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# looking for an equation...

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Intern
Joined: 13 Mar 2018
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13 Mar 2018, 02:17
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...
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Re: looking for an equation...  [#permalink]

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13 Mar 2018, 08:32
turlogh wrote:
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...

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.
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Re: looking for an equation...  [#permalink]

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14 Mar 2018, 00:44
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!
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Re: looking for an equation...  [#permalink]

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23 Mar 2018, 08:28
turlogh wrote:
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...

What you're looking for is information about consecutive integers.

Specifically, you want to know how many multiples of 5 there are from 80,005 to 200,000,000, inclusive. (Bonus question: why did I say 80,005 and not 80,000?)

To count how many numbers are in a range, you have to do four things:

1. Check your endpoints. Make sure both endpoints are divisible by the number you're counting up by. In this case, you want to make sure your endpoints are divisible by 5. Both 80,005 and 200,000,000 are divisible by 5, so you're good.

2. Find the difference. 200,000,000-80,005 = 199,919,995.

3. Divide the difference by the number you're counting up by. 199,919,995/5 = 39,983,999.

4. Add one, then you're done! 39,983,999 + 1 = 39,984,000.

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Re: looking for an equation...   [#permalink] 23 Mar 2018, 08:28
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