*a-128 must be factors of 2^7. Now use the fact that a and b must both be odd integers to determine which factors of 2^7 they can be. Don't forget that they could be negative! -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 09/24/97 at 09:59:41 From: Doctor Anthony Subject: Re: Very Challenging Integers Problem Express b in terms of a: a = b(3a-128) so a b = -------- 3a - 128 Now a and b are odd integers, and this equation shows us a > 3a-128 128 > 2a so a < 64 Also 3a-128 >= 1 3a >= 129 so a >= 43 Thus we have 43 <= a <= 63 If a = 43 then b = 43/(3x43 -128) = 43/1 = 43 So one pair of values is (43, 43) We must stick to odd integers so the next value up for a is 45. 45 45 b = --------- = ----- not an integer 3x45 -128 7 We can see that the numerator will increase by 2 each time and the denominator by 6 (since it is 3a in the denominator), so the next few calculations for b will have following results: b = 47/13, 49/19, 51/25, 53/31, 55/37, 57/43, 59/49, 61/55, 63/61 and it is unnecessary to go further since the calculated values for b will now become less than 1. From the above calculation we see that there is just one pair of odd integers satisfying the equation a + 128 = 3ab. and these are a = 43, b = 43. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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