If in a six-digit integer n, f(k) represents the value of the

From the question stem, F(k) gives us the k-th digit in the k-th place. Since F(4) is the hundredths digit, we are counting the digits from the left for the 6-digit number N.
So F(1) gives us the 1st digit, F(2) the 2nd digit and so on.

Stmt 1.
Let \(F(1) = x, F(2) = y, F(3) = z\), where x,y,z are any of the digits from 0 to 9.
So the first 3 digits are x,y,z.
From the statement, the next 3 digits are the same as these. So the number is the form xyzxyz (e.g. 123123 or 375375)
Now a number like 123123 can be written as \(123*1000 + 123\)
\(123123 = 123*1000 + 123 = 1001*123\)
So \(xyzxyz = 1001*xyz\)

The key thing here is that 1001 is divisible by 7. So that means the other side of the above equation is also divisible by 7 i.e. the number xyzxyz is divisible by 7.
So if the first 3 digits are the same as the last 3, it is divisible by 7.

SUFF

Stmt 2.
This says that all the digits are the same e.g. (111111,444444,55555 etc). This is just a special case of the same principle in Stmt 1. Here too, the first 3 digits are equal to last 3 digits. (in addition, all are equal)

SUFF

So D.

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This can lead into a nice test for checking whether a number is divisible by 7. Take the digits of the number from the right three at a time and take the alternating sum. If that sum is divisible, then the number is divisible too.

E.g. Is 12348763 divisible by 7?
Take numbers in groups of 3 from the right: 12, 348, 763
Alternately add and subtract each group \(= 12-348+763 = 427\), which is divisible by 7.
So 12348763 is divisible by 7.