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

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If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 00:36
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post Updated on: 25 Nov 2019, 03:11
1
Note that, the number of consecutive numbers can only be Odd, as sum of even consecutive integers is always even

\(75 = 3*5^2\)
--> Factors of \(75\) = {\(1, 3, 5, 15, 25, 75\)}

Possible cases:
1. If n = 2, Set of integers = {\(37, 38\)}

2. If n = 3, Set of integers = {\(24, 25, 26\)}

3. If n = 5, Set of integers = {\(13, 14, 15, 16, 17\)}

4. If n = 6, Set of integers = {\(10, 11, 12, 13, 14, 15\)}

5. If n = 10, Set of integers = {\(3, 4, 5, 6, 7, 8, 9, 10, 11, 12\)}

Sum of possible values of n = 2 + 3 + 5 + 6 + 10 = 26

IMO Option D

Originally posted by Dillesh4096 on 25 Nov 2019, 02:11.
Last edited by Dillesh4096 on 25 Nov 2019, 03:11, edited 1 time in total.
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 02:23
1
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
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If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post Updated on: 26 Nov 2019, 03:24
1
Quote:
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)

Originally posted by exc4libur on 25 Nov 2019, 04:40.
Last edited by exc4libur on 26 Nov 2019, 03:24, edited 1 time in total.
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 05:23
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

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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 06:41
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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.
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 09:55
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
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 09:58
1
75 = \(3 X 5^2\)
no of odd factors = (2+1)(1+1)-1=5
Total five ways we can write 75 as sum of consecutive numbers
But question ask the value of sum of n

\((n)*(2a + (n-1))\) = 150 150 can be written as

2 X 75
3 X 50
5 X 30
6 X 25
10 X 15
The first number indicates the number of terms in that sequence example 3 x 50 implies 150 /3 = 25 implies 24+25+26
Therefore 2 + 3 + 5 + 6 + 10 = 26
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 14:34
1
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.
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 25 Nov 2019, 15:47
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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

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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 26 Nov 2019, 00:13
CaptainLevi wrote:
75 = \(3 X 5^2\)
no of odd factors = (2+1)(1+1)-1=5
Total five ways we can write 75 as sum of consecutive numbers
But question ask the value of sum of n

\((n)*(2a + (n-1))\) = 150 150 can be written as

2 X 75
3 X 50
5 X 30
6 X 25
10 X 15
The first number indicates the number of terms in that sequence example 3 x 50 implies 150 /3 = 25 implies 24+25+26
Therefore 2 + 3 + 5 + 6 + 10 = 26




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.....????
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 26 Nov 2019, 01:08
madgmat2019 wrote:
CaptainLevi wrote:
75 = \(3 X 5^2\)
no of odd factors = (2+1)(1+1)-1=5
Total five ways we can write 75 as sum of consecutive numbers
But question ask the value of sum of n

\((n)*(2a + (n-1))\) = 150 150 can be written as

2 X 75
3 X 50
5 X 30
6 X 25
10 X 15
The first number indicates the number of terms in that sequence example 3 x 50 implies 150 /3 = 25 implies 24+25+26
Therefore 2 + 3 + 5 + 6 + 10 = 26




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.

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If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 26 Nov 2019, 01:15
eakabuah wrote:
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.

eakabuah
Can you shed more light on red part.
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 26 Nov 2019, 01:55
1
lnm87 wrote:
eakabuah wrote:
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.

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.
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Re: If the sum of n consecutive positive integers, where n is greater than  [#permalink]

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New post 26 Nov 2019, 05:33
eakabuah wrote:
lnm87 wrote:
eakabuah wrote:
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.

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.!!
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Re: If the sum of n consecutive positive integers, where n is greater than   [#permalink] 26 Nov 2019, 05:33
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