SOLUTION: A positive integer is 14 more than 21 times another. Their product is 6307. Find the two integers.

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Question 1008565: A positive integer is 14 more than 21 times another. Their product is 6307. Find the two integers.

Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
THE UNEXPECTED SOLUTION:
6307=7%2A17%2A53 has just 3 prime factors,
each with an exponent of green%281%29 in the prime factorization.
Therefore 6307 has
factors,
making the 8%2F2=4 pairs of factors listed below.
1%2A6307=6307
7%2A901=6307
17%2A371=6307
53%2A119=6307 .
Obviously 119 is a lot less than 21 times 53 ,
and 901 is a lot more than 14 more than 21 times 7 ,
so the answer must be highlight%2817%29 and highlight%28371%29 .
Let's check, just in case it could be a trick question:
21%2A17%2B14=357%2B14=371 , so it works.

THE EXPECTED SOLUTION:
x= the fist (smaller) number (positive integer) they are talking about
21x%2B14= the other number = 14 more than 21 times the first number
x%2821x%2B14%29=6307= the product of the two numbers
We need to solve the quadratic equation
x%2821x%2B14%29=6307<-->21x%5E2%2B14x=6307
It can be simplified by dividing both sides of the equal sign by 7 ;
21x%5E2%2B14x=6307<-->%2821x%5E2%2B14x%29%2F7=6307%2F7<-->3x%5E2%2B2x=901<-->3x%5E2%2B2x-901=0
Solving by factoring:
We look for factors of 3%2A%28-901%29=-3%2A17%2A53=-%2851%2A53%29 that add up to 2 .
53 and -51 add up to 2 , so we re-write the equation as
3x%5E2-51x%2B53x-901=0 and factor by parts
%283x%5E2-51x%29%2B%2853x-901%29=0-->3x%28x-17%29%2B53%283x-17%29=0-->%283x%2B53%29%28x-17%29=0-->system%283x%2B53=0%2C%22or%22%2Cx-17=0%29-->system%28x=-53%2F3%2C%22or%22%2Cx=17%29 .
Since x was a positive integer, x=-55%2F3 is not a solution to the problem.
x=highlight%2817%29 and 21x%2B14=21%2A17%2B14=357%2B14=highlight%28371%29 are the two integers.