The Number System forms the backbone of quantitative aptitude in competitive examinations. This chapter is crucial as questions from this topic appear consistently across all major competitive exams, often accounting for 15-20% of the quantitative section. Mastering number systems, divisibility rules, and related concepts not only helps solve direct questions but also builds a strong foundation for topics like HCF-LCM, averages, percentages, and algebra. A thorough understanding of this chapter can significantly boost your overall score and problem-solving speed.
The Hindu-Arabic number system uses ten digits: $0, 1, 2, 3, 4, 5, 6, 7, 8, 9$
A numeral is a group of digits representing a number, written using a place-value chart:
| Ten Crores | Crores | Ten Lakhs | Lakhs | Ten Thousands | Thousands | Hundreds | Tens | Ones |
|---|
"Nine crore four lakh six thousand two" is written as $9,04,06,002$
The face value of a digit is its own value, regardless of its position in the numeral.
The place value depends on the position of the digit:
Counting numbers: $\{1, 2, 3, 4, 5, \ldots\}$
Natural numbers including zero: $\{0, 1, 2, 3, 4, \ldots\}$
All counting numbers, zero, and their negatives: $\{\ldots, -3, -2, -1, 0, 1, 2, 3, \ldots\}$
Numbers divisible by $2$: $\{0, 2, 4, 6, 8, 10, \ldots\}$
Numbers not divisible by $2$: $\{1, 3, 5, 7, 9, 11, \ldots\}$
Numbers with exactly two factors ($1$ and itself).
All primes less than 100: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97$
Since $12^2 > 137$, test divisibility by primes less than $12$: $2, 3, 5, 7, 11$.
None divides $137$, so $137$ is prime.
Since $18^2 > 319$, test divisibility by primes less than $18$: $2, 3, 5, 7, 11, 13, 17$.
$11$ divides $319$, so $319$ is not prime.
Numbers with more than 2 factors. The least composite number is $4$.
A number equal to the sum of its factors (excluding itself).
Two numbers whose HCF is $1$.
Examples: $(2, 3)$, $(8, 9)$
Two prime numbers whose difference is $2$.
Examples: $(3, 5)$, $(5, 7)$, $(11, 13)$
Numbers expressible as $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$.
Examples: $\frac{1}{8}, \frac{2}{11}, -\frac{8}{3}, 0, 6, 5$
Numbers with non-terminating, non-repeating decimal expansions.
Examples: $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{7}, \pi, e, 0.231764735\ldots$
Using $a^2 - b^2 = (a + b)(a - b)$:
$$= (796 + 204)(796 - 204) = 1000 \times 592 = 592000$$
$$= (2a)^2 + b^2 + (-c)^2 + 2(2a)(b) + 2(b)(-c) + 2(2a)(-c)$$
$$= (2a + b - c)^2$$
$\therefore$ Square root = $2a + b - c$
A number is divisible by $2$ if its unit digit is $0, 2, 4, 6$, or $8$.
$58694$ is divisible by $2$; $86945$ is not.
A number is divisible by $3$ if the sum of its digits is divisible by $3$.
In $695421$, sum $= 6 + 9 + 5 + 4 + 2 + 1 = 27$, divisible by $3$
$\therefore$ $695421$ is divisible by $3$
A number is divisible by $9$ if the sum of its digits is divisible by $9$.
In $246591$, sum $ = 27$, divisible by $9$
$\therefore$ $246591$ is divisible by $9$
Sum $= 1 + 9 + 7 + x + 5 + 4 + 6 + 2 = 34 + x$
For divisibility by $9$, $x = 2$ (making sum $= 36$)
A number is divisible by $4$ if its last two digits form a number divisible by $4$.
$6879376$ is divisible by $4$ ($76$ is divisible by $4$)
A number is divisible by $8$ if its last three digits form a number divisible by $8$.
$16789352$ is divisible by $8$ ($352$ is divisible by $8$)
A number is divisible by $10$ if its unit digit is $0$.
A number is divisible by $5$ if its unit digit is $0$ or $5$.
A number is divisible by $11$ if the difference between the sum of digits at odd places and the sum at even places is $0$ or divisible by $11$.
Odd places sum: $8+7+3+4 = 22$
Even places sum: $1+2+8 = 11$
Difference: $22-11 = 11$ (divisible by $11$)
$\therefore$ $4832718$ is divisible by $11$
A number is divisible by $25$ if its last two digits form $00$ or a number divisible by $25$.
$63875$ is divisible by $25$ ($75$ is divisible by $25$)
Divide the number into groups of $3$ digits from right. Find the difference between sum of groups at odd and even positions. If this is $0$ or divisible by $7/13$, the number is divisible by $7/13$.
$(792 + 4) - 537 = 259$, divisible by $7$
$\therefore$ $4537792$ is divisible by $7$
A number is divisible by $16$ if its last four digits form a number divisible by $16$.
Important Note: If a number is divisible by $p$ and $q$ (co-primes), then it's divisible by $pq$. If $p$ and $q$ are not co-primes, this may not hold.
$24 = 3 \times 8$ (co-primes)
Sum of digits $= 36$ (divisible by $3$) ✓
Last $3$ digits $= 744$ (divisible by $8$) ✓
$\therefore$ $52563744$ is divisible by $24$
$$= 567958 \times (100000 - 1)$$
$$= 56795800000 - 567958$$
$$= 56795232042$$
Put $n$ zeros to the right and divide by $2^n$.
$$= \frac{9754360000}{16} = 609647500$$
$$= 1600^2 + 5^2 + 2(1600)(5)$$
$$= 2560000 + 25 + 16000 = 2576025$$
$$n! = n \times (n-1) \times (n-2) \times \cdots \times 3 \times 2 \times 1$$
Note: $0! = 1$
Examples: $|-5| = 5$, $|4| = 4$, $|-1| = 1$
$[x]$ denotes the greatest integer not exceeding $x$.
Examples: $[1.35] = 1$, $\left[\frac{11}{4}\right] = [2.75] = 2$
$$\text{Divisor} = \frac{\text{Dividend} - \text{Remainder}}{\text{Quotient}}$$
$$= \frac{15968 - 37}{89} = 179$$
Number $= 114k + 21 = 19(6k) + 19 + 2 = 19(6k+1) + 2$
$\therefore$ Remainder when divided by $19 = 2$
$(9^6 - 1)$ is divisible by $(9 - 1) = 8$
$\therefore$ $(9^6 - 1) + 8$ is divisible by $8$
$\therefore$ $(9^6 + 7)$ gives remainder $0$
$(397^{3589} + 1)$ is divisible by $(397 + 1) = 398$ [odd power]
$\therefore$ $(397^{3589} + 1) + 4$ gives remainder $4$
$\therefore$ Remainder $= 4$
To find zeros at the end of $N = 1 \times 2 \times 3 \times \cdots \times 100$:
Find highest power of $5$ (limiting factor with $2$)
Highest power of $5 = \left[\frac{100}{5}\right] + \left[\frac{100}{25}\right] = 20 + 4 = 24$ zeros
Unit digit = Unit digit of $7^{153} \times 1^{72}$
$7^4$ gives unit digit $1$
$\therefore$ $7^{152}$ gives unit digit $1$
$\therefore$ $7^{153}$ gives unit digit $7$
$1^{72}$ gives unit digit $1$
$\therefore$ Unit digit $= 7 \times 1 = 7$
Working backwards:
$z = 8(1) + 7 = 15$
$y = 5(15) + 4 = 79$
$x = 3(79) + 1 = 238$
$$\begin{align} & 8888 \\ & 888 \\ & 88 \\ + & 8 \\ \hline & 9872 \end{align}$$
Answer: $9872$
This is of form $a^2 + b^2 + 2ab = (a+b)^2$
$$= (387 + 113)^2$$
$$= 500^2$$
$$= 250000$$
Using $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$
$$= \frac{(789 + 211)(789^2 - 789 \times 211 + 211^2)}{789^2 - 789 \times 211 + 211^2}$$
$$= 789 + 211$$
$$= 1000$$
$24^2 > 571$
Prime numbers $< 24$: $2, 3, 5, 7, 11, 13, 17, 19, 23$
$571$ is not divisible by any of these
$\therefore$ $571$ is prime
Odd places: $2 + 8 + 6 + 7 = 23$
Even places: $2 + 3 + 4 + 5 = 14$
Difference: $23 - 14 = 9$ (not divisible by $11$)
$\therefore$ Not divisible by $11$
For divisibility by $8$: Last $3$ digits $58N$ must be divisible by $8$
$\therefore$ $N = 4$ ($584$ is divisible by $8$)
For divisibility by $11$:
$(8 + 4 + 4 + 9 + M) - (4+5+8+0+3) = 0$ or multiple of $11$
$(25 + M) - 20 = M+5$
For divisibility: $M = 6$
$\therefore$ $M = 6, N = 4$
Number $= 6k + 3$
Square $= (6k + 3)^2 = 36k^2 + 36k + 9$
$= 36k^2 + 36k + 6 + 3$
$= 6(6k^2 + 6k + 1) + 3$
$\therefore$ Remainder $= 3$
Smallest $5$-digit number $= 10000$
$10000 \div 476$ gives remainder $4$
Number to add $= 476 - 4 = 472$
$\therefore$ Required number $= 10472$
Highest power of $5$ in $100!$
$$= \left[\frac{100}{5}\right] + \left[\frac{100}{25}\right]$$
$$= 20 + 4$$
$$= 24 \text{ zeros}$$
Remainder = Remainder when last $4$ digits $(5859)$ divided by $16$
$5859 \div 16 = 366$ remainder $3$
$\therefore$ Remainder $= 3$
| 1. What is the Hindu-Arabic numeral system? | ![]() |
| 2. What is the difference between face value and place value? | ![]() |
| 3. What are the different types of numbers? | ![]() |
| 4. What are some important algebraic formulae related to numbers? | ![]() |
| 5. How can one determine the divisibility of a number? | ![]() |