Five numbers 10, 7, 5, 4 and 2 are to be arranged in a sequence from left to right following the directions given below:
1. No two odd or even numbers are next to each other.
2. The second number from the left is exactly half of the left-most number.
3. The middle number is exactly twice the right-most number.

Which is the second number from the right?

Question

Prepp Team · Accepted Answer

Analyzing the Constraints

The problem requires arranging the numbers {10, 7, 5, 4, 2} into a sequence $N_1, N_2, N_3, N_4, N_5$ based on three rules.

Rule 1: Alternating Parity No two odd or even numbers can be adjacent. The available numbers are {10, 4, 2} (Even) and {7, 5} (Odd). Since there are 3 even and 2 odd numbers, the sequence must follow the pattern: Even, Odd, Even, Odd, Even (E O E O E).
Rule 2: Left-Most Relation The second number from the left ($N_2$) is half the left-most number ($N_1$). $N_2 = N_1 / 2$.
Rule 3: Middle Number Relation The middle number ($N_3$) is twice the right-most number ($N_5$). $N_3 = 2 \times N_5$.

Applying Rule 2

We test possible pairs for ($N_1, N_2$) from the set {10, 7, 5, 4, 2} that fit the E O pattern:

If $N_1 = 10$ (E), then $N_2 = 10 / 2 = 5$ (O). Both numbers are available and fit the E O pattern. This is a valid possibility.
If $N_1 = 4$ (E), then $N_2 = 4 / 2 = 2$ (E). Both are available, but this violates the E O pattern (Rule 1).
If $N_1 = 2$ (E), then $N_2 = 2 / 2 = 1$. 1 is not in the set.

Therefore, the sequence must start with 10, 5. Sequence: 10, 5, _, _, _.

Applying Rule 3

The sequence pattern is E O E O E. We have $N_1=10$ and $N_2=5$. The remaining numbers are {7, 4, 2}. The positions $N_3$ and $N_5$ must be filled by the remaining even numbers {4, 2}, and $N_4$ must be the odd number {7}.

We need to check Rule 3 ($N_3 = 2 \times N_5$) using the available even numbers {4, 2} for $N_3$ and $N_5$:

If $N_5 = 2$, then $N_3 = 2 \times 2 = 4$. Both 4 and 2 are available and fit the required E O E pattern. This works.
If $N_5 = 4$, then $N_3 = 2 \times 4 = 8$. 8 is not available in the set.

Thus, $N_3 = 4$ and $N_5 = 2$. The only remaining number, 7, must be $N_4$. Sequence: 10, 5, 4, 7, 2.