TchrWill
Advanced Member
USA
79 Posts 
Posted  03/23/2013 : 18:03:19

quote: Originally posted by Ultraglide
576
Right on Ultraglide. Might you share your solution path with us?
One of mine is as follows:
1Lets express your 3 digit number as N = 100A + 10B + C 2The requirement is to have 100A + 10B + C = 32(A + B + C) 3This equality can be simplified to 31C + 22B = 68A 4For there to be any positive integer solutions of this equation, the g.c.d. of 31 and 22 must evenly divide 68. 5By inspection, the g.c.d.(31,22) = 1 which clearly divides 68. 6We first seek to find a solution to 31C + 22B = 68 assuming A = 1 for now. 7Dividing through by 22 yields 1C + 9C/22 + B = 3 + 2/22 or (9C  2)/22 = 2  C  B 8(9C  2)/22 must be an integer but we want to get the coefficient of C = 1. 9(9C  2)/22 x 5 = (45C  10)/22 10Dividing by 22 again yields 2C + 1/22  10/22 11Again, (C  10)/22 must be an integer k making C = 22k + 10 12 Substituting this back into (6) yields 31(22)k + 310 + 22B = 68 or B = 31k  11. 13A solution derives from k = 0 or C = 10 and B = 11. 14Other solutions derive from C = 10 + 22t and B = 11  31t 15If integer solutions exist, they must fall between (10/22) < t < (11/31) or .......4542 < t < .3548. 16Clearly no integers exist between these limits so A must be more than 1. 17Letting A increase in increments of 1 and establishing the new limits for t, we eventually reach 31(50) + 22(55) = 340 where A = 5. 18Then, C = 50 + 22t and B = 55  31t. 26The limits now become 50/22 < t < 55/31 or 2.272 < t < 1.774. 29Clearly, t = 2 fits between these limits. 30Therefore, C = 50  22(2) = 6 and B = 55  31(2) = 7. 31Therefore, our number is 576.


