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As there are 7 days,

62 = can be divided into 9 for each day -1 (Best possibility)

But we want Friday to be more than any other day hence 10,9,9,9,9,8,8

IMO B
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Looking at this problem, I need to find the minimum number of children born on Friday while ensuring Friday still has more births than any other day.

Step 1: Understanding the Constraint

Let \(F\) = number of children born on Friday
For Friday to have "more" children than any other day: \(F > M, F > T, F > W, F > Th, F > S, F > Su\)

Step 2: Strategy to Minimize F

To make \(F\) as small as possible, I need to make the other days have as many children as possible (while still being less than \(F\)).
The maximum any other day can have is \(F-1\) children.

Step 3: Setting Up the Equation
In the most "packed" distribution:

Friday: \(F\) children
Each of the other 6 days: up to \(F-1\) children

Total children: \(F + 6(F-1) \leq 62\)

For the minimum \(F\), we want this to equal 62:
\(F + 6(F-1) = 62\)
\(F + 6F - 6 = 62\)
\(7F = 68\)
\(F = \frac{68}{7} = 9.71...\)

Step 4: Finding the Integer Solution

Since \(F\) must be a whole number: \(F \geq 10\)

Step 5: Verification
Let's verify \(F = 10\) works:

Friday: 10 children
Other 6 days: Can have at most 9 each
Maximum total: \(10 + 6(9) = 64\)

Since \(64 > 62\), we can distribute 62 children with Friday having 10.

Example distribution: Friday = 10, five days with 9 children each, one day with 7 children

Check: \(10 + 5(9) + 7 = 10 + 45 + 7 = 62\) ✓

Let's verify \(F = 9\) doesn't work:

Friday: 9 children
Other 6 days: Can have at most 8 each
Maximum total: \(9 + 6(8) = 57 < 62\)

We can't fit all 62 children with this constraint!

Answer: B. 10

Key Insight: The problem asks for an optimization - minimizing one value while satisfying constraints. The strategy is to push the constraint to its limit by making other days as "full" as possible.
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