Understanding the Problem

Nitin counts down from 32: 32, 31, 30, ..., 1.

Sumit counts up odd numbers from 1: 1, 3, 5, 7, ...

They call numbers at the same speed (same time intervals), so we need to find a number that appears in both sequences at the same position (step).

Step 1: Define the Sequences

Let $n$ be the step number (starting from 1).

Nitin's number at step $n$: $a_n=32-n+1=33-n$

Sumit's number at step $n$: $b_n=2n-1$ (the nth odd number)

Step 2: Set the Numbers Equal

We need $a_n=b_n$:

$33-n=2n-1$

Step 3: Solve for $n$

$33+1=2n+n$

$34=3n$

$n=\frac{34}{3}\approx 11.333$

Since $n$ must be an integer (step number), there is no integer solution. This suggests they never call the same number at the same step.

Step 4: Verify with Integer Steps

Let's check a few steps around n=11:

• At n=11: Nitin: 33-11=22, Sumit: 2*11-1=21 → Not equal
• At n=12: Nitin: 33-12=21, Sumit: 2*12-1=23 → Not equal
• At n=10: Nitin: 33-10=23, Sumit: 2*10-1=19 → Not equal

No match found. They never call the same number simultaneously.

Final Answer

They will not call out the same number.

Related Topics

• Arithmetic Sequences: A sequence where each term is obtained by adding a constant difference to the previous term.
• Linear Equations: Equations of the first degree (e.g., ax + b = c).
• Integer Solutions: Solutions to equations that must be integers; often arise in discrete counting problems.

Key Formulae

• nth term of a decreasing arithmetic sequence: $a_n=a+(n-1)d$, where $d$ is negative. For Nitin: first term=32, d=-1, so $a_n=32-(n-1)=33-n$.
• nth odd number: $2n-1$.