Is It Really Possible?
How many unordered pairs of positive integers ( A , B ) are there such that
A + B + lcm ( A , B ) = 5 1 ?
The answer is 4.
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3 solutions
Can you please elaborate your answer? What are the pairs exactly?
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(1,25);(6,15);(3,24);(17,17)
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@Arulx Z – I too made the same assumption.
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@Janardhanan Sivaramakrishnan – Oh. In that case, I've added a report to fix the question.
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@Arulx Z – Sorry for inconvenience sir......I just fixed that
When you write ( A , B ) without comment, you are referring to an ordered pair (like a pair of coordinates), so there should be 7 solutions: (17,17), (3,24), (24,3), (6,15), (15,6), (1,25) and (25,1).
If you want the answer 4 , the question should probably read: "How many unordered pairs A , B of positive integers ..."
LCM (0, 51) = 0 by Microsoft Excel. Meaning to say not only starting with 1 but 0, both of them can have common 0 as a result. LCM (6, 15) = 30 as both started with 1 which does not include 0, but otherwise there could only be 0.
Starting only with multiplier of 1, 0 can never have common multiple with others. 0 suppose to be excluded in this sense.
Original definition of LCM does not include Non-Natural Numbers. But because "Least Common Multiple" was not mentioned, I thought I could just applied LCM() of Microsoft Excel. I didn't realize or otherwise I could have tried 4 as my third trials.
Microsoft Excel could have bugs I think. But please remove ambiguity of this question. Please consider to change the title. There is nothing surprising for having answers.
So we must find to (positive) integers a , b such that S = a + b + lcm ( a , b ) = 5 1 .
Write a = k m , b = k n , where m , n are coprime and k is the greatest common divisor of a , b . Then S = k ( m + n + m n ) = 5 1 , or ( m + 1 ) ( n + 1 ) = k 5 1 + 1 . This limits the possibilities to k = 1 , 3 , 1 7 , 5 1 . We solve the equation, keeping only the solutions where m and n have no common factors. k = 1 k = 3 k = 1 7 k = 5 1 ( m + 1 ) ( n + 1 ) = 5 2 ( m + 1 ) ( n + 1 ) = 1 8 ( m + 1 ) ( n + 1 ) = 4 ( m + 1 ) ( n + 1 ) = 1 = 2 ⋅ 2 6 4 ⋅ 1 3 = 2 ⋅ 9 3 ⋅ 6 = 2 ⋅ 2 m = 1 , n = 2 5 m = 3 , n = 1 2 m = 1 , n = 8 m = 2 , n = 5 m = 1 , n = 1 a = 1 , b = 2 5 invalid a = 3 , b = 2 4 a = 6 , b = 1 5 a = 1 7 , b = 1 7 no solutions Thus there are 4 solutions, with ( a , b , lcm ( a , b ) ) = ( 1 , 2 5 , 2 5 ) ; ( 3 , 2 4 , 2 4 ) ; ( 6 , 1 5 , 3 0 ) ; ( 1 7 , 1 7 , 1 7 ) .
My solution: x+y+xy=51 By Simon's favorite factoring trick, (x+1)(y+1)=52 we list all numbers which when multiplied equals 52 ~4(13), 13(4) ,2(2), 2(26), 52(1), 1(52).. Clearly 52(1) and 1(52) cant be qualified because either x or y will become zero, so.. There are 4 such pairs.
Let us suppose two numbers as kx and ky.
LCM=kxy
k(x+y+xy) = 51.
x+y+xy=51..(x+1)(y+1)=52. so (1,25)...if k=1
x+y+xy=17 (x+1)(y+1)=18 so (1,8), (2,5)...if k=3
x+y+xy = 3 (x+1)(y+1)=4 so (1,1)...if k=17.
So,the answer is 4.