In a set of 10 consecutive odd positive integers, the product ........

Question

In a set of 10 consecutive odd positive integers, the product of the least and the greatest integers is 63? What is the arithmetic mean(average) of the set?

A.  -12
 B.  0
 C.  8
 D.  12
 E.  30

To read all our articles: Must read articles to reach Q51

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Quant-workshop_Webinar-e1533628475738.jpg

Archit3110 · Answer

set of integers " ( 3,5,7,9,11,13,15,17,19,21)
product 21*3 = 63 and mean = 11+13/2 = 12 
IMO D

gracie · Answer

let x=least term
x(x+18)=63
x=3
x+18=21=greatest term
3+21=24
24/2=12 mean
D

hibobotamuss · Answer

Can someone break this down?

Darshi04 · Answer

Detailed Explanation and Approach: 
Problems like these where difference between 2 terms of any 2 elements is the same and we are asked to find the sum, can be solved using principles of sequences (arithmetic sequence)
Since we want the average, we can find the sum of the 10 numbers and divide that by 10.

The formula for sum of 'n' numbers with common difference 'd' =  \frac{n }{2} 
...where 'a' is first term, 'd' is common difference, 'n' is number of terms.
d=2 (consecutive odd integers have common difference 2)
n=10

How to find "a" (1st term): 
Go back to the question.
Product of 1st and last term is 63.
On factorizing 63, we get 3 * 3 * 7
Product of 1st and last term can be 63 ONLY WHEN first term is 3 and last term is 7*3=21...since with other combinations, the number of terms will never be 10.
So a=3

Plug values of a, d and n in the above formula
We get sum of 10 terms as 120.
Average will be  \frac{120}{10}  =12

Hope this helps.

EgmatQuantExpert · Answer

Solution 
  Given:  
 • A set of 10 consecutive odd positive integers
• The product of the least and the greatest integers = 63

To find:  
 • The arithmetic mean of the set

Approach and Working:   
Let us assume that the 10 consecutive integers are: {a, a + 2, a + 4, a + 6, … , a + 16, a + 18}
 • We are given that a * (a + 18) = 63
 o Implies,  a^2 + 18a – 63 = 0 
o (a – 3) * (a + 21) = 0
o Thus, a = 3 or -21

• Since a is a positive integer, a cannot be equal to -21

Therefore, average of the set = (first term + last term)/2 =  \frac{(a + a + 18)}{2} = a + 9 = 3 + 9 = 12

Hence, the correct answer is option D.

Answer: D

https://e-gmat.com/blogs/wp-content/uploads/2018/08/Free-Session-NP-e1533292073470.jpg

fskilnik · Answer

\left( {{a_1}\,;\,{a_2}\,;\, \ldots \,,\,{a_5}\,;\,{a_6}\,;\, \ldots \,;\,{a_9}\,;\,{a_{10}}} \right)\,\, = \,\,\left( {2M - 9\,;\,2M - 7\,;\, \ldots \,;\,2M - 1\,;\,2M + 1\,;\, \ldots \,;\,2M + 7\,;\,2M + 9} \right)

{\rm{where}}\,\,M \ge 5\,\,{\mathop{\rm int}} \,\,\,\,\left

? = {\rm{average}}\,\,\mathop = \limits^{\left( * \right)} \,\,{\rm{median}}\,\,\mathop = \limits^{\left( * \right)} \,\,{{{a_5} + {a_6}} \over 2} = {{\left( {2M - 1} \right) + \left( {2M - 1} \right)} \over 2} = 2M\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left

63 = \left( {2M - 9} \right)\left( {2M + 9} \right) = {\left( {2M} \right)^2} - {9^2}\,\,\,\,\, \Rightarrow \,\,\,\,{\left( {\,2M} \right)^2} = 144 = {12^2}

\Rightarrow \,\,\,\,\left| {2M} \right| = 12\,\,\,\,\,\mathop \Rightarrow \limits^{M\, > \,0} \,\,\,\,? = 2M = 12

We follow the notations and rationale taught in the  GMATH  method.

Regards, 
Fabio.

SenseiA · Answer

Here's how I solved it:

63 has only a few factors (1,3,7,9,21 and 63), and there are only 3 ways to derive 63 from the product of positive integers.
1*63
3*21
7*9

Only the 3*21 pairing,  \frac{21 - 3}{2}  +1 ,gives 10 integers. 
 Concept: To count the number of terms in an evenly-spaced sequence, the formula is  \frac{Last Term-First Term}{Increment}  +1

From this point, the mean is easy to derive. Formula is  \frac{Last Term + First Term}{2}  i.e.  \frac{21+3}{2}  = 12

Answer is D

MathRevolution · Answer

Total numbers: 10

Product of the least and the greatest of them is 63

Possible combination: 1 * 63  , 7 * 9 , 3 * 21

Average:  \frac{Sum}{Total}

Sum:  \frac{n}{ 2}  *

Sum=  \frac{10}{2}  *   =  \frac{10 }{2}  * 24 = 10 * 12 = 120

Average:  \frac{120 }{ 10}  = 12

Answer D

ScottTargetTestPrep · Answer

Solution:

Since 63 = 1 x 63 = 3 x 21 = 7 x 9, we see that the only possible pair for the least and greatest integers are 3 and 21, respectively, and the set is {3, 5, 7, 9, 11, 13, 15, 17, 19, 21}. Since this is an evenly spaced set, the mean of the set is equal to the average of the least and greatest integers of the set. Therefore, the mean is (3 + 21)/2 = 12.

Answer: D

bumpbot · Answer

Automated notice from GMAT Club BumpBot: 

A member just gave Kudos to this thread, showing it’s still useful. I’ve bumped it to the top so more people can benefit. Feel free to add your own questions or solutions.

 This post was generated automatically.

In a set of 10 consecutive odd positive integers, the product ........

Prep Toolkit

Top 5 GMAT Debrief Videos

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards

GMAT Problem Solving (PS) Questions

In a set of 10 consecutive odd positive integers, the product ........

Prep Toolkit

615 to 715 in 15 Days (Ria's Strategies)

More than Quant/Verbal Abilities: Speed vs. Accuracy

745 with Only Self-Prep and GMAT Club

From 555 to 765: 210-point Improvement

Perfect 805 Score Debrief by Julia

My Rewards