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fun with number properties

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by pathaniaus » Thu Jun 04, 2009 7:44 am
For any positive integer n, the sum of the first n positive integers equals (N(N+1))/2. What is the sum of all even integers between 99 and 301?

A) 10,100
B)20,200
C)22,650
D)40,200
E)45,150
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Source: — Problem Solving |

by p2pg » Thu Jun 04, 2009 10:44 am
Is it A?
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by Osirus@VeritasPrep » Thu Jun 04, 2009 11:10 am
Sum = N [(first + last)/2]

sum = 101 (100 + 300)/2 = 20,200

N = the total number in the series

first = the first number in the series

last = last number in the series

I started with 100 because its the first even number and ended with 300 because its the last even number. There are 101 total even numbers
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by pathaniaus » Thu Jun 04, 2009 12:55 pm
that's correct. but can u explain in more detail please? im looking at ur method and the book's but the book has got a completely different way of doing it and urs is making more sense to me..
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by Svedankae » Thu Jun 04, 2009 2:11 pm
osirus0830 wrote: There are 101 total even numbers

How do you know that? I thought there is a 100 ... then ended up at 20,000 and then realized that there have to be a 101 numbers. But how do you know that instantly?
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by Osirus@VeritasPrep » Thu Jun 04, 2009 4:17 pm
Svedankae wrote:
osirus0830 wrote: There are 101 total even numbers

How do you know that? I thought there is a 100 ... then ended up at 20,000 and then realized that there have to be a 101 numbers. But how do you know that instantly?
(301-99)/2
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by Osirus@VeritasPrep » Thu Jun 04, 2009 4:19 pm
pathaniaus wrote:that's correct. but can u explain in more detail please? im looking at ur method and the book's but the book has got a completely different way of doing it and urs is making more sense to me..
Its just memorization, believe me I'm not a math whiz or anything...lol. That is the formula for solving the sum of an arithmetic series.
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by jba05d » Thu Jun 04, 2009 6:25 pm
Pathaniaus and Svedankae,

When working with consecutive series problems (such as this one) there are a couple of important concepts that you should memorize. Firstly, in any sequence of consecutive numbers or consecutive multiples the number of the digits in the series is equal to (First Term - Last Term)/Increment + 1. Therefore, if you want to know how many consecutive numbers are from 1-100 inclusive simply calculate the value using the formula (100 - 1)/1 + 1 = 100. Similarly, if you want to figure out how many even numbers are from 1-100 simply use the formula (100-2)/2 + 1 = 50. As another example if you wanted to know, say, how many multiples of 15 there were from 1-100 simply identify the first multiple of 15 in that series (i.e 15) and the last multiple in that series (i.e 90) and use the formula (90-15)/15 + 1 = 6. Therefore, there are 6 multiples of 15 from 1-100. This method works for counting every series of consecutive numbers and multiples.

Secondly, in any series of consecutive numbers or consecutive multiples the average/mean of the series is easily determined by taking the average of the first and last numbers. For example, if you want the average of all the numbers from 1 to 100 inclusive simply take the average of 100 and 1 (i.e. (100+1)/2 = 50.5). Similarly, the average of all the even integers from 1-100 inclusive is (2+100)/2 = 51

Using these two principles and your knowledge of averages you can essentially answer any question involving the summation of consecutive numbers and consecutive multiples. For example, we know the average of any series is equal to the sum of the series divided by the number of items in the series (i.e. Sum of Items/Number of Items = Average). Conversely, the the sum of items in a series is equal to the number of items in the series by the average of the series. Therefore, using the two principles that were previously discussed we can find the sum of any consecutive series fairly rapidly. For example, if we wanted to determine the sum of all multiples of 3 between 300 and 900 we would calculate the numbers of numbers in the series using the counting formula i.e. ((900-300)/3 + 1 = 201). Then we would calculate the average by taking the average of the first and last term (i.e. (300+900)/2 = 600. Then to determine the sum we would simply multiply the number of items in series (i.e 201) by the average (i.e. 600), doing so yields 120,600. Therefore, the sum of all multiples of 3 from 300-900 is 120,600.

Hope that helps!
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