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Let the three integers be \( A, B, C \). Let \( n \) be a natural number to be added.
The initial average of \( A, B, C \) is 13. Therefore, \( A + B + C = 3 \times 13 = 39 \).
When \( n \) is added, there are four numbers: \( A, B, C, n \). The new average is \( \frac{A + B + C + n}{4} \).
This new average must be an odd integer. Let the odd integer be \( 2k + 1 \), where \( k \) is an integer.
So, \( \frac{39 + n}{4} = 2k + 1 \). Multiplying by 4 gives \( 39 + n = 4(2k + 1) = 8k + 4 \).
Rearranging for \( n \), we get \( n = 8k + 4 - 39 = 8k - 35 \).
We require \( n \) to be a natural number, meaning \( n>0 \). We test integer values of \( k \) to find the smallest \( n>0 \).
The smallest natural number \( n \) that satisfies the condition is 5.
Correct Option: C. 5
The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64.Then, the largest number in the original set of three numbers is