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

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In a set of 10 consecutive odd positive integers, the product ........  [#permalink]

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New post 09 Jan 2019, 01:20
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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

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Re: In a set of 10 consecutive odd positive integers, the product ........  [#permalink]

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New post 09 Jan 2019, 02:18
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

EgmatQuantExpert wrote:
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

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In a set of 10 consecutive odd positive integers, the product ........  [#permalink]

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New post 09 Jan 2019, 18:54
EgmatQuantExpert wrote:
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



let x=least term
x(x+18)=63
x=3
x+18=21=greatest term
3+21=24
24/2=12 mean
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Re: In a set of 10 consecutive odd positive integers, the product ........  [#permalink]

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New post 10 Jan 2019, 01:07
Can someone break this down?
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In a set of 10 consecutive odd positive integers, the product ........  [#permalink]

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New post 10 Jan 2019, 01:36
hibobotamuss wrote:
Can someone break this down?


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' = \(\frac{n[2a + (n-1)d]}{2}\)
...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.
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Re: In a set of 10 consecutive odd positive integers, the product ........  [#permalink]

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New post 11 Jan 2019, 06:05

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

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Number Properties – Even Odd | LCM GCD | Statistics-1 | Statistics-2 | Remainders-1 | Remainders-2
Word Problems – Percentage 1 | Percentage 2 | Time and Work 1 | Time and Work 2 | Time, Speed and Distance 1 | Time, Speed and Distance 2
Advanced Topics- Permutation and Combination 1 | Permutation and Combination 2 | Permutation and Combination 3 | Probability
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Re: In a set of 10 consecutive odd positive integers, the product ........ &nbs [#permalink] 11 Jan 2019, 06:05
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