Advanced Concepts in Numbers

These are special tools and observations that help you crack seemingly tough problems in seconds — especially useful in speed-based aptitude tests.


Digital Root (also called Digit Sum)

Definition: Repeatedly add digits of a number until a single digit remains.

Example:

Digital root of 9876=9+8+7+6=303+0=3\text{Digital root of 9876} = 9 + 8 + 7 + 6 = 30 \Rightarrow 3 + 0 = 3

Why it's Useful

  • Helps in checking divisibility by 9 and 3
  • Used in verifying calculation errors
  • Reduces big number calculations for options-based questions

Properties

  • Digital root of a number is same as numbermod9\text{number} \mod 9, except when divisible by 9 (result is 9, not 0)
  • DR(a+b)=DR(DR(a)+DR(b))\text{DR}(a + b) = \text{DR}(\text{DR}(a) + \text{DR}(b))
  • DR(a×b)=DR(DR(a)×DR(b))\text{DR}(a \times b) = \text{DR}(\text{DR}(a) \times \text{DR}(b))

Cyclicity of Last Digits

Definition: Repeating pattern in the last digit of powers of numbers.

BasePattern of last digit
22, 4, 8, 6 (length = 4)
33, 9, 7, 1
44, 6
77, 9, 3, 1
99, 1

How to Use:
To find the last digit of aba^b:

  1. Find the cyclic pattern of base
  2. Compute bmodcycle lengthb \mod \text{cycle length}
  3. Pick the corresponding digit in the cycle

Example: Last digit of 72227^{222}
Cycle for 7: 7, 9, 3, 1 → Length = 4
222 mod 4 = 2 → Answer = 9


Remainder Theorems

1. Remainder when a number is divided by 9 or 3

Use digital root or sum of digits.

Remainder of 7382 \div9=Sum of digits=2020mod9=2\text{Remainder of 7382 \div 9} = \text{Sum of digits} = 20 → 20 mod 9 = 2

2. abmodma^b \mod m

Use modular arithmetic + cyclicity (or Euler’s theorem for high level).

Example: 2100mod52^{100} \mod 5
Cycle of 2nmod52^n \mod 5: 2, 4, 3, 1
100 mod 4 = 0 → 4th element → Answer = 1


3. Fermat's Little Theorem (Advanced)

If pp is a prime, then:

ap11modpa^{p-1} \equiv 1 \mod p

Used when:

  • aa is not divisible by pp
  • Needed to reduce large exponents

Number of Zeroes in Factorial

Trailing Zeroes in n!n!

Count of zeroes = Number of times 5 is a factor (2 is always in excess):

Zeroes in n!=n5+n25+n125+\text{Zeroes in } n! = \left\lfloor \frac{n}{5} \right\rfloor + \left\lfloor \frac{n}{25} \right\rfloor + \left\lfloor \frac{n}{125} \right\rfloor + \dots

Example: Zeroes in 100!
= 20 + 4 = 24


Factorial and Units Digit

To find the last non-zero digit of n!n!, eliminate 0s (remove 5s and matching 2s) and find the digit modulo 10. These are usually asked indirectly.


Highest Power of a Prime in a Factorial

In n!n!, the exponent of prime pp is:

np+np2+np3+\left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \left\lfloor \frac{n}{p^3} \right\rfloor + \dots

Finding Number of Digits in a Number

To find number of digits in base 10:

Digits in n=log10(n)+1\text{Digits in } n = \left\lfloor \log_{10}(n) \right\rfloor + 1

For aba^b:

Digits=blog10a+1\text{Digits} = \left\lfloor b \log_{10} a \right\rfloor + 1

Common Mistakes

MistakeClarification
Assuming digital root works for 7 or 11Only valid for 3 and 9
Ignoring mod cycle lengthsMust match base with correct cycle length
Forgetting to handle zero case in factorial problemsAlways watch for trailing zero traps

Example Problems

Q1. What is the digital root of 85328532?

8 + 5 + 3 + 2 = 18 → 1 + 8 = 9


Q2. Find last digit of 3453^{45}

Cycle: 3, 9, 7, 1
45 mod 4 = 1 → Answer = 3


Q3. Remainder when 2345623^456 is divided by 5?

Cycle of 23 mod 5 = same as 3 mod 5 = 3 → Cycle: 3, 4, 2, 1
456 mod 4 = 0 → Answer = 1


Q4. How many trailing zeroes in 200!?

2005+20025+200125=40+8+1=49\left\lfloor \frac{200}{5} \right\rfloor + \left\lfloor \frac{200}{25} \right\rfloor + \left\lfloor \frac{200}{125} \right\rfloor = 40 + 8 + 1 = **49**

Visual Summary

→ Digital Root = mod 9 trick
→ Cyclicity = last digits of powers (mod 10)
→ Factorials = zeroes, prime powers
→ Mod Arithmetic = use for remainder-based exponent problems