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Number Theory

Number Theory

MODULES

Basics of Numbers
Types of Numbers
Fractions
Arithmetic Operations
Other Numerical Operations
Algebraic Expansion
Prime Numbers
Counting Integers
Past Questions

CONCEPTS & CHEATSHEET

Concept Revision Video

SPEED CONCEPTS

Number Theory 1
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Number Theory 2
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Algebraic Expansion 1
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PRACTICE

Number Theory : Level 1
Number Theory : Level 2
Number Theory : Level 3
ALL MODULES

CAT 2025 Lesson : Number Theory - Other Numerical Operations

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3. Other Numerical Operations

3.1 Square Root

Calculating square roots, cube roots, etc. by applying the prime factorisation method is detailed in the Surds & Indices lesson. This section will cover the Long division method for square roots only.

The steps to be followed can be best explained with an example. Let's find the square root of
556965569655696, i.e. 55696\sqrt{55696}55696​

Step 1: Starting from the right end, insert commas after every 2 digits.
556965569655696 is split into three parts and written as 5,56,96\bm{5 , 56 , 96}5,56,96

Step 2: Write the largest number whose square is less than the first part as the quotient (on the top) and divisor (on the left).

Step 3: After subtracting this square from the dividend, bring down the next pair of digits.



Step 4: Add the units digit of the divisor to itself. Find the largest possible digit which can then be added to the divisor such that the product of the digit and the divisor is less than the first part of the dividend brought down.

Step 5: Add this digit to the quotient. Subtract the product from the dividend and bring down the next pair of digits.



Steps 4 and 5 continue till we get a remainder of zero or the decimal at which we wish to approximate.



In the above example,
55696=236\sqrt{55696} = 23655696​=236.

As shown in the example below, in case of irrational square roots, we bring down zeroes in pairs after the decimal point.

Example 10

Approximate 70\sqrt{70}70​ to two decimal places.

Solution



We stop at three decimal places so that we can approximate.

70=8.366..∼8.37\sqrt{70} = 8.366.. \sim 8.3770​=8.366..∼8.37

Answer:
8.378.378.37

3.2 Modulus

The modulus function of a number provides the non-negative value of the number. To define it as a function,

∣x∣=x|x| = x∣x∣=x if x≥0x \ge 0x≥0
      =−x\space \space \space \space \space \space= -x      =−x if x<0x \lt 0x<0

Example 11

How many integral values of xxx satisfy ∣x−2∣<4|x - 2 | \lt 4∣x−2∣<4?

Solution

If x>2\bm{x \gt 2}x>2, then
x−2<4x - 2 \lt 4x−2<4
⇒
x<6 x \lt 6x<6
Possible values of
xxx are 333, 444 and 555.

If
x<2\bm{x \lt 2}x<2, then
−(x−2)<4-(x - 2) \lt 4−(x−2)<4
⇒
−x<2 -x \lt 2−x<2
x>−2 x \gt -2x>−2
Possible values of
xxx are −1-1−1, 000 and 111.

If
x=2\bm{x = 2}x=2, then condition is satisfied. ∴x=2\therefore x = 2∴x=2 is also a possible value.

∴\therefore∴ Total of 777 possible values.

Answer:
777


3.3 Factorial

A Factorial is represented by an exclamation mark or '
!!!'. 'n!\bm{n!}n!' is read as 'n\bm{n}n factorial'.

n!\bm{n!}n! is the product of all natural numbers less than or equal to n\bm{n}n.
∴n!=1×2....×n\therefore n! = 1 \times 2.... \times n∴n!=1×2....×n

Exception: Although 0 is not a natural number,
0!=1\bm{0! = 1}0!=1

To enhance your speed, you should memorise the values of the following commonly used factorials.

0!=10! = 10!=1 1!=11! = 11!=1 2!=22! = 22!=2 3!=63! = 63!=6 4!=244! = 244!=24
5!=1205! = 1205!=120 6!=7206! = 7206!=720 7!=50407! = 50407!=5040 8!=403208! = 403208!=40320


Note the following properties with factorials.

1)
n!×(n+1)=(n+1)!n! \times (n + 1) = (n + 1)!n!×(n+1)=(n+1)!

2)
(n+1)!−n!=n×n!(n + 1)! - n! = n \times n!(n+1)!−n!=n×n!

Example 12

Which of the following equals 10!10!10!?

(1)
9!×99! \times 99!×9            (2) 11!10\dfrac{11!}{10}1011!​            (3) 8!×1008! \times 1008!×100            (4) 8!×908! \times 908!×90           

Solution

8!×90=8!×9×10=10!8! \times 90 = 8! \times 9 \times 10 = 10!8!×90=8!×9×10=10!

Answer: (4)
8!×908! \times 908!×90

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