Egyptian Fraction?

Number Theory Level 5

Let x , y , x, \space y, and z z be positive real integer, with 1000 < x < y < z < 2000 1000< x <y <z<2000 and satisfy the condition 1 2 + 1 3 + 1 7 + 1 x + 1 y + 1 z + 1 45 = 1 \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} + \dfrac{1}{45} = 1 .

Given further that x x is divisible by 42, find x + y + z x+y+z .


The answer is 5678.

Fidel Simanjuntak by Fidel Simanjuntak , 19, Indonesia

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1 solution

Fidel Simanjuntak
Apr 23, 2017

First, notice that 1 2 + 1 3 + 1 7 + 1 42 = 1 \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{42} = 1

1 2 + 1 3 + 1 7 + 1 42 − 1 43 + 1 43 − 1 44 + 1 44 − 1 45 + 1 45 = 1 1 2 + 1 3 + 1 7 + ( 1 42 − 1 43 ) + ( 1 43 − 1 44 ) + ( 1 44 − 1 45 ) + 1 45 = 1 1 2 + 1 3 + 1 7 + 1 42 × 43 + 1 43 × 44 + 1 44 × 45 + 1 45 = 1 1 2 + 1 3 + 1 7 + 1 1806 + 1 1892 + 1 1980 + 1 45 = 1 \begin{aligned} \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{42} - \dfrac{1}{43} + \dfrac{1}{43} - \dfrac{1}{44} + \dfrac{1}{44} - \dfrac{1}{45} + \dfrac{1}{45} & =1 \\ \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \left( \dfrac{1}{42} - \dfrac{1}{43}\right) + \left( \dfrac{1}{43} - \dfrac{1}{44}\right) + \left( \dfrac{1}{44} - \dfrac{1}{45} \right) + \dfrac{1}{45} & =1 \\ \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{42 \times 43} + \dfrac{1}{43 \times 44} + \dfrac{1}{44 \times 45} + \dfrac{1}{45} & =1 \\ \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{7} + \dfrac{1}{1806} + \dfrac{1}{1892} + \dfrac{1}{1980} + \dfrac{1}{45} & =1 \end{aligned}

Hence, ( x , y , z ) = ( 1806 , 1892 , 1980 ) (x,y,z) = (1806,1892,1980) ; then the answer is 5678 5678

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This shows that 5678 is a possible answer. How do you know that's the unique answer?

Calvin Lin Staff - 4 years, 1 month ago

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I don't know.. I was just using the identity 1 a − 1 a + 1 = 1 a ( a + 1 ) \dfrac{1}{a} - \dfrac{1}{a+1} = \dfrac{1}{a(a+1)} to find x , y , z x,y,z .

Fidel Simanjuntak - 4 years, 1 month ago

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If you look at the reports, you will see that there are multiple integer solutions to 1 x + 1 y + 1 z = 1 − 1 2 − 1 3 − 1 7 − 1 45 \frac{1}{x} + \frac{1}{y} + \frac{1}{z} = 1 - \frac{1}{2} - \frac{1}{3} - \frac{1}{7} - \frac{ 1}{45} .

Thus, we have to be careful with such problems. Simply finding one possible solution doesn't mean that's the only solution.

Calvin Lin Staff - 4 years, 1 month ago

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@Calvin Lin – I see what you mean. Thanks!

Fidel Simanjuntak - 4 years, 1 month ago

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