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Number System & Divisibility

Mathematics · Arithmetic Foundations

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Overview

Numbers are classified (natural, whole, integers, rational, irrational, prime, composite) and divisibility rules let you test factors without long division.

Key points

Natural numbers start at 1; whole numbers include 0; integers include negatives.
A prime has exactly two factors; 1 is neither prime nor composite; 2 is the only even prime.
Divisible by 4: last two digits divisible by 4; by 8: last three digits; by 3 or 9: digit sum.
Divisible by 11: (sum of odd-place digits − sum of even-place digits) is 0 or a multiple of 11.
If N = p^a × q^b × r^c, the number of factors is (a + 1)(b + 1)(c + 1).
Unit digits of powers repeat in cycles of 4 (2: 2,4,8,6; 3: 3,9,7,1; 7: 7,9,3,1; 8: 8,4,2,6).

Explanation

Divisibility by a composite number needs its co-prime factors: by 12 means by 3 and 4, not by 2 and 6.
For the unit digit of a^n, use n mod 4; if the remainder is 0, take the 4th term of the cycle.
Sum of factors of p^a q^b = (1 + p + … + p^a)(1 + q + … + q^b).
Between 1 and 100 there are 25 primes; memorise them up to 100 for elimination.

Step by step

Identify what is asked (factor, digit, divisibility).
Factorise or apply the rule.
Verify with one option before marking.

Examples

Is 7,32,152 divisible by 8? Last three digits 152 ÷ 8 = 19 → Yes.
Number of factors of 360: 360 = 2³ × 3² × 5 → (3+1)(2+1)(1+1) = 24.
Unit digit of 7²³: cycle 7, 9, 3, 1; 23 mod 4 = 3 → third term → 3.

Traps and advanced points

Trap: Treating 1 as prime, or forgetting that 2 is prime.
Trap: Testing divisibility by 12 using 2 and 6 (not co-prime) — e.g. 18 passes both but is not divisible by 12.
Trap: Using n mod 4 = 0 as "first term" instead of the fourth term of the cycle.

Quick revision

Factors count: add 1 to each prime power and multiply.
Co-prime pairs: 12 = 3 × 4, 18 = 2 × 9, 24 = 3 × 8.
Unit-digit cycle length = 4.

Practice MCQs

What is the remainder when 575 is divided by 11?
1. 8
2. 4
3. 5
4. 3
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Answer: (d) 3. 575 = 11×52 + 3.
What is the remainder when 670 is divided by 11?
1. 1
2. 0
3. 11
4. 10
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Answer: (d) 10. 670 = 11×60 + 10.
Find the LCM of 15 and 25.
1. 85
2. 70
3. 80
4. 75
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Answer: (d) 75. LCM × HCF = product of the two numbers.
Find the LCM of 20 and 60.
1. 100
2. 60
3. 40
4. 80
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Answer: (b) 60. LCM × HCF = product of the two numbers.
Find the LCM of 26 and 91.
1. 169
2. 182
3. 208
4. 195
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Answer: (b) 182. LCM × HCF = product of the two numbers.
What is the unit digit of 7^59?
1. 7
2. 9
3. 1
4. 3
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Answer: (d) 3. 7 ke powers ka unit-digit cycle: 7,9,3,1 (cyclicity 4).
The largest 4-digit number exactly divisible by 88 is:
1. 9988
2. 9944
3. 9999
4. 9856
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Answer: (b) 9944. Sabse badi 4-digit = 9999.
If a number is divisible by both 3 and 8, it must be divisible by:
1. 11
2. 24
3. 12
4. 16
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Answer: (b) 24. 3 aur 8 co-prime hain (koi common factor nahi except 1).
The number 4^61 + 4^62 + 4^63 is divisible by:
1. 11
2. 13
3. 21
4. 10
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Answer: (c) 21. Common factor 4^61 bahar nikaalo: 4^61 × (1 + 4 + 16) = 4^61 × 21.
What least number must be subtracted from 3000 to make it a perfect square?
1. 84
2. 61
3. 81
4. 91
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Answer: (a) 84. 54^2 = 2916 aur 55^2 = 3025.
The sum of first 20 natural numbers is:
1. 190
2. 220
3. 200
4. 210
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Answer: (d) 210. Formula: n(n+1)/2 = 20×21/2 = 210.
If the 9-digit number 8175x3268 is divisible by 9, the value of x is:
1. 6
2. 2
3. 8
4. 5
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Answer: (d) 5. 9 se divisible → digit-sum 9 ka multiple.
The unit digit of the product 1×2×3×...×10 is:
1. 0
2. 5
3. 2
4. 4
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Answer: (a) 0. Product me 10 (aur 2×5) present hai → factor of 10 → unit digit hamesha 0.
How many prime numbers are there between 1 and 50?
1. 15
2. 16
3. 14
4. 12
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Answer: (a) 15. Primes 1-50: 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47 = 15 primes.
The remainder when 2^31 is divided by 5 is:
1. 2
2. 1
3. 4
4. 3
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Answer: (d) 3. 2 ke powers mod 5 ka cycle: 2^1=2, 2^2=4, 2^3=3, 2^4=1 (cyclicity 4).