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

Step 1: Work out why A7A itself must be a perfect square.
Let \(m = A7A\), the three digit number with hundreds digit A, tens digit 7, and units digit A, so \(m = 101A + 70\). We are told \(N = m^{17}\) is a perfect square. Since 17 is an odd power, write \(m^{17} = (m^8)^2 \times m\). The part \((m^8)^2\) is already a perfect square, so N is a perfect square only if the leftover factor, m itself, is also a perfect square.

Step 2: Find the digit A.
We need \(m = 101A + 70\) to be a perfect square, where A is a digit from 1 to 9 (A is a leading digit, so it cannot be 0). Testing each value:
\(A=1: 171\), \(A=2: 272\), \(A=3: 373\), \(A=4: 474\), \(A=5: 575\), \(A=6: 676\), \(A=7: 777\), \(A=8: 878\), \(A=9: 979\).
Checking these against known squares near them, only \(676 = 26^2\) is a perfect square. So \(A = 6\) and \(m = A7A = 676\).

Step 3: Check statement (1), A is an even digit.
\(A = 6\), which is even. This statement is TRUE.

Step 4: Check statement (2), A is divisible by 3.
\(A = 6 = 3 \times 2\), which is divisible by 3. This statement is TRUE.

Step 5: Check statement (3), the remainder when N is divided by 13 is 3.
Notice that \(676 = 26^2 = (2 \times 13)^2 = 4 \times 169\), so 676 is itself divisible by 13 (in fact by \(13^2\)). Since \(m = 676\) is divisible by 13, and \(N = m^{17}\), N is also divisible by 13, because any power of a multiple of 13 stays a multiple of 13. So dividing N by 13 leaves a remainder of 0, not 3. This statement is FALSE.

Step 6: Check statement (4), None of these.
Since statement (3) is already false, None of these is not the right choice, because a false statement does exist among (1) to (3).

Final Answer:
Statement (3) is false, since N leaves remainder 0, not 3, when divided by 13. \[ \boxed{\text{Statement (3) is FALSE}} \]