We have 52 positive integers arranged in increasing order: $a_1 < a_2 < a_3 < ... < a_{52}$. The key condition is: The average of all 52 numbers is exactly 1 less than the average of the last 51 numbers. We call: - Average of all 52 numbers = $M$ - Average of last 51 numbers = $M + 1$

From the averages: - Sum of all 52 numbers = $52M$ - Sum of last 51 numbers = $51(M + 1) = 51M + 51$ Since $a_1 +$ (sum of last 51 numbers) = sum of all 52 numbers: $a_1 + (51M + 51) = 52M$ $a_1 = 52M - 51M - 51 = M - 51$ This is our key relationship! To maximize $a_1$, we need to maximize $M$.

To maximize $M$, we need to maximize the sum of all 52 numbers, which means maximizing the sum of the last 51 numbers (since $a_1 = M - 51$). The constraint: We need $a_1 < a_2 < a_3 < ... < a_{52} = 100$ The strategy: To maximize the sum $a_2 + a_3 + ... + a_{52}$, we should make these numbers as large as possible while maintaining strict inequality. The optimal arrangement is to make them consecutive integers ending at 100: - $a_{52} = 100$ - $a_{51} = 99$ - $a_{50} = 98$ - ... - $a_2 = 50$ So $a_2, a_3, ..., a_{52}$ become: 50, 51, 52, ..., 100

Sum of consecutive integers from 50 to 100: Using the formula: Sum = $\frac{\text{(first term + last term)} \times \text{number of terms}}{2}$ Sum = $\frac{(50 + 100) \times 51}{2} = \frac{150 \times 51}{2} = 3825$

Since sum of last 51 numbers = $51M + 51 = 3825$: $51M = 3825 - 51 = 3774$ $M = \frac{3774}{51} = 74$ Therefore: $a_1 = M - 51 = 74 - 51 = 23$

Therefore, the largest possible value of $a_1$ is 23.

Key Insight: The secret to solving this problem is recognizing that to maximize $a_1$, we need to maximize the sum of the other 51 numbers, which is achieved by making them consecutive integers ending at the given maximum value.

CAT (QA) PYQs

Keyboard Shortcuts