An algebra problem by Paola Ramírez

Algebra Level 3

Given that:

a b = 1 b c = 2 c d = 3 d e = 4 e a = 5 b = ? \begin{array}{lcr} ab=1 & bc=2 & cd= 3 \\ de=4 & ea=5 & b=? \\ \end{array}

with a , b , c , d a,b,c,d and d d positive numbers.Find b b

If b = x ! y b=\dfrac{\sqrt{x!}}{y} where x < 6 x<6 is a prime,submit the answer as x + y x+y .


The answer is 20.

Paola Ramírez by Paola Ramírez , 24, Mexico

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1 solution

Hung Woei Neoh
May 25, 2016

a b = 1 , b c = 2 , c d = 3 , d e = 4 , e a = 5 ab=1,\;bc=2,\;cd=3,\;de=4,\;ea=5

Multiply them all:

a b × b c × c d × d e × e a = 1 × 2 × 3 × 4 × 5 a 2 b 2 c 2 d 2 e 2 = 5 ! ( a b c d e ) 2 = 5 ! a b c d e = 5 ! ab \times bc \times cd \times de \times ea = 1 \times 2 \times 3 \times 4 \times 5\\ a^2b^2c^2d^2e^2 = 5!\\ (abcde)^2 = 5!\\ abcde = \sqrt{5!}

b = a b c d e a c d e = a b c d e ( c d ) ( e a ) = 5 ! ( 3 ) ( 5 ) = 5 ! 15 b=\dfrac{abcde}{acde}\\ =\dfrac{abcde}{(cd)(ea)}\\ =\dfrac{\sqrt{5!}}{(3)(5)}\\ =\dfrac{\sqrt{5!}}{15}

Therefore, the solution required is 5 + 15 = 20 5+15 = \boxed{20}

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