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TargetMBA007
Joined: 22 Nov 2019
Last visit: 07 Dec 2024
Posts: 271
Given Kudos: 216
Schools: Stanford (S)
GPA: 4.0
12 Mar 2020, 21:04
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Hi Guys,
With the equation like this, if we are looking to find the value of N, we can always prime factorize and test different combinations to see what would give us 1890 when multiplied and 3 when subtracted, but I find with bigger numbers this process can take a bit of time as it requires some trial and error. I wondered if there is a more "efficient" approach when dealing with bigger numbers like this, that would quickly tell us which factors are to be used?
\(N^2+3n-1890\)
Thanks
With the equation like this, if we are looking to find the value of N, we can always prime factorize and test different combinations to see what would give us 1890 when multiplied and 3 when subtracted, but I find with bigger numbers this process can take a bit of time as it requires some trial and error. I wondered if there is a more "efficient" approach when dealing with bigger numbers like this, that would quickly tell us which factors are to be used?
\(N^2+3n-1890\)
Thanks
13 Mar 2020, 00:20
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TargetMBA007
Hi Guys,
With the equation like this, if we are looking to find the value of N, we can always prime factorize and test different combinations to see what would give us 1890 when multiplied and 3 when subtracted, but I find with bigger numbers this process can take a bit of time as it requires some trial and error. I wondered if there is a more "efficient" approach when dealing with bigger numbers like this, that would quickly tell us which factors are to be used?
\(N^2+3n-1890\)
Thanks
With the equation like this, if we are looking to find the value of N, we can always prime factorize and test different combinations to see what would give us 1890 when multiplied and 3 when subtracted, but I find with bigger numbers this process can take a bit of time as it requires some trial and error. I wondered if there is a more "efficient" approach when dealing with bigger numbers like this, that would quickly tell us which factors are to be used?
\(N^2+3n-1890\)
Thanks
I deal with them as follows
\(N^2+3n-1890\)
I.e. \(N^2+3n = 1890\)
I.e. \(N(N+3) = 1890\)
I.e. Product of two numbers separated by 3 resulting in 1890
Also, One of them must be a multiple of 5 because result is divisible by 5
Let's take approximate square root of 1890 which is approx 42
so the possibility may be 42*45 which fits well
So N may be 42 or -45
I hope this helps!!!
18 Mar 2020, 08:35
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I found this to be useful for solving equations with coefficients, just wanted to share it incase anyone is struggling with this like me
