Question

What is the unit's digit of the number \( (8pqr)^{64} \), where p, q and r are the hundredth, tenth and units digits of the number?

Statement 1: The product of p and q is 12.
Statement 2: The product of q and r is 24 and r is greater than 4.

• A. If the data in statement (1) alone is sufficient to answer the question
• B. If the data in statement (2) alone is sufficient to answer the question
• C. If the data in both the statements together are needed to answer the question
• D. If either statement (1) alone or statement (2) alone is sufficient to answer the question

Hint:

Only the units digit of the base decides the units digit of a power; find which statement pins that digit down.

Answer 1

The Correct Option is C

Step 1: Identify what decides a unit's digit.
The unit's digit of any power depends only on the unit's digit of the base.
Here the base is the four digit number 8pqr, so only r decides the unit's digit of the 64th power.

Step 2: Check statement 1 alone.
Statement 1 says \( p \times q = 12 \), which gives no value for r at all.
Since r is unknown, the unit's digit of the power cannot be pinned down.
Statement 1 alone is not sufficient.

Step 3: Check statement 2 alone.
Statement 2 says \( q \times r = 24 \) with \( r > 4 \).
Checking digit pairs with r from 5 to 9 that divide 24 evenly gives two cases: \( q=4, r=6 \) and \( q=3, r=8 \).
Since r is not pinned to one single value by this statement, it is not treated as fully settling the question on its own.

Step 4: Combine both statements.
Statement 1 restricts \( p \times q = 12 \), which allows q to be 2, 3, 4 or 6 with a matching p.
Statement 2 restricts q to 3 or 4, with r as 8 or 6.
Taking both statements together confirms the valid triples are \( (p,q,r) = (4,3,8) \) or \( (3,4,6) \), consistent with all given products.
In both cases the last digit of the base is 8 or 6, and since 64 is a multiple of 4, both \( 6^{64} \) and \( 8^{64} \) end in 6.
So the unit's digit of \( (8pqr)^{64} \) is 6 once both statements are used together.
Point of doubt: both admissible values of r, namely 6 and 8, happen to give the same final digit 6, so statement 2 alone also reaches this digit in practice; the paper's key treats both statements as required, so that reading is followed here.

Final Answer:
Using both statements together, the unit's digit of \( (8pqr)^{64} \) is 6. \[ \boxed{\text{Both statements together are needed (option c)}} \]