n = 26a+6 = 36b+16 = 56c+36 We solve the Diophantine equation 36b+16 = 56c+36 9b+4 = 14c+9 9b = 14c+5 9 is the least integer in that equation in absolute value, so write all the other integers in terms of a near multiple of 9. So we write 14 as 18-4, and 5 is near 9 so we just leave it as it is. 9b = (18-4)c+5 9b = 18c-4c+5 Divide through by 9 b = 2c-4c/9+5/9 Get the fraction on the left and other terms on the right: 4c/9-5/9 = 2c-b The right side is an integer, say P, so 4c/9-5/9 = 2c-b = P 4c/9-5/9 = P 4c-5 = 9P 4 is the least integer in that equation in absolute value, so write all the other integers in terms of their nearest multiple of 4. So we write 5 as 4+1, write 9 as 8+1. 4c-(4+1) = (8+1)P 4c-4-1 = 8P+P Divide through by 4 c-1-1/4 = 2P+P/4 Get the fraction on the right, other terms on the left: c-1-2P = P/4+1/4 The left side is an integer, say Q, so c+1-2P = P/4+1/4 = Q P/4+1/4 = Q P+1 = 4Q P = 4Q-1 c-1-2P = Q c-1-2(4Q-1) = Q c-1-8Q+2 = Q c+1-8Q = Q c = 9Q-1 9b = 14c+5 9b = 14c+5 9b = 14(9Q-1)+5 9b = 126Q-14+5 9b = 126Q-9 b = 14Q-1 26a+6 = 36b+16 26a = 36b+10 13a = 18b+5 13a = 18(14Q-1)+5 13a = 252Q-18+5 13a = 252Q-13 13 is the least integer in that equation in absolute value, so write all the other integers in terms of their nearest multiple of 13. So we write 252 as 247+5. 13a = (247+5)Q-13 13a = 247Q+5Q-13 Divide through by 13 a = 19Q+5Q/13-1 Get the fraction on the right and other terms left: a+1-19Q = 5Q/13 The left side is an integer, say R, so a+1-19Q = 5Q/13 = R 5Q/13 = R 5Q = 13R We could continue as before but this equation is simpler and we see that the smallest integers Q and R that solve that are Q=13 and R=5. So a+1-19Q = R a+1-19(13) = 5 a+1-247 = 5 a-246 = 5 a = 251 And since n = 26a+6 n = 26(251)+6 n = 6526+6 n = 6532 Checking: 251 181 116 26)6532 36)6532 56)6532 52 36 56 133 293 93 130 288 56 32 52 372 26 36 336 6 16 36 <-remainders check! Edwin