Number System
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Solved Examples
Study MaterialSolved Examples – Number System
Solved examples help students understand the practical application of Number System concepts in competitive examinations. These examples are designed from basic to advanced level and cover important topics frequently asked in SSC, Banking, Railway, CAT, CDS, NDA, UPSC, and placement aptitude tests.
Topics Covered in Solved Examples
- Face Value and Place Value
- Even and Odd Numbers
- Prime and Composite Numbers
- Divisibility Rules
- Remainder Problems
- Recurring Decimals
- Algebraic Identities
- Unit Digit Questions
- Series Formula Problems
- Simplification Techniques
Example 1: Face Value and Place Value
Question: Find the place value of 7 in the number 572943.
Solution:
In the number 572943, digit 7 is at the ten-thousands place.
Place Value = 7 × 10000
= 70000
Example 2: Even and Odd Number
Question: Determine whether 984562 is even or odd.
Solution:
The last digit is 2.
Numbers ending with 0, 2, 4, 6, or 8 are even numbers.
Therefore, 984562 is an even number.
Example 3: Prime Number Check
Question: Check whether 97 is a prime number.
Solution:
97 is divisible only by 1 and 97.
Therefore, 97 is a prime number.
Example 4: Divisibility by 9
Question: If 481d673 is divisible by 9, find the smallest value of d.
Solution:
Sum of known digits:
4 + 8 + 1 + 6 + 7 + 3 = 29
The next multiple of 9 after 29 is 36.
Therefore:
d = 36 − 29
= 7
Example 5: Number Divisible by 6
Question: How many 3-digit numbers are completely divisible by 6?
Solution:
Smallest 3-digit number divisible by 6 = 102
Largest 3-digit number divisible by 6 = 996
These numbers form an AP:
a = 102, d = 6, l = 996
Using:
l = a + (n − 1)d
996 = 102 + (n − 1)6
996 = 102 + 6n − 6
900 = 6n
n = 150
Example 6: Algebraic Identity Shortcut
Question: Find the value of 297 × 297.
Solution:
297 × 297 = (300 − 3)²
Using identity:
(a − b)² = a² + b² − 2ab
= 300² + 3² − 2 × 300 × 3
= 90000 + 9 − 1800
= 88209
Example 7: Remainder Theorem
Question: Find the remainder when 6767 + 67 is divided by 68.
Solution:
67 ≡ -1 (mod 68)
Therefore:
6767 ≡ (-1)67
= -1
So:
6767 + 67
= -1 + 67
= 66
Example 8: Recurring Decimal to Fraction
Question: Convert 0.232323... into a fraction.
Solution:
Repeating digits = 23
Number of repeating digits = 2
Fraction = 23/99
Therefore:
0.232323... = 23/99
Example 9: Smallest Number Divisible by 111
Question: Find the smallest 6-digit number exactly divisible by 111.
Solution:
Smallest 6-digit number = 100000
100000 ÷ 111 gives remainder 100.
Required number:
100000 + (111 − 100)
= 100000 + 11
= 100011
Example 10: Sum of Natural Numbers
Question: Find the value of:
1 + 2 + 3 + ... + 100
Solution:
Using formula:
1 + 2 + 3 + ... + n = n(n + 1)/2
= 100 × 101 / 2
= 5050
Therefore, the answer is 5050.
Example 11: Sum of Squares
Question: Find:
1² + 2² + 3² + ... + 10²
Solution:
Using formula:
1² + 2² + 3² + ... + n² = n(n + 1)(2n + 1)/6
= 10 × 11 × 21 / 6
= 385
Therefore, the answer is 385.
Example 12: Unit Digit Problem
Question: Find the unit digit of 723.
Solution:
Pattern of unit digits for powers of 7:
7, 9, 3, 1
Cycle length = 4
23 ÷ 4 leaves remainder 3.
Third number in the cycle = 3
Example 13: Co-prime Numbers
Question: Are 15 and 28 co-prime numbers?
Solution:
Factors of 15 = 1, 3, 5, 15
Factors of 28 = 1, 2, 4, 7, 14, 28
Common factor = 1
Therefore, HCF = 1
Hence, 15 and 28 are co-prime numbers.
Important Exam Tips
- Use divisibility rules before performing long calculations.
- Apply algebraic identities for square and cube calculations.
- Remember recurring decimal conversion shortcuts.
- Practice remainder and unit digit problems regularly.
- Memorize standard series formulas for quick solving.
- Use approximation techniques wherever possible.
Practicing solved examples regularly improves speed, accuracy, and confidence in solving Number System questions in competitive examinations.