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Question

How many factors of $2^{2} \times 3^{1} \times 5^{2} \times 7^{1}$ are divisible by 50 but not by 100?

This question was previously asked in
RRB NTPC 2019 CBT 1 Question Paper (8-Mar-2021) (Shift 2)
The correct answer is
4

1. Analyze the Number's Prime Factorization:

The given number is $N = 2^{2} \times 3^{1} \times 5^{2} \times 7^{1}$.

A factor of $N$ will have the general form $F = 2^a \times 3^b \times 5^c \times 7^d$, where the exponents must be within the following ranges:

  • $0 \le a \le 2$
  • $0 \le b \le 1$
  • $0 \le c \le 2$
  • $0 \le d \le 1$

2. Define Conditions for Divisibility:

  • Divisible by 50: $50 = 2^1 \times 5^2$. A factor $F$ is divisible by 50 if its exponents satisfy $a \ge 1$ and $c \ge 2$.
  • Divisible by 100: $100 = 2^2 \times 5^2$. A factor $F$ is divisible by 100 if its exponents satisfy $a \ge 2$ and $c \ge 2$.

3. Combine Conditions:

We need factors that are divisible by 50 BUT NOT by 100.

Combining the conditions:

  • From $N$: $0 \le a \le 2$, $0 \le b \le 1$, $0 \le c \le 2$, $0 \le d \le 1$.
  • Divisible by 50 requires: $a \ge 1$ and $c \ge 2$.
  • Not divisible by 100 requires that it's NOT the case that ($a \ge 2$ and $c \ge 2$). This means $a < 2$ OR $c < 2$.

To satisfy all requirements simultaneously:

  • We need $a \ge 1$ (from divisible by 50) and $a < 2$ (from not divisible by 100). This implies $a = 1$.
  • We need $c \ge 2$ (from divisible by 50). Combined with $c \le 2$ from $N$, this implies $c = 2$.
  • The exponents $b$ and $d$ are constrained only by the original number $N$: $b \in \{0, 1\}$ and $d \in \{0, 1\}$.

4. Calculate the Number of Factors:

The required factors must have the form $2^a \times 3^b \times 5^c \times 7^d$ with the specific exponents:

  • $a = 1$ (1 possibility)
  • $b \in \{0, 1\}$ (2 possibilities)
  • $c = 2$ (1 possibility)
  • $d \in \{0, 1\}$ (2 possibilities)

The total number of such factors is the product of the number of possibilities for each exponent:

$ \text{Number of factors} = (\text{options for } a) \times (\text{options for } b) \times (\text{options for } c) \times (\text{options for } d) $ $ \text{Number of factors} = 1 \times 2 \times 1 \times 2 = 4 $

Therefore, there are 4 factors of $2^{2} \times 3^{1} \times 5^{2} \times 7^{1}$ that are divisible by 50 but not by 100.

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