An algebra puzzle

Algebra Level 3

We know that

( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 4 2 ) ( 1 1 n 2 ) = k (1-\frac{1}{2^2})*(1-\frac{1}{3^2})*(1-\frac{1}{4^2})*\dots*(1-\frac{1}{n^2})=k

where n n is a positive integer.

If k k can't be 1 2 \frac{1}{2} and can't be 1008 2017 \frac{1008}{2017} , then choose 0.

If k k can't be 1 2 \frac{1}{2} , but it can be 1008 2017 \frac{1008}{2017} , then choose 1.

If k k can't be 1008 2017 \frac{1008}{2017} , but it can be 1 2 \frac{1}{2} , then choose 2.

If k k can be 1 2 \frac{1}{2} and 1008 2017 \frac{1008}{2017} too, then choose 3.

3 2 1 0

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2 solutions

Max Willich
Jun 5, 2017

Plugging in some numbers, we get:

For n = 2 : k = 3 4 n = 2: k = \frac{3}{4}

For n = 3 : k = 2 3 n = 3: k = \frac{2}{3}

For n = 4 : k = 5 8 n = 4: k = \frac{5}{8}

For n = 5 : k = 3 5 n = 5: k = \frac{3}{5}

Expanding some fractions, we get the pattern 3 4 , 4 6 , 5 8 , 6 10 . . . n + 1 2 n \frac{3}{4}, \frac{4}{6}, \frac{5}{8}, \frac{6}{10} ... \frac{n+1}{2n} for n = 2 , 3 , 4 , . . . , n n = {2, 3, 4, ..., n} .

To prove that the series above is equal to n + 1 2 n \frac{n+1}{2n} , we test it for n = 2, which is 2 + 1 4 \frac{2+1}{4} , therefore true.

Assuming that ( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 3 2 ) . . . ( 1 1 n 2 ) = n + 1 2 n (1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{3^2})...(1 - \frac{1}{n^2})=\frac{n+1}{2n} , we try to prove that the same is true for n + 1 n+1 :

( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 3 2 ) . . . ( 1 1 ( n + 1 ) 2 ) = n + 2 2 n + 2 (1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{3^2})...(1 - \frac{1}{(n+1)^2})=\frac{n+2}{2n + 2}

Notice that on the left side, since we assume that ( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 3 2 ) . . . ( 1 1 n 2 ) = n + 1 2 n (1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{3^2})...(1 - \frac{1}{n^2})=\frac{n+1}{2n} , we can exchange all the factors infront of 1 ( n + 1 ) 2 \frac{1}{(n+1)^2} with n + 1 2 n \frac{n+1}{2n} :

n + 1 2 n ( 1 1 ( n + 1 ) 2 = n + 2 2 n + 2 \frac{n+1}{2n}\cdot (1 - \frac{1}{(n+1)^2} = \frac{n+2}{2n + 2}

Simplyfing the second factor on the left hand side yields:

n + 1 2 n n 2 + 2 n ( n + 1 ) 2 = n + 2 2 n + 2 \frac{n+1}{2n}\cdot \frac{n^2 + 2n}{(n+1)^2} = \frac{n+2}{2n + 2}

1 2 n n 2 + 2 n n + 1 = n + 2 2 n + 2 \iff \frac{1}{2n}\cdot \frac{n^2 + 2n}{n+1} = \frac{n+2}{2n + 2}

1 2 n n n + 1 = 1 2 n + 2 \iff \frac{1}{2n}\cdot \frac{n}{n+1} = \frac{1}{2n + 2}

1 2 1 n + 1 = 1 2 n + 2 \iff \frac{1}{2}\cdot \frac{1}{n+1} = \frac{1}{2n + 2}

1 2 n + 2 = 1 2 n + 2 \iff \frac{1}{2n+2} = \frac{1}{2n + 2}

, which is obviously true. Therefore, the assumption ( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 3 2 ) . . . ( 1 1 n 2 ) = n + 1 2 n (1 - \frac{1}{2^2})(1 - \frac{1}{3^2})(1 - \frac{1}{3^2})...(1 - \frac{1}{n^2})=\frac{n+1}{2n} is true.

Testing for n + 1 2 n = 1 2 \frac{n+1}{2n} = \frac{1}{2} solves to 0 = 1 0 = 1 , which is false. Similarily, solving n + 1 2 n = 1008 2017 \frac{n+1}{2n} = \frac{1008}{2017} yields n = 2017 n = -2017 , but n 2 n \geq 2 , therefore this is also no possible solution. Therefore the correct answer is 0 0

We will show that ( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 4 2 ) ( 1 1 n 2 ) > 1 2 (1-\frac{1}{2^2})*(1-\frac{1}{3^2})*(1-\frac{1}{4^2})*\dots*(1-\frac{1}{n^2})>\frac{1}{2} .

We know that x 2 = ( x 1 ) ( x + 1 ) + 1 x^2=(x-1)(x+1)+1 , so 1 1 y 2 = ( y 1 ) ( y + 1 ) y 2 1-\frac{1}{y^2}=\frac{(y-1)(y+1)}{y^2} . From that we can see that

( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 4 2 ) ( 1 1 n 2 ) = 1 3 2 2 2 4 3 3 ( n 1 ) ( n + 1 ) n 2 = 1 2 3 2 2 3 4 3 3 4 4 3 n n 1 n 1 n n + 1 n (1-\frac{1}{2^2})*(1-\frac{1}{3^2})*(1-\frac{1}{4^2})*\dots*(1-\frac{1}{n^2})=\frac{1*3}{2*2}*\frac{2*4}{3*3}*\dots*\frac{(n-1)(n+1)}{n^2}=\frac{1}{2}*\frac{3}{2}*\frac{2}{3}*\frac{4}{3}*\frac{3}{4}*\frac{4}{3}*\dots*\frac{n}{n-1}*\frac{n-1}{n}*\frac{n+1}{n} .

Since a b b a = 1 \frac{a}{b}*\frac{b}{a}=1 , ( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 4 2 ) ( 1 1 n 2 ) = 1 2 1 1 1 1 n + 1 n (1-\frac{1}{2^2})*(1-\frac{1}{3^2})*(1-\frac{1}{4^2})*\dots*(1-\frac{1}{n^2})=\frac{1}{2}*1*1*1*\dots*1*\frac{n+1}{n} . Now we can see that 1 2 n + 1 n > 1 2 \frac{1}{2}*\frac{n+1}{n}>\frac{1}{2} (because n + 1 n > 1 \frac{n+1}{n}>1 ), so ( 1 1 2 2 ) ( 1 1 3 2 ) ( 1 1 4 2 ) ( 1 1 n 2 ) > 1 2 (1-\frac{1}{2^2})*(1-\frac{1}{3^2})*(1-\frac{1}{4^2})*\dots*(1-\frac{1}{n^2})>\frac{1}{2} .

The answer is 0, because 1 2 = 1 2 > 1008 2017 \frac{1}{2}=\frac{1}{2}>\frac{1008}{2017} .

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