Question

N = aebfcg is a six digit number where a, b, c, e, f, g are its six digits. If \(a = e\), \(b = f\) and \(c = g\), which of the given statements is NOT correct?

• A. If g = 4 then N is divisible by 44.
• B. If a + b + c = 6 then N is divisible by 33.
• C. If g = 8, then for different values of a, b, c and e, N may be a perfect square.
• D. If b = c = 0, then N is not a perfect square.

Hint:

Write N as 11 times (10000a + 100b + c) since the digit pattern is a, a, b, b, c, c, then check each option's claim using this factored form.

Answer 1

The Correct Option is C

Step 1: Write N in terms of a, b and c.
The digits of N in order are a, e, b, f, c, g. Since \(e = a\), \(f = b\) and \(g = c\), the digit string is really a, a, b, b, c, c. So
\[ N = a \times 10^5 + a \times 10^4 + b \times 10^3 + b \times 10^2 + c \times 10 + c \]
\[ N = 110000a + 1100b + 11c = 11(10000a + 100b + c) \]
So N is always a multiple of 11, whatever a, b and c are.

Step 2: Check option (1): if g = 4, N is divisible by 44.
Here \(c = g = 4\), so \(N = 11 \times (10000a + 100b + 4)\). For N to be divisible by \(44 = 4 \times 11\), the bracket \((10000a + 100b + 4)\) must be divisible by 4. Both \(10000a\) and \(100b\) are already divisible by 4, so the bracket is divisible by 4 exactly when 4 is, which it always is. So N is always divisible by 44 when g = 4. This statement is correct.

Step 3: Check option (2): if a + b + c = 6, N is divisible by 33.
For N to be divisible by \(33 = 3 \times 11\), the bracket \((10000a + 100b + c)\) must be divisible by 3, since N already carries a factor of 11. Since \(10000\) and \(100\) both leave remainder 1 on division by 3, the bracket has the same remainder on division by 3 as \(a + b + c\). If \(a + b + c = 6\), a multiple of 3, the bracket is divisible by 3, so N is divisible by 33. This statement is correct.

Step 4: Check option (3): if g = 8, N may be a perfect square for some a, b, c.
Here \(c = g = 8\), so \(N = 11 \times (10000a + 100b + 8)\). For N to be a perfect square, since 11 is prime, N needs an even total count of the factor 11. As N already carries exactly one factor of 11 from the formula, the bracket \((10000a + 100b + 8)\) would itself need to supply one more factor of 11, meaning the bracket must equal \(11t^2\) for some whole number t, so that the whole thing becomes \(121t^2 = (11t)^2\), a perfect square. But the bracket's last two digits are always fixed at 0 and 8, since \(10000a\) and \(100b\) never change the last two digits. Checking every possible case, no value of \(11t^2\) ever has last two digits equal to 0 and 8. So there is no choice of a, b, c that makes N a perfect square when g = 8. This statement is FALSE, so it is the one that is NOT correct.

Step 5: Check option (4): if b = c = 0, N is not a perfect square.
Here \(N = 11 \times 10000a = 110000a\), for a from 1 to 9. Since \(110000 = 11 \times 10^4\) and \(10^4\) is already a perfect square, N is a perfect square only if \(11a\) is a perfect square. As a is a single digit from 1 to 9, \(11a\) ranges from 11 to 99 and can never contain a matching pair of the factor 11 needed to make it a perfect square, since a itself would have to be a multiple of 11, which is impossible for a single digit. So N is never a perfect square in this case. This statement is correct.

Final Answer:
Option (3) is the one that is not correct, because N can never be a perfect square when g = 8. \[ \boxed{\text{Option (3) is not correct}} \]