Question

N = \((A7A)^{17}\) is a perfect square, where A7A is the three digit number with hundreds digit A, tens digit 7 and units digit A. Which of the following statements is FALSE?

• A. A is an even digit.
• B. A is divisible by 3
• C. When N is divided by 13 we get remainder 3.
• D. None of these

Hint:

Since 17 is odd, N is a perfect square only if A7A itself is one. Find A, then check each statement, especially whether 676 carries the factor 13.

Answer 1

The Correct Option is C

Here's a different way to pin down A: instead of plugging in every digit from 1 to 9 into $101A+70$, we search directly among the three digit perfect squares for one that has the shape A7A.

First, since $N = (A7A)^{17}$ is a perfect square and 17 is an odd power, the base $m = A7A$ must itself be a perfect square. An odd power of a number is a perfect square only when the number itself is one, because $m^{17} = (m^8)^2 \times m$ and $(m^8)^2$ is already a perfect square on its own.

Now list the three digit perfect squares: $100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961$.

The shape A7A fixes the tens digit at 7. Scanning the list, only two squares have a tens digit of 7: $576$ and $676$.

1. 576: hundreds digit 5, units digit 6. These are not equal, so it does not have the form A7A, which needs the hundreds digit to match the units digit.
2. 676: hundreds digit 6, units digit 6. These match, so this is our number: $m = 676 = 26^2$, giving $A = 6$.

Now check the three statements about $N = 676^{17}$:

• $A = 6$ is even, so statement (1) holds.
• $A = 6$ is divisible by 3, so statement (2) holds.
• For statement (3), notice $676 = 26^2$ and $26 = 2 \times 13$, so $676$ is a multiple of 13. Raising a multiple of 13 to any power keeps it a multiple of 13, so $N = 676^{17}$ is a multiple of 13 and leaves remainder 0, not 3, when divided by 13. Statement (3) does not hold.

Let's summarize:

• Searching perfect squares with tens digit 7 quickly gives $m = 676$, so $A = 6$.
• A being even and divisible by 3 both check out.
• N is always a multiple of 13, since 676 already is, so remainder 3 is impossible; the true remainder is 0.

So the false statement is (3): the remainder is not 3, it is 0.