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13 Mar 2018, 02:17
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I need to know to figure out how many times you can increase a number by itself+5 until you reach a certain total
lets say you start at x=80,000 how many times would it take to reach y=200,000,000 by 80,000+80,005+80,010 etc
so I want to know how many additions of x=x+5 it takes to reach y so I can do x=x+(z*5) if z is the total count of these additions
hope I typed this out clearly enough...
lets say you start at x=80,000 how many times would it take to reach y=200,000,000 by 80,000+80,005+80,010 etc
so I want to know how many additions of x=x+5 it takes to reach y so I can do x=x+(z*5) if z is the total count of these additions
hope I typed this out clearly enough...
13 Mar 2018, 08:32
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turlogh
Let's say you need to go from X to Y. With additions of (x+5) at a time... You are asking how many such additions you need . I hope I got it right.
First of all... You need to add an amount = Y-X
Let's say you need n additions... What is n equal to?
NX + 5 + 10 + ... + 5N
NX + 5( 1 + 2 + ... + N) = Y - X
X ( N+1) + 5 ( an AP who's sum equals N * ( N+1) /2) = Y
X(N+1) + 5*N*(N+1)/2 =Y
(N+1) (2X + 5N) /2 =Y
I suppose you have the values of Y and X.
So this becomes a quadratic in N.
Solve for N.
Hope this helped.... Please give Kudos if you liked my post.
14 Mar 2018, 00:44
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Hi turlogh
So, basically what you are talking about is an AP and you need to find the
the total number of elements in the AP
Formula for an AP: An = A1 + (n – 1)d
where
The first term is A1
the common difference is d
the number of terms is n
For this problem(as per your description)
An = 200,000,000
A1 = 80000
d = 5, you substitute the values and find the value of n!
Hope this helps you!
So, basically what you are talking about is an AP and you need to find the
the total number of elements in the AP
Formula for an AP: An = A1 + (n – 1)d
where
The first term is A1
the common difference is d
the number of terms is n
For this problem(as per your description)
An = 200,000,000
A1 = 80000
d = 5, you substitute the values and find the value of n!
Hope this helps you!
