Finding Patterns in DigitsDate: 04/29/98 at 08:18:33 From: Nathan Spriggs Subject: Alg II There are certain numbers that when put in the calculator, are equal. For example: 2^5*9^2 = 2592 and 11^2 * 9 1/3 = 1129 1/3 How does this work? Is there a particular process involved to be able to find more examples of this occurrence? Date: 04/29/98 at 16:08:48 From: Doctor Rob Subject: Re: Alg II If you pick a relationship you wish to hold on some number of digits, you will be able to solve a Diophantine equation (that is, one whose solutions must be integers) to find all solutions, if any. Example 1: a^2*b = 100*a + 2*10 + b, where 1 <= a <= 9 and 1 <= b <= 9 This is equivalent to: (a^2-1)*b = 20*(5*a+1) Case 1: a is even, so a^2-1 is odd, and 5*a+1 is odd, so b must be a multiple of 4, but not a multiple of 8, so b = 4. Then a^2 - 1 = 25*a + 5, a^2 - 25*a - 6 = 0, a = [25 + sqrt(649)]/2, so a is not an integer. This case is impossible. Case 2: a = 4*A+1, with 0 <= A <= 2. Then a^2-1 = 8*A*(2*A+1) and 5*a+1 = 20*A+6 are both even. Thus, when we substitute a = 4*A+1 into: (a+1)*(a-1)*b = 20*(5*a+1) we find: 8*A*(2*A+1)*b = 40*(10*A+3) A*(2*A+1)*b = 50*A+15 Clearly A = 0 is impossible. A = 1 gives 3*b = 65, b is not an integer. A = 2 gives 10*b = 115, so b is not an integer. This case is impossible. Case 3: a = 4*A+3, with 0 <= A <= 1. Then a^2-1 = 16*A^2+24*A+8, 5*a+1 = 20*A+16, and: 8*(A+1)*(2*A+1)*b = 80*(5*A+4) (A+1)*(2*A+1)*b = 10*(5*A+4) Now A = 0 yields b = 40, an integer, but too large. A = 1 gives 6*b = 90, b = 15, an integer, but also too large. This case is impossible. This proves that the desired relationship cannot hold. Example 2: a*b^3 = 100*a + 10*b + 3 with 1 <= a and 1 <= b <= 9 Here we will allow a to be as big as we want. Clearly a and b must both be odd. Substituting a = 2*A + 1, 0 <= A, and b = 2*B + 1, 0 <= B <= 4, then: (2*A+1)*(2*B+1)^3 = 200*A + 20*B + 113 16*A*B^3+8*B^3+24*A*B^2+12*B^2+12*A*B+6*B+2*A+1 = 200*A+20*B+113 16*A*B^3+8*B^3+24*A*B^2+12*B^2+12*A*B-14*B-198*A-112 = 0 Trying B = 0, we get A = -56/99, not an integer. Trying B = 1, we get A = -53/73, not an integer. Trying B = 2, we get A = 14/25, not an integer. Trying B = 3, we get A = -85/243, not an integer. Trying B = 4, we get A = -268/629, not an integer. Thus there are no solutions. Example 3: a^5*b^2 = 1000*a + 100*5 + 10*b + 2 with 1 <= a and 1 <= b <= 9 Case 1: b is even, so b = 2*B. Thus a^5*4*B^2 = 1000*a + 502 + 20*B, so the left side is divisible by 4, and the right side is not. This is impossible. Case 2: b odd, so b = 2*B+1. Then a is even, so a = 2*A. Dividing out a 4 from both sides, we find that B must be a multiple of 4, so B = 4*C and b = 8*C+1. Subcase 2a: C = 0, then 2*A^2 - 125*A - 32 = 0, which has no integer roots. This is impossible. Subcase 2b: C = 1, then 162*A^5 - 125*A - 37 = 0. This has only one integer root, A = 1. Then a = 2, b = 9. Thus the only solution is 2^5*9^2 = 2592. Every form you can dream up can be subjected to this kind of analysis. The lucky part is finding one with any solutions at all. Another problem of a similar kind is 19/95 = 1/5 (erroneous canceling of equal digits yields a correct answer). What other two-digit numerators and denominators satisfy this property? How about three- digit ones? -Doctor Rob, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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