Let the 4 odd integers be of the form 2n-3, 2n-1, 2n+1, 2n+3. Then the sum = 8n. divide it with 10 and you get 4n/5.

From the question, 4n/5 = Square of a number X

Since 4 is a square of a number, n/5 should also be a square of a number i.e n/5 = a^2 (where a is an integer)

n = 5 * a^2

Substitute

a = 1, n = 5, 2n-3 = 7, 2n-1 = 9, 2n+1 = 11, 2n+3 =13
a = 2, n = 20, 2n-3 = 37, 2n-1 = 39, 2n+1 = 41, 2n+3 = 43
a = 3, n = 45, 2n-3 = 87, 2n-1 = 89, 2n+1 = 91, 2n+3 = 93

so the answer is 41, Option C

Correct me if I am wrong !

GmatKiss wrote:The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73

Perfect squares = 1,4,9,16,25,36...

Each multiplied by a factor of 10:
10,40,90,160,250,360...

The sum of the 4 integers is contained in the list above.
The sum of consecutive odd integers is a multiple of 4.
Of the values in the list above, only the following are multiples of 4:
40,160,360.

If the sum = 40, then the average = 40/4 = 10.
If the sum = 160, then the average = 160/4 = 40.
If the sum = 360, then the average = 360/4 = 90.

The numbers in the answer choices are too great to yield an average value of 10 and too small to yield an average value of 90.

Thus, the needed average is 40, implying that the correct answer is C:
37+39+41+43 = 160.
160/10 = 16, a perfect square.

The correct answer is C.

I used a crude method to this one. Started directly with the options ..

21 and 25 can be tested quickly because they are very close

the key thing is that the sum of odd numbers is divisible by 10. That means the sum of units digits should give us a ZERO.

I basically used the units digits only and listed the numbers till 45. [30 secs]

19 21 23 25 27... 35 37 39 41 43..

luckily, I was able to find the numbers -- sum of unis digits of 37 39 41 43 [ 7+3+9+1 = 20 ] hence I selected this set. Sum = 160 and 16 is a perfect square.

Hence, correct answer OPTION C

Let the no be : 10 X +Y , {10 X+Y+2}, {10 X+Y+4}, {10 X+Y+6}

Sum of 4 consecutive 4 odd integers will be 40 X+ 4Y +12.

As per the stem, sum divided by 10 will be a perfect square.

Let's list down the perfect squares : 1,4,9,16,25 ...

SO the sum will be one among 10,40,90,160,250..... { plz note I have multiplied the perfect sqaures with 10).

We know that Y can be of 1,3,5,7,9 ( as Y is a odd integer)

when X = 3, Y = 7 , the sum becomes 160.

So the odd integer is 37.

Remaining 3 integers will be 39,41,43.

We can see that 41 is one among the options pick C.

GMATGuruNY wrote:

GmatKiss wrote:The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73

Perfect squares = 1,4,9,16,25,36...

Each multiplied by a factor of 10:
10,40,90,160,250,360...

The sum of the 4 integers is contained in the list above.
The sum of consecutive odd integers is a multiple of 4.
Of the values in the list above, only the following are multiples of 4:
40,160,360.

If the sum = 40, then the average = 40/4 = 10.
If the sum = 160, then the average = 160/4 = 40.
If the sum = 360, then the average = 360/4 = 90.

The numbers in the answer choices are too great to yield an average value of 10 and too small to yield an average value of 90.

Thus, the needed average is 40, implying that the correct answer is C:
37+39+41+43 = 160.
160/10 = 16, a perfect square.

The correct answer is C.

Thanks for your explanation. I follow you except for this statement: "The sum of consecutive odd integers is a multiple of 4." I thought that the sum of n consecutive integers was not divisible by n if n is even. Since 4 is even, shouldn't the sum of the consecutive odd integers not be divisible (and therefore be a multiple) of 4?

Or does it have something to do with the statement 2n+1 = odd integer? Sorry I am not grasping this easily

Let the four 2-digit odd numbers be
n-3 n-1 n+1 n+3

Sum of the 4 numbers ==> 4n

acc to qn, when the sum is divided by 10
we get a perfect square...
perfect squares include==>1,4,9,16,25,36,49,.....

Possible values of 4n/10 ==> 4, 16,36...

If 4n/10=4
n=10
Hence, the corresponding nos are 7,9,11,13(all of which are NOT 2-digit nos)


If 4n/10=16
n=40
Hence the corresponding nos are 37, 39, 41, 43

If 4n/10=36
n=90
Hence the corresponding nos are 87, 89, 91, 93


The answer to the ques therefore is Option C

qisma wrote:

GMATGuruNY wrote:

GmatKiss wrote:The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73

Perfect squares = 1,4,9,16,25,36...

Each multiplied by a factor of 10:
10,40,90,160,250,360...

The sum of the 4 integers is contained in the list above.
The sum of consecutive odd integers is a multiple of 4.
Of the values in the list above, only the following are multiples of 4:
40,160,360.

If the sum = 40, then the average = 40/4 = 10.
If the sum = 160, then the average = 160/4 = 40.
If the sum = 360, then the average = 360/4 = 90.

The numbers in the answer choices are too great to yield an average value of 10 and too small to yield an average value of 90.

Thus, the needed average is 40, implying that the correct answer is C:
37+39+41+43 = 160.
160/10 = 16, a perfect square.

The correct answer is C.

Thanks for your explanation. I follow you except for this statement: "The sum of consecutive odd integers is a multiple of 4." I thought that the sum of n consecutive integers was not divisible by n if n is even. Since 4 is even, shouldn't the sum of the consecutive odd integers not be divisible (and therefore be a multiple) of 4?

Or does it have something to do with the statement 2n+1 = odd integer? Sorry I am not grasping this easily

Let n and n+2 = consecutive odd integers.
Sum = n + (n+2) = 2n+2 = 2(n+1) = even*even = (multiple of 2)(multiple of 2) = multiple of 4.
Thus, the sum of ANY two consecutive odd integers is a multiple of 4.

GMATGuruNY wrote:

qisma wrote:

GMATGuruNY wrote:

GmatKiss wrote:The sum of 4 consecutive two-digit odd integers, when divided by 10, becomes a perfect square. Which of the following can possibly be one of these 4 integers?

A. 21
B. 25
C. 41
D. 67
E. 73

Perfect squares = 1,4,9,16,25,36...

Each multiplied by a factor of 10:
10,40,90,160,250,360...

The sum of the 4 integers is contained in the list above.
The sum of consecutive odd integers is a multiple of 4.
Of the values in the list above, only the following are multiples of 4:
40,160,360.

If the sum = 40, then the average = 40/4 = 10.
If the sum = 160, then the average = 160/4 = 40.
If the sum = 360, then the average = 360/4 = 90.

The numbers in the answer choices are too great to yield an average value of 10 and too small to yield an average value of 90.

Thus, the needed average is 40, implying that the correct answer is C:
37+39+41+43 = 160.
160/10 = 16, a perfect square.

The correct answer is C.

Thanks for your explanation. I follow you except for this statement: "The sum of consecutive odd integers is a multiple of 4." I thought that the sum of n consecutive integers was not divisible by n if n is even. Since 4 is even, shouldn't the sum of the consecutive odd integers not be divisible (and therefore be a multiple) of 4?

Or does it have something to do with the statement 2n+1 = odd integer? Sorry I am not grasping this easily

Let n and n+2 = consecutive odd integers.
Sum = n + (n+2) = 2n+2 = 2(n+1) = even*even = (multiple of 2)(multiple of 2) = multiple of 4.
Thus, the sum of ANY two consecutive odd integers is a multiple of 4.

I see now. Thank you!

I like GMATGuruNY and qisma's methods because they are simple and quick. Algebra is effective, but requires more steps and takes longer for me.