Question

P, Q, R, S, T are five consecutive integers (not necessarily in that order), such that the smallest of these is greater than 60 and the greatest is less than 70. It is known that: (i) A and B both are prime numbers. (ii) T is a multiple of 9. (iii) Both the digits of P are same. (iv) The average of R and S is 63 and the difference between R and S is 2. What is the sum of the digits of the number Q?

• A. 10
• B. 11
• C. 12
• D. 9

Hint:

In logic puzzles with multiple constraints, start with the most definitive clues. Clues that give specific numerical values (like the average and difference for R and S) are the best starting points to narrow down the possibilities.

Answer 1

The Correct Option is B

Step 1: Determine R and S.

- The average of R and S is 63. So \( \frac{R+S}{2} = 63 \Rightarrow R+S = 126 \).
- The difference between R and S is 2. So \( R-S = 2 \) (assuming \(R > S\)). 
- Solving these two equations: \( 2R = 128 \Rightarrow R = 64 \). Then \( S = 62 \). So, two of the numbers are 62 and 64.

Step 2: Determine the set of five consecutive integers.

- Since 62 and 64 are part of a set of five consecutive integers, the set must be one of these: {60,61,62,63,64}, {61,62,63,64,65}, {62,63,64,65,66}. 
- The problem states the smallest is \(> 60\). This eliminates the first set. 
- The greatest is \(< 70\). This condition is met by both remaining sets.

Step 3: Use the remaining clues to identify the correct set and assign numbers.

- (iii) "Both the digits of P are same". In the range 61–66, the only number with identical digits is 66. So \(P=66\). This means the set must be {62, 63, 64, 65, 66}. 
- (ii) "T is a multiple of 9". In this set, the only multiple of 9 is 63. So \(T=63\). 
- We have \(P=66, T=63, R=64, S=62\). 
- The remaining number in the set is 65, so \(Q=65\). 
- (i) "A and B both are prime numbers". This seems to be a typo and should refer to two of the numbers P,Q,R,S,T. Let's check the primality of our numbers. None of 62, 63, 64, 65, 66 are prime. This indicates a contradiction in the problem statement.

Let's reconsider the set {61,62,63,64,65}. 
- \(P=66\) is not in this set. This set is not possible. 
Let's re-read clue (i). "A and B both are prime numbers." This is likely a typo for two of P,Q,R,S,T. If no numbers in the set are prime, the question is flawed. 
Let's check the primes between 60 and 70. They are 61 and 67. 
If the set contains two primes, it must contain 61 and 67. The only set of 5 consecutive integers that could contain one of them is {61,62,63,64,65}. It doesn't contain 67. A set of 5 cannot contain both. 
This is a major contradiction. Let's ignore clue (i) and see if we can proceed.

Let's assume the set is {62, 63, 64, 65, 66}. 
\(P=66, T=63, R=64, S=62, Q=65\).

Step 4: Find the sum of the digits of Q.

\[ Q = 65 \quad \Rightarrow \quad 6+5 = 11 \]

Final Answer:

\[ \boxed{11} \]