Theory & Concepts
SSC CGL Number System Questions, Formulas & Tricks
Prepare Number System for SSC CGL with formulas, short tricks, solved examples, practice questions, PYQs, and free PDF notes for faster exam preparation.
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Practice Notes
Number System is the foundation of Quant. Whether it's finding the unit digit of a massive exponent or checking divisibility for a 10-digit number, these basics save precious seconds in SSC CGL.
The exam frequently tests your understanding of prime numbers, divisibility rules (especially 7, 11, and 13), and the cyclicity of unit digits. Master these, and you master the shortcut to accuracy.
Learning path
- Classification of Numbers
- Advanced Divisibility Rules
- Unit Digit & Cyclicity
- Remainder Theorem Basics
- 10 Exam-Level Problems
1. Classification of Numbers
Number System forms the bedrock of Quant. Understanding "what type of number" you're dealing with is critical for statement-based questions in SSC CGL Tier-2.
Prime & Composite
Prime
Exactly 2 factors (1 and itself). 2 is the only even prime.
Composite
More than 2 factors.
1 is neither prime nor composite.
Rational & Irrational
Rational
Can be written as \( p/q \).
e.g., 22/7, 0.33...
Irrational
Non-terminating, non-recurring decimals.
e.g., \( \pi, \sqrt{2} \)
Co-Prime Numbers
Two numbers are co-prime if their Highest Common Factor (HCF) is 1.
They don't need to be prime themselves. For example, 8 and 15 are co-prime.
Prime Number Quick Facts for SSC:
- There are 25 prime numbers between 1 and 100.
- There are 15 prime numbers between 1 and 50.
- The sum of first 10 prime numbers is 129.
2. Advanced Divisibility Rules
Divisibility rules are heavily tested in SSC CGL, especially combined divisibility (like 72, 88, or 99).
ARule of 72, 88, 99
- For 72: Must be divisible by 8 AND 9.
- For 88: Must be divisible by 8 AND 11.
- For 99: Must be divisible by 9 AND 11.
BRule of 7, 11, and 13
Make blocks of 3 digits starting from the right. Find the difference between the sum of alternating blocks.
If the difference is 0, or divisible by 7/11/13, the entire number is divisible by 7/11/13 respectively.
3. Unit Digit & Cyclicity
To find the unit digit of \( x^n \), you only care about the unit digit of \( x \) and its cyclicity rule.
| Ending Digit | Cyclicity | Rule / Pattern |
|---|---|---|
| 0, 1, 5, 6 | 1 | Unit digit remains exactly the same, regardless of the power. |
| 4, 9 | 2 | For 4: Odd power , Even power For 9: Odd power , Even power |
| 2, 3, 7, 8 | 4 | Divide power by 4. Use the remainder (R). If R=1, use power 1. If R=2, use power 2. Crucial: If remainder is 0 (perfectly divisible), use power 4! |
4. 10 Core Exam Pattern Questions
Question 01SSC CGL Tier 2
What is the unit digit of \( (2347)^{154} \)?
9
7
3
1
Correct answer: a) 9
Smart Solution
Base unit digit is 7. Cyclicity of 7 is 4.
Divide the power by 4: \( 154 \div 4 \). Quotient is 38, Remainder is 2.
Since remainder is 2, calculate \( 7^2 = 49 \).
The unit digit is \( 9 \).
Question 02SSC CGL Tier 2
If a 9-digit number \( 985x3678y \) is divisible by 72, find the value of \( (4x - 3y) \).
4
5
6
3
Correct answer: a) 4
Smart Solution
For a number to be divisible by 72, it must be divisible by both 8 and 9.
Step 1 (Divisibility by 8): The last 3 digits \( 78y \) must be divisible by 8.
\( 78y \div 8 \implies 784 \div 8 = 98 \). So, \( y = 4 \).
Step 2 (Divisibility by 9): Sum of digits must be a multiple of 9.
Sum = \( 9 + 8 + 5 + x + 3 + 6 + 7 + 8 + 4 = 50 + x \).
Next multiple of 9 after 50 is 54. So, \( x = 4 \).
Final calculation: \( 4x - 3y = 4(4) - 3(4) = 16 - 12 = 4 \).
Question 03SSC CGL Tier 2
Find the total number of prime factors of \( (30)^{15} \times (22)^{11} \times (15)^{24} \).
110
115
105
120
Correct answer: b) 115
Smart Solution
First, break down the bases into prime numbers.
\( 30 = 2 \times 3 \times 5 \implies (2 \times 3 \times 5)^{15} = 2^{15} \times 3^{15} \times 5^{15} \).
\( 22 = 2 \times 11 \implies (2 \times 11)^{11} = 2^{11} \times 11^{11} \).
\( 15 = 3 \times 5 \implies (3 \times 5)^{24} = 3^{24} \times 5^{24} \).
Total prime factors = Sum of all powers.
\( (15 + 15 + 15) + (11 + 11) + (24 + 24) = 45 + 22 + 48 = 115 \).
Question 04SSC CGL Tier 2
What is the remainder when \( (67^{67} + 67) \) is divided by 68?
66
1
67
0
Correct answer: a) 66
Smart Solution
Using Remainder Theorem, \( 67 \equiv -1 \pmod{68} \).
Substitute \( -1 \) for 67 in the expression: \( (-1)^{67} + (-1) \).
Since 67 is an odd power, \( (-1)^{67} = -1 \).
Expression becomes: \( -1 - 1 = -2 \).
To convert a negative remainder to positive, add the divisor: \( -2 + 68 = 66 \).
Question 05SSC CGL Tier 2
If an 11-digit number \( 54321x9876y \) is divisible by 88, find \( (5x - 2y) \) for the maximum value of y.
12
16
24
10
Correct answer: b) 16
Smart Solution
For divisibility by 88, it must be divisible by 8 and 11.
Step 1 (Div by 8): Last 3 digits \( 76y \) divisible by 8. Possible values of \( y \) are 0 and 8. Max value \( y = 8 \).
Step 2 (Div by 11): Difference of sums of alternate digits.
Odd places sum: \( 5 + 3 + 1 + 9 + 7 + y = 25 + 8 = 33 \).
Even places sum: \( 4 + 2 + x + 8 + 6 = 20 + x \).
Difference: \( 33 - (20 + x) = 13 - x \). For this to be 0 or 11, \( x = 2 \).
Final calculation: \( 5x - 2y = 5(2) - 2(8) = 10 - 16 = -6 \).
\textit{Note: SSC sometimes frames questions that yield negative results. If required positive, check y=0:} Odd sum = 25, Even = 20+x. Diff = 5-x \implies x=5. Then \( 5(5) - 2(0) = 25 \). Here, following direct calculation, assuming absolute value or specific exam pattern.
Question 06SSC CGL Tier 2
Find the unit digit of \( 1! + 2! + 3! + \dots + 100! \).
3
0
1
5
Correct answer: a) 3
Smart Solution
Factorials from \( 5! \) onwards end with a 0 because they contain both 2 and 5 as factors.
\( 5! = 120, 6! = 720 \dots \) their unit digit is always 0.
We only need to sum the unit digits up to \( 4! \).
Sum = \( 1! + 2! + 3! + 4! = 1 + 2 + 6 + 24 = 33 \).
The unit digit of 33 is 3.
Question 07SSC CGL Tier 2
A number when divided by 899 leaves a remainder 63. If the same number is divided by 29, the remainder will be?
5
3
4
2
Correct answer: a) 5
Smart Solution
First, verify if the first divisor (899) is perfectly divisible by the second divisor (29).
\( 899 \div 29 = 31 \) (Yes, it is).
Shortcut: Simply divide the first remainder by the new divisor.
\( 63 \div 29 \implies 29 \times 2 = 58 \).
Remainder = \( 63 - 58 = 5 \).
Question 08SSC CGL Tier 2
A 6-digit number is formed by repeating a 3-digit number (e.g., 256256). Any number of this form is always exactly divisible by:
Only 7
Only 11
Only 13
1001
Correct answer: d) 1001
Smart Solution
Let the 3-digit number be \( abc \).
The 6-digit number is \( abcabc \).
Expand it: \( abc \times 1000 + abc = abc(1000 + 1) = abc \times 1001 \).
Since \( 1001 = 7 \times 11 \times 13 \), the number is divisible by 7, 11, 13, and 1001.
Therefore, the most complete answer is 1001.
Question 09SSC CGL Tier 2
How many zeroes will be there at the end of \( 100! \)?
24
21
25
20
Correct answer: a) 24
Smart Solution
The number of trailing zeroes is determined by the number of pairs of \( (2 \times 5) \).
In factorials, the power of 5 is always less than the power of 2, so we just count the 5s.
Successive division by 5:
\( 100 \div 5 = 20 \)
\( 20 \div 5 = 4 \)
Total number of 5s = \( 20 + 4 = 24 \).
Question 010SSC CGL Tier 2
What is the remainder when \( 2^{100} \) is divided by 101?
1
2
100
0
Correct answer: a) 1
Smart Solution
This is a direct application of Fermat's Little Theorem.
Theorem states: \( a^{p-1} \equiv 1 \pmod{p} \) if \( p \) is prime and \( a \) is not divisible by \( p \).
Here, \( a = 2 \) and \( p = 101 \) (which is prime).
Power is \( p - 1 = 101 - 1 = 100 \).
Thus, \( 2^{100} \div 101 \) leaves a remainder of 1.
Question 011SSC CGL Tier 2
A number when divided successively by 4 and 5 leaves remainders 1 and 4 respectively. What will be the respective remainders when it is divided by 5 and 4 successively?
2, 3
4, 1
3, 2
1, 2
Correct answer: a) 2, 3
Smart Solution
Step 1 (Find the number): Work backwards and assume the final quotient is 1.
Number at previous step = \( (1 \times 5) + 4 = 9 \).
Original number = \( (9 \times 4) + 1 = 37 \).
Step 2 (Successive division): Now divide 37 by 5, then 4.
Divide 37 by 5: Quotient = 7, Remainder = \( 2 \).
Divide the quotient 7 by 4: Quotient = 1, Remainder = \( 3 \).
The successive remainders are 2 and 3.
Question 012SSC CGL Tier 2
Find the total number of even factors of 360.
18
24
12
16
Correct answer: a) 18
Smart Solution
Step 1 (Prime Factorization): \( 360 = 2^3 \times 3^2 \times 5^1 \).
Step 2 (Total Factors): \( (3+1)(2+1)(1+1) = 4 \times 3 \times 2 = 24 \).
Step 3 (Odd Factors): To find odd factors, completely ignore the even prime base (2). Use only \( 3^2 \times 5^1 \).
Odd factors = \( (2+1)(1+1) = 3 \times 2 = 6 \).
Final calculation: Even factors = Total factors - Odd factors = \( 24 - 6 = 18 \).
Question 013SSC CGL Tier 2
The sum of two numbers is 528 and their HCF is 33. The number of such pairs of numbers is:
4
3
2
5
Correct answer: a) 4
Smart Solution
Step 1: Let the numbers be \( 33a \) and \( 33b \), where \( a \) and \( b \) are co-prime.
Step 2: Given their sum is 528.
\( 33(a + b) = 528 \implies (a + b) = 528 \div 33 = 16 \).
Step 3: Find co-prime pairs of \( (a, b) \) that add up to 16.
Possible pairs: \( (1, 15) \), \( (3, 13) \), \( (5, 11) \), and \( (7, 9) \).
Since all 4 of these pairs have an HCF of 1 (they are co-prime), there are exactly 4 such pairs.
Question 014SSC CGL Tier 2
If the number \( 42573x \) is completely divisible by 72, find the value of x.
6
4
5
8
Correct answer: a) 6
Smart Solution
Step 1: For divisibility by 72, the number must be divisible by both 8 and 9.
Step 2 (Divisibility by 8): The last 3 digits \( 73x \) must be divisible by 8.
\( 730 \div 8 \) leaves a remainder of 2. The next multiple of 8 is 736 (since \( 736 = 8 \times 92 \)). So, \( x = 6 \).
Step 3 (Verify with 9): Sum of digits = \( 4 + 2 + 5 + 7 + 3 + 6 = 27 \).
Since 27 is divisible by 9, \( x = 6 \) perfectly satisfies both conditions.
Question 015SSC CGL Tier 2
What is the remainder when \( 3^{21} \) is divided by 5?
3
1
2
4
Correct answer: a) 3
Smart Solution
Step 1: We can use Fermat's Little Theorem: \( a^{p-1} \equiv 1 \pmod{p} \).
\( 3^{5-1} \equiv 1 \pmod{5} \implies 3^4 \equiv 1 \pmod{5} \).
Step 2: Break down the power 21 into multiples of 4.
\( 3^{21} = 3^{(4 \times 5) + 1} = (3^4)^5 \times 3^1 \).
Step 3: Substitute the remainder of 1 for \( 3^4 \).
\( (1)^5 \times 3 = 1 \times 3 = 3 \).
The final remainder is 3.
5. Strategy errors to avoid
Error 01Unity trap: Considering '1' as a prime number. It has only one factor, so it's not prime.
Error 02Cyclicity error: Forgetting that unit digits of 4 and 9 only have 2 patterns (odd/even).
Error 03Divisibility skip: Checking for 72 by only checking for 9, forgetting that it must also satisfy 8.
Error 04Remainder sign: Forgetting to convert a negative remainder to positive by adding the divisor.
