Divisibility problem
Number Theory
Level
pending
13 divides (a+11b) , and 11 divides ( a + 13b) . Then the least value of (a+b) is ( for positive integers a,b)
The answer is 28.
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1 solution
Since 11 divides a+13b = 11b+(a+2b), so 11 divides a+2b. Thus (1) 143 = 13 11 divides 13(a+2b) = 13a+26b. Since 13 divides a+11b = 13b+(a-2b), so 13 divides a-2b. So (2) 143 = 11 13 divides 11(a-2b) = 11a-22b. By (1) and (2), 143 divides both 13a+26b and 11a-22b, or their difference (13a+26b)+(11a-22b) = 24a+4b = 4(6a+b). Since 143 does not divide 4, so 143 divides 6a+b. To minimize a+b, we can set a = floor(143/6) = 23 and b = 143-6a = 143-138 = 5. Hence the least possible value of a+b is 23+5 = 28. This problem came in Romanian Mathematical Olympiad 2006, and RMO West Bengal 2006.