If the sum of n consecutive positive integers, where n is greater than

possible values = ( 3-12) ; ( 10-15) ; (13-17) ; (24-26); ( 37-38)
total = 10+6+5+3+2 ; 26
IMO D


If the sum of n consecutive positive integers, where n is greater than 1, is 75. What is the sum of all possible values of n?

A. 5
B. 10
C. 16
D. 26
E. 50

Sum of the First Consecutive Integers from 1 to n: n(n+1)/2
Sum of Consecutive Integers: a+(a+1)+(a+2)…

a=2: 75=a+(a+1)…74=2a…a=37…valid
a=3: 75=a+(a+1)+(a+2)…3a=72…a=24…valid
a=4: 75=4a+3(4)/2…4a=75-6…a≠integer
a=5: 75=5a+4(5)/2…5a=65…a=integer…valid
a=6: 75=6a+5(6)/2…6a=60…a=integer…valid
a=7: 75=7a+6(7)/2…7a=75-21=54≠integer
a=8: 75=8a+7(8)/2…8a=75-even=odd≠integer
a=9: 75=9a+8(9)/2…9a=75-36=39≠integer
a=10: 75=10a+9(10)/2…a=integer…valid
a=11: 75=11a+10(11)/2…a≠integer
a=12: 75=12a+11(12)/2…12a=75-even=odd≠integer
a=13: 75=13a+12(13)/2…13a=negative

Possible a's = 2+3+5+6+10 = 26

Ans (D)

2,3,5,6 are the possible consecutive integers whose sum equals to 75

2- 37,38
3- 24,25,26
5- 13,14,15,16,17
6- 10,11,12,13,14,15

so 2+3+5+6=16

OA:C

If the sum of n consecutive positive integers, where n is greater than 1, is 75. What is the sum of all possible values of n?

A. 5
B. 10
C. 16
D. 26
E. 50
For any consecutive integers n has to be at least two or more.
n >= 2 where all numbers are positive and sum of all integers is 75.

Sum of 2 consecutive numbers = x + (x+1) = 2x + 1, x = 37
Sum of 3 consecutive numbers = x + (x+1) + (x+2) = 3x + 3, x = 24
Sum of 4 consecutive numbers = x + (x+1) + (x+2) + (x+3) = 4x + 6, x = 17.25
Sum of 5 consecutive numbers = x + (x+1) + (x+2) + (x+3) + (x+4) = 5x + 10, x = 13
Sum of 6 consecutive numbers = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) = 6x + 15, x = 10
Sum of 7 consecutive numbers = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) = 7x + 21, x = 7.2
Sum of 8 consecutive numbers = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) = 8x + 28, x = 4.5
Sum of 9 consecutive numbers = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) + (x+8) = 9x + 36, x = 4.3
Sum of 10 consecutive numbers = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) + (x+8) + (x+9) = 10x + 45, x = 3
Sum of 11 consecutive numbers = x + (x+1) + (x+2) + (x+3) + (x+4) + (x+5) + (x+6) + (x+7) + (x+8) + (x+9) + (x+10) = 11x + 55, x = 1.8

Since x can't be less than 1, 12 consecutive integers are not possible as it gives x < 1.

Leaving aside the non-integers values of x possible values of n are 2,3,5,6,10 which sum to 26.

Answer D.

hit & trial:
1) n=2, yes(37&38)
2) n=3, yes(24,25&26)
3) n=4, lets check n+n+1+n+2+n+3=75, 4n+6=75, 4n=69, n has to be integer.
4) n=5, 5n+6+4=75, 5n=75-10=65, yes
5) n=6, 6n+10+5=75, 6n=60, yes
6) n=7, 7n+15+6=75, 7n=75-21=54, no
7) n=8, 8n+21+7=75, 8n=75-28=47, no
8) n=9, 9n+28+9=75, 9n=75-36=39, no
no need to go further, possible values of n are 2,3 5, &6: total : 16.

Please suggest a better method to do this

We are given that the sum of n consecutive positive integers greater than 1 is 75. We are to determine the sum of all possible values of n.

We have an arithmetic progression with a common difference of 1, and the first term greater than 1

Generically, the sum, Sn = (a1+an)*n/2
Now Sn=75, and Let a be the first term, then the last term is a+n, where n is the number of terms
so 75=(a+a+n)n/2
150=2an+n^2
Hence n^2 + 2an - 150 = 0 ..........................(1)
We need sets of two pairs of positive integers whose product is 150
These pairs are: 2,75; 3,50; 5,30; 6,25; 10,15.
Substituting each of these pairs into (1) above, we are able to come out with the positive roots of the equation and these positive roots correspond to the smaller value of the integer pairs.
Hence the possible values of n=2, 3, 5, 6, and 10.
Summing the possible values of n=2+3+5+6+10=26

The answer is, therefore, option D.

If the sum of n consecutive positive integers, where n is greater than 1, is 75. What is the sum of all possible values of n?
—> 37+38= 75 (2)

—> a+a+1+a+2= 75
3a = 72
a= 24–> 24,25,26 (3)

—> 4a+ 3+3= 75
4a= 69
a should be integer

—> 5a+ 6+ 4= 75
5a= 65
a= 13 —> 13,14,15,16,17 (5)

—> 6a+ 10+5= 75
6a= 60
a= 10 —> 10,11,12,13,14,15 (6)

—> 7a+ 15+6= 75
7a= 54
a should be integer

—> 8a+ 21+ 7= 75
8a = 47
a should be integer

—> 9a+ 28+ 8= 75
9a= 39
a should be integer

—> 10a + 36+9= 75
10a = 30
a= 3 —> 3,4,5,6,7,8,9,10,11,12 (10)

—> 11a+ 45+ 10= 75
11a= 20
a should be integer

In total, 2+3+5+6+10= 26
The answer is. D

Posted from my mobile device

The first number indicates the number of terms in that sequence ........ how can we say that the first number indicates so.......why not the other factors 15,25.....????

The best way to test is to use the least of the second numbers.
In this case, that number is 15.
The sum of 15 consecutive integers from 1to 15 = 15*16/2= 15*8 = 120.
Automatically all other second numbers cannot yield 75 since even the smallest is more than 75. That’s the easiest way.

Alternatively, these number pairs are the root of the equation n^2 + an - 150=0
In order to get an>0, in the formulation of the equation, you will negate the smaller number and keep the larger number as positive. Doing so will yield two roots of which the smaller number will be positive while the larger will be negative. We know that n cannot be negative, hence n must therefore be the positive root which in this case is the smaller number. I will illustrate with 10,15
n^2 + 15n - 10n - 150 = 0
n(n+15) - 10(n+15) = 0
(n-10)(n+15)=0
n=10 and n=-15
Hence n has to be 10.

Posted from my mobile device

eakabuah
Can you shed more light on red part.

You would realize that by the use of the sum of n terms in an arithmetic progression, we came up with a quadratic equation: n^2 + 2an - 150 = 0
What is interesting about this equation is that it is expected to yield two roots, and you will notice that the nature of these roots depends on the value of 2a. Our focus is not to determine a, which is the first term in the series but rather to determine n. So, for simplicity, you can even replace 2a with b or any variable.
You are now left with the task of guessing the values whose sum equal b and whose product equals -150. The best way to take guesses is to find find the factors of 150.
Factors of 150 are: {1, 2, 3, 5, 6, 10, 15, 25, 30, 50, 75, 150}.
The two numbers whose sum equals b and product equal -150 are definitely among the set of numbers above.
Note that n>1, hence 1, 150 pair is not part of these pairs. So, you are left with 2,75; 3, 50; 5,30; 6,25; and 10,15. So that's how the pairs came about.

Thanks for your prompt reply.
I now understand it and its better than how i solved. How do i give more kudos..!!
While solving the thought that i am doing it wrong always grappled me. Initially, i took few steps in this manner but on the next step faltered, may be because i didn't pay much attention.

So, i switched to time taking method. First, since 75 is sum of consecutive integers n must be two or more(n>1 given also) as per average formula. If i divide 75 by 2 then i shall reach those consecutive integers and thus 37,38 resulted. For three numbers 75/3 = 25 and thus 24,25,26. Next for for consecutive integers 75/4 = 18.75 but 0.75 distributed among four integers would not give integers, hence four consecutive integers series not possible. Similarly, i arrived at 5, 6 and 10. However, for 11 consecutive non-integers result and thus series not possible.

This way i did it but midway i had second thoughts about going ahead.

Once again the way you did it is awesome.. Kudos.!!

Hi,

You have written that number of consecutive numbers can only be odd where as n takes even values as well here.

Also, why did you factorize 75 when apparently it is not being used.

Hi,

Why did you find the number of odd factors when we also have n=2, 6 and 10 i.e even number of consecutive terms.

altairahmad

The first thing the poster did was find the total count of ways that the Number 75 can be expressed as the Sum of N Consecutive Integers.

Generally, the number of ways of writing some constant M (here 75) as the SUM of N Consecutive Integers can be found by finding the Number or Odd Factors of M (here 75) EXCEPT the odd factor of 1

75 = (3) (5)^2

Total no. of odd factors = (1 + 1) * (2 + 1) = 2 * 3 = 6

6 - (1) = 5

There should be 5 ways to write 75 as the SUM of 2 or more consecutive positive integers.


Further, a good approach is the following:

M (here 75) can be expressed as the SUM of N consecutive integers if and only if either:

(1) the N consecutive integers is an ODD count, then: (M) / (N) must = Integer

Or

(2) if the N consecutive integers is an EVEN Count then: (M) / (N) + (1/2) must = Integer

The key constraint to avoid having zero or negative integers in your SUM is:

(M/N) >/= (N/2)—— otherwise we still start involving negative numbers along with zero

The cut off is when N = 13 consecutive positive integers

(75/13) < (13/2)

So we are looking for N to be any number from 2 through and including 12


Ignoring the “mathy” approach, I think the best way of approaching this problem is to form an organized “hit and run” approach.


I believe some people expressed this method above, but say you want to express 75 as the sum of 2 consecutive positive integers:

(75/2) = 37.5

37.5 would be the Median of the 2 consecutive integers. The 2 consecutive integers would then “sandwich” this value of 37.5 in the middle as the Median.

36 + 37 = 75

If you combine this approach with the knowledge that there are 5 ways to write 75 as the sum of consecutive positive integers (as found above), you can keep an organized trial approach until you find the 5th way.

Also, we know that in a set of consecutive positive integers, if there is an Even Count of terms, the Median must take the form of ——> XXX . 5

If there is an Odd count of terms in a set of consecutive positive integers, then the median must be one of the terms in the set and thus an Integer.

75/3 = 25 = Integer —— can do it

24 + 25 + 26 = 75

So far N = 2 and 3 works


75/4 = not of the form XX.5, so we won’t be able to make a sequence of positive consecutive integers around the result


75/5 = 15 —— Integer, so will work

13 + 14 + 15 + 16 + 17 = 75

N = 2, 3, and 5 work thus far - we have two more to find


75/6 = 12.5

10 + 11 + 12 + 13 + 14 + 15 = 75

N = 2, 3, 5, and 6 work - we have one more to find


75/7 ——> not an integer

75/8 ——> not of form XX . 5

75/9 —-> not an integer

75/10 = 7.5

3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 = 75

You can see at this point even if we went much further, we will start having to involve 0 and/or negative numbers in the set of consecutive positive integers.

The five ways to write 75 as the sum of consecutive positive integers are:

2—3—5—6—10

Sum of these = 2 + 3 + 5 + 6 + 10 = 26

Answer: 26





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