All Exams Test series for 1 year @ ₹349 only
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.

Was this answer helpful?

Similar Questions

  1. What is the total number of odd and even divisors of 120, respectively?
  2. The Government of India Constituted Narcotics Control Bureau in _____.
  3. The least number by which 294 must be multiplied to make it a perfect square, is:
  4. How many factors does the number 12288 have?
  5. The number of factors of 4200 are:
  6. Which of the following expresses the prime factorisation of 54?

Important Questions from Multiples and Factors

  1. If 847 × 385  × 675 × 3025 = 3 a × 5 b × 7 c × 11 d, then the value of ab – cd is:

  2. (mx + n) is a factor of:

  3. If 7-digit number 678p37q is divisible by 75 and p is not a composite, then the values of p and q are:

  4. Which of the following numbers will completely divide 412 + 413 + 414 + 415?

  5. Which of the following numbers Is divisible by 24?

Need Expert Advice?
Upcoming Exams
RRB ALP
July 28, 2026
RRB Group D
August 03, 2026
Test Series
RRB NTPC img
Railways
RRB NTPC Under Graduate 2026 New Mock Test Series
1459 Tests 2 Tests Free
215 Attempts
4.3(512)
English, Hindi, Telugu +7 More

Start Your Preparation with Prepp Mobile App

Download the app from Google Play & App Store
Download the app from Google Play & App Store
Prepp Mobile App