What Number System Questions Test
- Divisibility rules and finding missing digits
- Remainder problems and cyclicity of unit digits
- HCF and LCM — direct and word problems
- Classification of numbers (prime, rational, integers)
- Factors, multiples and counting zeros
Question bank coverage by sub-pattern
How Number System Appears in SSC CGL
Number System is part of the Quantitative Aptitude section (25% of the SSC CGL syllabus). See the full SSC CGL Tier-I breakdown.
Number System Solved Examples for SSC CGL
12 exam-pattern number system questions from the practice bank, each with a full solution.
Find the greatest number 23a 68b, which is divisible by 3 but NOT divisible by 9 .
- A.238689
- B.239685
- C.239688Correct
- D.237687
Solution
A number is divisible by 3 if the sum of its digits is divisible by 3. A number is divisible by 9 if the sum of its digits is divisible by 9. Check each option.
The HCF of two numbers is 12 and their LCM is 360. If one of the numbers is 60, find the other number.
- A.72Correct
- B.84
- C.48
- D.96
Solution
Using the relation: HCF x LCM = Product of two numbers. 12 x 360 = 60 x other number 4320 = 60 x other number Other number = 4320 / 60 = 72. Verification: HCF(60, 72) = 12, LCM(60, 72) = 360.
Which of the following numbers is a divisor of (4915 – 1)?
- A.46
- B.14
- C.8
- D.50Correct
Solution
49^15 - 1 = (7^2)^15 - 1 = 7^30 - 1. Using a^n - b^n = (a-b)(a^(n-1) + ... + b^(n-1)), 7^30 - 1 = (7-1)(...) = 6(...). Also, 7^30 - 1 = (7^2)^15 - 1 = 49^15 - 1 = (49-1)(...) = 48(...). So, it's divisible by both 6 and 48. Also, 49^15 - 1 = (49^3 - 1)(...) = (49-1)(49^2 + 49 + 1)(...) = 48(...). 49^15 - 1 = (49+1)(...) = 50(...). So, it's divisible by 50.
Which of the following can be the value of 'k' so that the number 217924k is divisible by 6?
- A.2Correct
- B.6
- C.0
- D.4
Solution
For divisibility by 6, the number must be divisible by 2 and 3. For divisibility by 2, k must be even. For divisibility by 3, the sum of digits must be divisible by 3. 2+1+7+9+2+4+k = 25+k. If k=2, 27 is divisible by 3.
What should replace * in the number 94 × 2357, so that number is divisible by 11 ?
- A.1Correct
- B.7
- C.8
- D.3
Solution
For divisibility by 11, (9+*+3+7) - (4+2+5) = 0 or multiple of 11. 19+* - 11 = 0 or 11 or 22. * = -8 or 3 or 14. So * = 3.
If the number of eight digits is 789x531y, the number is divisible by 72, then the value of (5x – 3y) will be:
- A.0
- B.–1
- C.1Correct
- D.2
Solution
Divisible by 72 means divisible by 8 and 9. For 8, 31y should be divisible by 8, so y = 2. For 9, 7+8+9+x+5+3+1+2 = 35+x should be divisible by 9, so x = 1. 5x-3y = 5-6 = -1.
A 6-digit number has digits as consecutive natural numbers. The number is always divisible by ______.
- A.4
- B.5
- C.2
- D.3Correct
Solution
The sum of consecutive natural numbers is always divisible by 3. Therefore, the 6-digit number is always divisible by 3.
What is the unit digit in the expansion of (33)^313?
- A.7
- B.3Correct
- C.9
- D.5
Solution
The unit digit of base is 3. The cyclicity of powers of 3 is 4. Power 313 mod 4 = 1. Therefore, unit digit = 3¹ = 3.
What is the unit digit in the expansion of (88)^397?
- A.2
- B.4
- C.8Correct
- D.0
Solution
The unit digit of base is 8. The cyclicity of powers of 8 is 4. Power 397 mod 4 = 1. Therefore, unit digit = 8¹ = 8.
What is the unit digit in the expansion of
- A.3
- B.1
- C.7Correct
- D.9
Solution
The unit digit of base is 7. The cyclicity of powers of 7 is 4. Power 309 mod 4 = 1. Therefore, unit digit = 7¹ = 7.
Find the HCF and LCM of 480 and 1280.
- A.HCF = 158, LCM = 3840
- B.HCF = 320, LCM = 3840
- C.HCF = 160, LCM = 3840Correct
- D.HCF = 160, LCM = 7680
Solution
To find HCF and LCM of 480 and 1280: HCF is the greatest common factor = 160. LCM = (Product of numbers) / HCF = (480 × 1280) / 160 = 3840.
Find the HCF and LCM of 1260 and 2940.
- A.HCF = 840, LCM = 8820
- B.HCF = 420, LCM = 8820Correct
- C.HCF = 418, LCM = 8820
- D.HCF = 420, LCM = 17640
Solution
To find HCF and LCM of 1260 and 2940: HCF is the greatest common factor = 420. LCM = (Product of numbers) / HCF = (1260 × 2940) / 420 = 8820.
Number System — Frequently Asked Questions
Is Number System worth preparing separately?
Yes. Divisibility, remainders and HCF/LCM appear as direct questions, and the same rules speed up simplification and algebra questions elsewhere in the paper.
What is the fastest way to improve at Number System questions?
Memorise divisibility rules for 3, 4, 7, 8, 9, 11 and 13, and the unit-digit cycles of 2–9. Most exam-pattern questions reduce to applying one rule correctly.
How many Number System practice questions does Parikshala have for SSC CGL?
21 exam-pattern Number System questions for SSC CGL Quantitative Aptitude, each with a step-by-step solution. Practice is free — no sign-up needed to start.