Nitin was counting down from 32. Sumeet was counting upwards the numbers…

Concept

When two counters move through arithmetic sequences — one decreasing by a fixed amount each turn and one increasing by a fixed amount each turn — they call the same number at the same turn only if setting their two turn-indexed expressions equal to each other gives a whole-number turn. If solving that equation gives a fraction, the two sequences pass by each other without ever coinciding.

Step-by-step

1. Let n be the turn number (n = 1, 2, 3, ...).

2. Nitin's number on turn n is 33 − n (he starts at 32 on turn 1 and drops by 1 each turn).

3. Sumeet's number on turn n is 2n − 1, the n-th odd number (he starts at 1 and calls odd numbers in order).

4. For them to call the same number on the same turn: 33 − n = 2n − 1, which gives 34 = 3n, so n = 34/3 (about 11.3).

5. Since a turn number must be a whole number, no turn n satisfies the equation — so they never call the same number at the same time.

Cross-check

Checking the turns on either side of n = 34/3 confirms this directly: on turn 11 Nitin is at 22 while Sumeet is at 21 (Nitin still ahead); on turn 12 Nitin drops to 21 while Sumeet moves to 23 (Sumeet now ahead) — the two sequences cross between these turns without ever landing on the identical value at the identical turn.

So the correct choice is that they never call out the same number simultaneously.