75th Problem 2016

Algebra Level 1

Find the sum of 3 consecutive positive odd integers such that the product of the first and third is 4 less than 7 times the second.

Check out the set: 2016 Problems


The answer is 21.

Angela Fajardo by Angela Fajardo , 19, Philippines

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1 solution

Angela Fajardo
Mar 23, 2016

Let:

x = 1 s t n u m b e r x + 2 = 2 n d n u m b e r x + 4 = 3 r d n u m b e r x=1st\quad number\\ x+2=2nd\quad number\\ x+4=3rd\quad number\\

( x ) ( x + 4 ) = 7 ( x + 2 ) − 4 x 2 + 4 x = 7 x + 14 − 4 x 2 + 4 x = 7 x + 10 x 2 − 3 x = 10 x 2 − 3 x − 10 = 0 ( x − 5 ) ( x + 2 ) = 0 x = 5 ∣ x = − 2 \left( x \right) \left( x+4 \right) =7\left( x+2 \right) -4\\ { x }^{ 2 }+4x=7x+14-4\\ { x }^{ 2 }+4x=7x+10\\ { x }^{ 2 }-3x=10\\ { x }^{ 2 }-3x-10=0\\ \left( x-5 \right) \left( x+2 \right) =0\\ x=5\quad |\quad x=-2

Since we are only looking for positive integers:

x = 5 x + 2 = 5 + 2 = 7 x + 4 = 5 + 4 = 9 5 + 7 + 9 = 21 \\ x=5\\ x+2=5+2=7\\ x+4=5+4=9\\ \\ 5+7+9=21\\ \\

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