Question

Let a, b, c be distinct digits. Consider a two digit number \( 'ab' \) and a three digit number \( 'ccb' \), both defined under the usual decimal number system. If \( (ab)^2 = ccb \) and \( ccb > 300 \) then the value of b is

• A. 5
• B. 0
• C. 1
• D. 6

Hint:

The square of a two-digit number is a three-digit number above 300, so the two-digit number lies between 18 and 31. The last digit of the square must equal b, and the first two digits must be equal.

Answer 1

The Correct Option is C

Step 1: Range for \( ab \).
\( 300 < (ab)^2 \le 999 \), so \( 18 \le ab \le 31 \).

Step 2: Use the last digit.
The units digit of \( (ab)^2 \) is the units digit of \( b^2 \), and it must equal b. This holds only for \( b = 0, 1, 5, 6 \), so the candidates are 20, 21, 25, 26, 30, 31.

Step 3: Test the candidates (first two digits must be equal).
\( 20^2 = 400 \) (digits 4, 0 differ), \( 21^2 = 441 \) (first two digits 4, 4: valid), \( 25^2 = 625 \), \( 26^2 = 676 \), \( 30^2 = 900 \), \( 31^2 = 961 \) (not of the form ccb).

Step 4: Check the conditions.
\( 21^2 = 441 \), \( 441 > 300 \), and \( a = 2,\ b = 1,\ c = 4 \) are distinct digits. This is the only solution, so \( b = 1 \).

Final Answer:
The correct option is (3): 1