Let m and n be two positive integers such that m + n + m n = 1 1 8
Then the value of m + n is
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How did you get to know the values of (m+1) and (n+1)?
I did the same
Great approach!!!
Let's restate the question in terms of a quadratic. Let x = m and a = m + n ⟹ n = a − x .
a + x ( a − x ) = 1 1 8
x 2 − a x + 1 1 8 − a = 0
x = 2 a ± a 2 − 4 ( 1 1 8 − a )
We know that x must be an integer, so we need to check the determinant a 2 − 4 ( 1 1 8 − a ) to see whether a = { 1 8 , 2 0 , 2 2 } lead to square numbers.
1 8 2 − 4 ( 1 1 8 − 1 8 ) = − 7 6
2 0 2 − 4 ( 1 1 8 − 2 0 ) = 8
2 2 2 − 4 ( 1 1 8 − 2 2 ) = 1 0 0 = 1 0 2
Only 2 2 leads to an integer solution for x , therefore our solution must be a = 2 2 .
m and n are not exactly variables, they cannot vary!
x is a variable in f ( x ) = x 2 + 7 x − 2 0 but when we write x 2 + 7 x − 2 0 = 0 the x here is not a 'variable' but an 'unknown'. This is the most basic distinction between the different ways that mathematicians use letters in place of numbers, there are many! One of the most common causes of confusion for kids who aren't so good at maths is that they are used to unknowns only because the teacher always asks them to "find x ", then when they advance through school they treat all letters as unknowns, even variables. When I tutor kids I often find they ask "but what is x ?" when I write down an expression (with x as a variable). I say "it's a variable, it can be any number". "Yeah, but what is it now?". "Any number". "so.. 7?". "well it could be 7, but we want to consider the expression, we don't want to consider particular values for x ". "so can we just say x = 7 ?" etc. etc. They find it very hard to keep an open mind with regards to variables. For many kids, all letters in maths are just "numbers in disguise", place holders waiting to be substituted for a number.
Sorry this is not a solution. I hope this is interesting to some people, particularly non-native english speakers. Language is very important in explaining mathematics.
for(int i=0;i<100;i++) { for(int j=0;j<100;j++) { if((i+j+(i*j))==118) { System.out.println("i = "+i+"j = "+j); } }
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m + n + m n = 1 1 8
m ( 1 + n ) + 1 ⋅ ( 1 + n ) = 1 1 8 + 1
( m + 1 ) ( n + 1 ) = 1 1 9
Since m and n are positive integers { m + 1 = 1 7 n + 1 = 7 ⟹ m = 1 6 , n = 6
{ m + 1 = 7 n + 1 = 1 7 ⟹ m = 6 , n = 1 6
Other cases will not yield positive solutions
So in either case m + n = 1 6 + 6 = 2 2