Question

The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is

• A. 3
• B. 4
• C. 5
• D. 1

Answer 1

The Correct Option is C

The correct answer is C :5
Let's denote the three integers as A, B, and C, and the natural number as n. We are given that the average of these three integers is 13, so we have:
\(\frac{(A+B+C)}{3}=13\)
A+B+C = 39
Now, we want to find the minimum value of n such that when it is included, the average becomes an odd integer. When we add n to the sum and
divide by 4 (since we now have four integers), we want the result to be an odd integer. An odd integer can be represented as 2k+1, where k is an
integer.
So, we have:
\(\frac{(A+B+C+n)}{4}=2k+1\)
Substitute the value of A + B + C from the first equation:
\(\frac{(39+n)}{4}=2k+1\)
Now, we want to find the minimum value of n that satisfies this equation. Let's try different values of k and see what n values work:
For k = 0:
\(\frac{(39 + n)}{4}=1\)
39+n = 4
n=-35 (which is not a natural number)
For k = 1:
\(\frac{(39+n)}{4}=3\)
39+n=12
n=-27 (which is not a natural number)
For k = 2:
\(\frac{(39 + n)}{4} = 5\)
39+n=20
n=-19 (which is not a natural number)
For k = 3:
\(\frac{(39 + n)}{4}=7\)
39+n=28
n=-11 (which is not a natural number)
For k = 4:
\(\frac{(39 + n)}{4} = 9\)
39+n=36
n=-3 (which is not a natural number)
For k = 5:
\(\frac{(39+n)}{4}=11\)
39+n=44
n=5
Since we're looking for a natural number n, the minimum possible value of n is 5. When n is 5, the average of the four integers becomes an odd integer (11).

Answer 2

Given that average of three numbers is \(13\).
Total sum \(= 3\times13 = 39\)
Average of four integers = \(\frac {39+n}{4}\)
It is given that the average of four integers remains an odd integer.
Minimum value of \(\frac {39+n}{4}\) can take such that n is a natural number i.e. 11
 \(\frac {39+n}{4}= 11\)
\(39+n = 44\)
\(n=44-39\)
\(n=5\)

So, the correct option is (C): \(5\)