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The sum of the first 'n' consecutive positive integers is a perfect 20 Sep 2019, 10:56
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!

Difficulty:

95% (hard)

Question Stats:

35% (02:38) correct 65% (02:31) wrong based on 157 sessions

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The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6
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A
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chetan2u
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The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 06:13
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Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6
Show more


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
49 and 50
Thus answer is 3, when n is 1, 8 and 49.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)
\(\frac{49*50}{2}=49*25=(7*5)^2\)

B
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The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 11:20
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chetan2u
Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
Thus answer is 2.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)

A
Show more

Just a little doubt here Sir, if n=49, sum will be 49*50/2=49*25 which is a perfect square(square of 35), shouldn't the answer be 3(total 3 values of n).

Please let me know where am I wrong?
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The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 21:38
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DonVito
chetan2u
Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
Thus answer is 2.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)

A

Just a little doubt here Sir, if n=49, sum will be 49*50/2=49*25 which is a perfect square(square of 35), shouldn't the answer be 3(total 3 values of n).

Please let me know where am I wrong?
Show more

I was also in a fix what I am getting wrong about.
I marked A since I missed a little insight that its not Sum(n) < 100 but n < 100.
The catch is that after dividing n(n+1) by 2 both the digits(even one) should be a perfect square

However, after the expert solution by chetan2u i figured out that 49 and 50 do make up the solution but why 'A'. He corrected it now.
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The sum of the first 'n' consecutive positive integers is a perfect 21 Sep 2019, 22:02
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DonVito
chetan2u
Hovkial
The sum of the first 'n' consecutive positive integers is a perfect square where n < 100. How many values of 'n' are possible?

(A) 2

(B) 3

(C) 4

(D) 5

(E) 6


Sum of the first 'n' consecutive positive integers = \(\frac{n(n+1)}{2}\), so this should be a perfect square...

So, let us see what can we make out of term \(\frac{n(n+1)}{2}\) --- This means product of TWO consecutive integers, n and n+1, should be twicw of a perfect square..
BUT two consecutive term are co-prime, that is they do not have any common factors, and one of the two n or n+1 will be even.

This further tells us that one of the two n or n+1 will be a PERFECT square and other will be TWO times of another perfect square.
1) Possible values of n or n+1, which is a perfect square, are 1, 4, 9, 16, 25, 36, 49, 64, 91....
2) And possible values of the second, which is twice of the perfect square, are 2, 8, 18, 32, 50, 72 or 98....

Let us check if there are two consecutive numbers in both the list..
1 and 2
8 and 9
Thus answer is 2.
\(\frac{1*2}{2}=1=1^2\)
\(\frac{8*9}{2}=4*9=36=6^2\)

A

Just a little doubt here Sir, if n=49, sum will be 49*50/2=49*25 which is a perfect square(square of 35), shouldn't the answer be 3(total 3 values of n).

Please let me know where am I wrong?
Show more

Yes, you are correct.
Although my solution did include 49 and 50 as one of possibilities, i did not include that by mistake..
Thanks
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The sum of the first 'n' consecutive positive integers is a perfect 11 Jul 2024, 02:51
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