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Let the three integers be \( A, B, C \), and let the natural number to be added be \( n \).
Given average of the three numbers: \[ \frac{A + B + C}{3} = 13 \Rightarrow A + B + C = 39 \]
New average becomes: \[ \frac{A + B + C + n}{4} \] which must be an odd integer.
Let an odd integer be represented as \( 2k + 1 \). Then: \[ \frac{39 + n}{4} = 2k + 1 \] Multiply both sides by 4: \[ 39 + n = 4(2k + 1) = 8k + 4 \Rightarrow n = 8k + 4 - 39 = 8k - 35 \]
We want \( n > 0 \). Try increasing integer values of \( k \):
So, the smallest natural number \( n \) such that the new average is odd is: \[ \boxed{5} \]
Correct Option: C. 5
Average of three numbers = \( 13 \)
Total sum of the three numbers: \[ 3 \times 13 = 39 \]
New average with four numbers: \[ \frac{39 + n}{4} \] It is given that this average must be an odd integer.
The smallest odd integer greater than \(\frac{39}{4} = 9.75\) is 11.
Set: \[ \frac{39 + n}{4} = 11 \] Multiply both sides: \[ 39 + n = 44 \Rightarrow n = 44 - 39 = 5 \]
Minimum value of n is: \(\boxed{5}\)
Correct Option: (C)
The number of patients per shift (X) consulting Dr. Gita in her past 100 shifts is shown in the figure. If the amount she earns is ₹1000(X - 0.2), what is the average amount (in ₹) she has earned per shift in the past 100 shifts?
When $10^{100}$ is divided by 7, the remainder is ?