Signed Sum Of Reciprocals

Logic Level 2

$\large 1 \, \square \, \dfrac12 \, \square \, \dfrac13 \, \square \, \dfrac14 \, \square \,\dfrac15 \, \square \, \cdots \square \, \dfrac1{31} =\dfrac1{32}$

There are $2^{30}$ ways in which we can fill the squares with the mathematical operators $+$ and $-$ .

How many ways would make the equation true?

The answer is 0.

1 solution

Pi Han Goh
Apr 4, 2016

Answer : It is impossible to make the equation true.

Move the RHS term to the LHS to get:

$1 \; \pm \; \dfrac12 \; \pm \; \dfrac13 \; \pm \; \cdots \; \pm \; \dfrac1{31}- \dfrac1{32}= 0$

Let's rewrite the entire expression as a single fraction by multiplying it $\dfrac{32!}{32!}$ :

$\dfrac{32! \; \pm \; \dfrac{32!}{2} \; \pm \; \dfrac{32!}3 \; \pm \; \cdots \; \pm \; \color{#D61F06}{\dfrac{32!}{31}} - \dfrac{32!}{32} }{32!}$

Notice that all the terms in the numerator except for the term in red is divisible by 31 (a prime number ), and the denominator is divisible by 31 (Because $31 | 32\cdot 31!$ ).

So no matter what signs we assign these boxes to be (whether it be $+$ or $-$ ), LHS is a fraction consisting of a numerator that is not divisible by 31, and a denominator that is divisible by 31. Thus, LHS is not equal to 0.

Hence, there is no solution.

Nice solution. +1

Sharky Kesa - 5 years, 2 months ago

Thanks! To be honest, this question was inspired by one of the Morrocan Olympiad questions.

Pi Han Goh - 5 years, 2 months ago

Or we may take another approach and say that multiply both sides by 32!/31, a whole number. 32!/31 is divisible by all the numbers from 1 to 32 except by 31, since 31 is a prime number. On multiplication, all the numbers on LHS will be whole numbers except 32!/(31×31), which is a fraction and the resultant answer will be a fraction irrespective of number of (+) and (-) in the equation. But on RHS, we will have a whole number. Hence the equation will never match.

Auro Light - 3 years, 9 months ago

Signed Sum Of Reciprocals

The answer is 0.

1 solution

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