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The arithmetic mean of 17 consecutive integers

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The arithmetic mean of 17 consecutive integers  [#permalink]

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New post 17 May 2017, 13:00
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A
B
C
D
E

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Question Stats:

52% (02:02) correct 48% (02:12) wrong based on 78 sessions

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The arithmetic mean of 17 consecutive integers is an odd number. Which of the following must be true?

I. Largest integer is even.
II. Sum of all integers is odd.
III. Difference between largest and smallest integer is even.

(A) I
(B) II
(C) III
(D) I, II
(E) II, III

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Re: The arithmetic mean of 17 consecutive integers  [#permalink]

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New post 21 May 2017, 03:31
1
Here is the approach for the problem.

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17

Average of the 17 consecutive numbers is the middle number which is the 9th number. If 9th number is odd then 17th number is odd and the first number is also odd.

Sum = There are 9 odd numbers and even numbers which means the sum of all will be odd

I. Largest integer is even: Not true

II. Sum of all integers is odd: True

III. Difference between largest and smallest integer is even. Odd – Odd = Even (True)

II and III are true.

Answer option is E.
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Re: The arithmetic mean of 17 consecutive integers  [#permalink]

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New post 21 May 2017, 05:16
Kritesh wrote:
The arithmetic mean of 17 consecutive integers is an odd number. Which of the following must be true?

I. Largest integer is even.
II. Sum of all integers is odd.
III. Difference between largest and smallest integer is even.

(A) I
(B) II
(C) III
(D) I, II
(E) II, III


Here is another way to look at this problem.

Given:
17 integers (odd number of integers)
"consecutive" integers, one after another.
Average is odd => SUM / 17 = ODD => SUM = ODD * 17 ==> SUM is odd. Already means II is always true.

Lets look at other choices:
I) Lets say largest number is even, since there are 17 numbers, the smallest number will also be even. Also, there will be 9 even and 8 odd numbers (you can do this with an example, but its pretty intuitive since every other number is odd)
If you want to take an example:
Lets say largest is 18
Smallest will be 2
2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 ==> As you can see there are 9 even and 8 odd.

Sum of 9 even numbers => even
Sum of 8 odd numbers => even
Sum of all numbers => even + even = even. So (1) can never be true actually. since the sum is even and we already established sum of numbers should be odd.

This also means that the largest number must be ODD (since it cannot be even). This means smallest number will also be ODD.
Largest - smallest = Odd - Odd = even.

Hence (III) is also true.

Hence I and III are both true.

(E).

Try my number theory workshop I to learn more about even /odd tips and tricks.
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Re: The arithmetic mean of 17 consecutive integers  [#permalink]

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New post 21 May 2017, 05:23
Kritesh wrote:
The arithmetic mean of 17 consecutive integers is an odd number. Which of the following must be true?

I. Largest integer is even.
II. Sum of all integers is odd.
III. Difference between largest and smallest integer is even.

(A) I
(B) II
(C) III
(D) I, II
(E) II, III


17 consecutive integers are \(\{x,x+1,x+2,...,x+15,x+16\}\)

The average of these 17 numbers is
\(\frac{x+(x+1)+...+(x+16)}{17}=\frac{17x+(1+2+...+16)}{17}=\frac{17x+\frac{17 \times 16}{2}}{17}=\frac{17x+17 \times 8}{17}=x+8\).

\(x+8\) is odd so \(x\) is odd.

\(x+16\) is odd so (I) is wrong.

Sum of these 17 integers is \(17x+17 \times 8 = 17(x+8)\). Since \(x\) is odd, \(17(x+8)\) is odd. Thus (II) is true.

\((x+16)-x=16\) is even. Hence, (III) is true.

The answer is E.
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Re: The arithmetic mean of 17 consecutive integers  [#permalink]

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New post 30 Jul 2018, 09:30
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Re: The arithmetic mean of 17 consecutive integers &nbs [#permalink] 30 Jul 2018, 09:30
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