๐ Master the Number System: Complete Guide for Competitive Exams
Welcome, aspirants! The Number System is the backbone of Quantitative Aptitude. Whether you are preparing for SSC, Banking, Railways, or State PSCs, mastering this topic is non-negotiable. Let's break it down step-by-step!
1. Classification of Numbers
[attachment_0](attachment)To solve complex problems, you must first understand the basic building blocks of mathematics.
- ๐น Natural Numbers (N): Counting numbers starting from 1. (Examples: 1, 2, 3, 4...)
- ๐น Whole Numbers (W): Natural numbers including zero. (Examples: 0, 1, 2, 3...)
- ๐น Integers (Z): All positive and negative whole numbers, including zero. (Examples: ...-3, -2, -1, 0, 1, 2, 3...)
- ๐น Rational Numbers (Q): Numbers that can be expressed in the form of p/q, where p and q are integers and q ≠ 0. (Examples: 1/2, -5/4, 7)
- ๐น Irrational Numbers: Numbers that cannot be expressed as a simple fraction. They have non-terminating and non-repeating decimals. (Examples: √2, √3, ฯ)
- ๐น Real Numbers (R): The combination of both Rational and Irrational numbers. All numbers on the number line are real numbers.
2. Special Categories of Numbers
- Even Numbers: Divisible by 2. (0, 2, 4, 6, 8...)
- Odd Numbers: Not divisible by 2. (1, 3, 5, 7, 9...)
- Prime Numbers: Numbers greater than 1 that have exactly two factors: 1 and themselves. (2, 3, 5, 7, 11...)
Note: 2 is the only even prime number. - Composite Numbers: Numbers with more than two factors. (4, 6, 8, 9, 10...)
- Co-Prime Numbers: Two numbers whose Highest Common Factor (HCF) is 1. (Examples: 4 & 5, 8 & 15)
๐ฏ Let's Practice: Questions & Solutions
From Basic Concepts to Advanced Shortcuts!
๐ข Beginner Level
Q1. What is the difference between the place value and face value of 7 in the numeral 297538?
Solution:
The face value of a digit is the digit itself. Face value of 7 = 7.
The place value depends on its position. 7 is in the thousands place. Place value = 7000.
Difference = 7000 - 7 = 6993.
The face value of a digit is the digit itself. Face value of 7 = 7.
The place value depends on its position. 7 is in the thousands place. Place value = 7000.
Difference = 7000 - 7 = 6993.
Q2. Find the sum of the first 50 natural numbers.
Solution:
Formula for the sum of the first 'n' natural numbers = n(n + 1) / 2
Here, n = 50.
Sum = 50 × (50 + 1) / 2 = 50 × 51 / 2 = 25 × 51 = 1275.
Formula for the sum of the first 'n' natural numbers = n(n + 1) / 2
Here, n = 50.
Sum = 50 × (50 + 1) / 2 = 50 × 51 / 2 = 25 × 51 = 1275.
๐ก Intermediate Level
Q3. Find the unit digit in the product of (4387)245 × (621)72.
Solution:
Rule for unit digit 1: Any power of a number ending in 1 will always end in 1. So, unit digit of (621)72 is 1.
Rule for unit digit 7: The cyclicity of 7 is 4. Divide the power (245) by 4.
245 ÷ 4 gives a remainder of 1.
Therefore, the unit digit of (4387)245 is the same as 71, which is 7.
Final Product Unit Digit = 7 × 1 = 7.
Rule for unit digit 1: Any power of a number ending in 1 will always end in 1. So, unit digit of (621)72 is 1.
Rule for unit digit 7: The cyclicity of 7 is 4. Divide the power (245) by 4.
245 ÷ 4 gives a remainder of 1.
Therefore, the unit digit of (4387)245 is the same as 71, which is 7.
Final Product Unit Digit = 7 × 1 = 7.
Q4. Find the number of trailing zeros at the end of 100! (100 factorial).
Solution:
Trailing zeros are formed by pairs of (2 × 5). In any factorial, the number of 5s is always less than the number of 2s, so we just count the total number of 5s.
Shortcut Trick: Successively divide the number by 5 and add the quotients.
100 / 5 = 20
20 / 5 = 4
(4 cannot be divided by 5 anymore).
Total trailing zeros = 20 + 4 = 24.
Trailing zeros are formed by pairs of (2 × 5). In any factorial, the number of 5s is always less than the number of 2s, so we just count the total number of 5s.
Shortcut Trick: Successively divide the number by 5 and add the quotients.
100 / 5 = 20
20 / 5 = 4
(4 cannot be divided by 5 anymore).
Total trailing zeros = 20 + 4 = 24.
๐ด Advanced Level
Q5. Find the total number of divisors (factors) of the number 360.
Solution:
Step 1: Express 360 as a product of its prime factors.
360 = 36 × 10 = (22 × 32) × (2 × 5) = 23 × 32 × 51.
Step 2: Add 1 to each of the powers and multiply them together.
Number of divisors = (3 + 1) × (2 + 1) × (1 + 1)
= 4 × 3 × 2 = 24.
Step 1: Express 360 as a product of its prime factors.
360 = 36 × 10 = (22 × 32) × (2 × 5) = 23 × 32 × 51.
Step 2: Add 1 to each of the powers and multiply them together.
Number of divisors = (3 + 1) × (2 + 1) × (1 + 1)
= 4 × 3 × 2 = 24.
Q6. A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. When the same number is successively divided by 5 and 4, what will be the respective remainders?
Solution:
Let's find the smallest such number. Work backwards.
Let the final quotient be 0.
Number before dividing by 5 = (Divisor × Quotient) + Remainder = (5 × 0) + 4 = 4.
Original Number before dividing by 4 = (4 × 4) + 1 = 17.
Now, reverse the successive division for 17:
Divide 17 by 5: Quotient = 3, Remainder = 2.
Divide the quotient 3 by 4: Quotient = 0, Remainder = 3.
The respective remainders are 2 and 3.
Let's find the smallest such number. Work backwards.
Let the final quotient be 0.
Number before dividing by 5 = (Divisor × Quotient) + Remainder = (5 × 0) + 4 = 4.
Original Number before dividing by 4 = (4 × 4) + 1 = 17.
Now, reverse the successive division for 17:
Divide 17 by 5: Quotient = 3, Remainder = 2.
Divide the quotient 3 by 4: Quotient = 0, Remainder = 3.
The respective remainders are 2 and 3.
Keep practicing, and numbers will become your best friends in the exam hall! Best of luck! ๐✍️
No comments:
Post a Comment