Episode 8 — Aptitude and Reasoning / 8.6 — Number System

8.6.c Number System -- Solved Examples

Each problem is categorized as Basic, Medium, or Advanced. Attempt the problem yourself before reading the solution.

Basic Level

Problem 1: Divisibility Check

Q: Is 4,83,126 divisible by 6?

Solution:

A number is divisible by 6 if it is divisible by BOTH 2 and 3.

Check divisibility by 2:
  Last digit = 6 (even) --> divisible by 2  ✓

Check divisibility by 3:
  Sum of digits = 4 + 8 + 3 + 1 + 2 + 6 = 24
  24 / 3 = 8 --> divisible by 3  ✓

Since divisible by both 2 and 3:
Answer: Yes, 4,83,126 is divisible by 6.

Problem 2: Number of Factors

Q: Find the number of factors of 180.

Solution:

Step 1: Prime factorization
180 = 2 x 90
    = 2 x 2 x 45
    = 2 x 2 x 3 x 15
    = 2 x 2 x 3 x 3 x 5
180 = 2^2 x 3^2 x 5^1

Step 2: Apply formula
Number of factors = (2+1)(2+1)(1+1)
                  = 3 x 3 x 2
                  = 18

Answer: 180 has 18 factors.

Problem 3: Unit Digit

Q: Find the unit digit of 3^2024.

Solution:

Unit digit of 3 follows cycle: {3, 9, 7, 1} with cycle length 4.

Divide power by cycle length:
2024 / 4 = 506 remainder 0

Remainder 0 --> take the LAST digit in the cycle = 1

Answer: Unit digit of 3^2024 is 1.

Problem 4: Place Value vs Face Value

Q: Find the difference between the place value and face value of 5 in the number 7,05,328.

Solution:

The digit 5 is in the thousands place.

Place value of 5 = 5 x 1000 = 5000
Face value of 5 = 5

Difference = 5000 - 5 = 4995

Answer: 4995

Problem 5: Sum of Factors

Q: Find the sum of all factors of 72.

Solution:

Step 1: Prime factorization
72 = 2^3 x 3^2

Step 2: Apply the sum of factors formula
Sum = [(2^4 - 1)/(2 - 1)] x [(3^3 - 1)/(3 - 1)]
    = [(16 - 1)/1] x [(27 - 1)/2]
    = 15 x 13
    = 195

Answer: Sum of all factors of 72 is 195.

Problem 6: Trailing Zeros

Q: How many trailing zeros are in 50! ?

Solution:

Trailing zeros = floor(50/5) + floor(50/25) + floor(50/125) + ...
               = 10 + 2 + 0
               = 12

Answer: 50! has 12 trailing zeros.

Problem 7: Even and Odd Factors

Q: Find the number of even and odd factors of 240.

Solution:

Step 1: Prime factorization
240 = 2^4 x 3^1 x 5^1

Step 2: Total factors = (4+1)(1+1)(1+1) = 5 x 2 x 2 = 20

Step 3: Odd factors (ignore the factor of 2)
Odd factors = (1+1)(1+1) = 2 x 2 = 4
The odd factors are: 1, 3, 5, 15

Step 4: Even factors = Total - Odd = 20 - 4 = 16

Answer: 4 odd factors, 16 even factors.

Medium Level

Problem 8: Remainder Using Properties

Q: Find the remainder when 37 x 43 is divided by 8.

Solution:

Rem(37/8) = 5     (since 37 = 8 x 4 + 5)
Rem(43/8) = 3     (since 43 = 8 x 5 + 3)

Rem(37 x 43 / 8) = Rem(5 x 3 / 8) = Rem(15 / 8) = 7

Answer: Remainder = 7

Problem 9: Divisibility by 11

Q: Check if 8,17,382 is divisible by 11.

Solution:

Number: 8  1  7  3  8  2
Pos:    1  2  3  4  5  6

Sum of odd-position digits  = 8 + 7 + 8 = 23
Sum of even-position digits = 1 + 3 + 2 = 6

Difference = 23 - 6 = 17

17 is not 0 and not a multiple of 11.

Answer: 8,17,382 is NOT divisible by 11.

Problem 10: Highest Power of a Prime in Factorial

Q: Find the highest power of 7 that divides 300!

Solution:

Highest power = floor(300/7) + floor(300/49) + floor(300/343) + ...
              = 42 + 6 + 0
              = 48

Answer: 7^48 is the highest power of 7 that divides 300!

Problem 11: Unit Digit of a Sum

Q: Find the unit digit of 7^75 + 3^75.

Solution:

Unit digit of 7^75:
  Cycle of 7: {7, 9, 3, 1}, length = 4
  75 / 4 = 18 remainder 3
  3rd position in cycle = 3

Unit digit of 3^75:
  Cycle of 3: {3, 9, 7, 1}, length = 4
  75 / 4 = 18 remainder 3
  3rd position in cycle = 7

Unit digit of sum = unit digit of (3 + 7) = 0

Answer: Unit digit = 0

Problem 12: Largest/Smallest n-Digit Multiple

Q: Find the largest 4-digit number divisible by 13.

Solution:

Largest 4-digit number = 9999

9999 / 13 = 769.15...

floor(769.15) = 769

Largest 4-digit multiple of 13 = 13 x 769 = 9997

Verification: 9997 / 13 = 769  ✓

Answer: 9997

Problem 13: Number of Multiples in a Range

Q: How many multiples of 8 lie between 100 and 400 (inclusive)?

Solution:

Multiples of 8 from 1 to 400 = floor(400/8) = 50
Multiples of 8 from 1 to 99  = floor(99/8)  = 12

Multiples in range [100, 400] = 50 - 12 = 38

Verification: First multiple >= 100 is 104 (8 x 13)
              Last multiple <= 400 is 400 (8 x 50)
              Count = 50 - 13 + 1 = 38  ✓

Answer: 38 multiples.

Problem 14: Co-prime Count (Euler's Totient)

Q: How many numbers less than 30 are co-prime to 30?

Solution:

30 = 2 x 3 x 5

phi(30) = 30 x (1 - 1/2) x (1 - 1/3) x (1 - 1/5)
        = 30 x 1/2 x 2/3 x 4/5
        = 30 x 8/30
        = 8

Answer: 8 numbers less than 30 are co-prime to 30.
They are: {1, 7, 11, 13, 17, 19, 23, 29}

Problem 15: Perfect Square Factors

Q: How many factors of 360 are perfect squares?

Solution:

360 = 2^3 x 3^2 x 5^1

For a factor to be a perfect square, each prime exponent must be even.

For 2: even exponents from 0 to 3 --> {0, 2} --> 2 choices
For 3: even exponents from 0 to 2 --> {0, 2} --> 2 choices
For 5: even exponents from 0 to 1 --> {0}    --> 1 choice

Number of perfect square factors = 2 x 2 x 1 = 4

They are: 2^0 x 3^0 x 5^0 = 1
          2^2 x 3^0 x 5^0 = 4
          2^0 x 3^2 x 5^0 = 9
          2^2 x 3^2 x 5^0 = 36

Answer: 4 perfect square factors.

Problem 16: Remainder of a Power

Q: Find the remainder when 2^100 is divided by 7.

Solution:

Method: Using Fermat's Little Theorem
Since 7 is prime and 2 is not divisible by 7:
  2^6 mod 7 = 1

2^100 = 2^(6 x 16) x 2^4
      = (2^6)^16 x 16
      
Rem = 1^16 x Rem(16/7)
    = 1 x 2
    = 2

Answer: Remainder = 2

Verification: 2^6 = 64, 64/7 = 9 remainder 1  ✓

Problem 17: Product of Factors

Q: Find the product of all factors of 24.

Solution:

24 = 2^3 x 3^1

Number of factors = (3+1)(1+1) = 8

Product of all factors = 24^(8/2) = 24^4

24^2 = 576
24^4 = 576^2 = 331,776

Answer: 331,776

Verification:
Factors of 24: {1, 2, 3, 4, 6, 8, 12, 24}
Product = 1 x 2 x 3 x 4 x 6 x 8 x 12 x 24
        = 6 x 4 x 6 x 8 x 12 x 24
        = 24 x 48 x 288
        = 24 x 13824
        = 331,776  ✓

Advanced Level

Problem 18: Remainder of Large Power

Q: What is the remainder when 17^200 is divided by 18?

Solution:

17 = 18 - 1, so 17 mod 18 = -1

17^200 mod 18 = (-1)^200 mod 18 = 1

Answer: Remainder = 1

Problem 19: Number of Trailing Zeros in a Product

Q: Find the number of trailing zeros in the product 10 x 20 x 30 x ... x 1000.

Solution:

This product = (10 x 1)(10 x 2)(10 x 3)...(10 x 100)
             = 10^100 x (1 x 2 x 3 x ... x 100)
             = 10^100 x 100!

Trailing zeros = 100 (from 10^100) + trailing zeros in 100!

Trailing zeros in 100!:
  floor(100/5) + floor(100/25) + floor(100/125)
  = 20 + 4 + 0
  = 24

Total trailing zeros = 100 + 24 = 124

Answer: 124 trailing zeros.

Problem 20: Divisibility Problem

Q: Find the largest number that divides 245, 1029, and 1813 leaving remainders 5, 9, and 13 respectively.

Solution:

If the number d divides:
  245 leaving remainder 5  --> d divides (245 - 5) = 240
  1029 leaving remainder 9 --> d divides (1029 - 9) = 1020
  1813 leaving remainder 13 --> d divides (1813 - 13) = 1800

We need HCF(240, 1020, 1800).

240  = 2^4 x 3 x 5
1020 = 2^2 x 3 x 5 x 17
1800 = 2^3 x 3^2 x 5^2

HCF = 2^2 x 3 x 5 = 60

Answer: The largest such number is 60.

Verification:
245 / 60 = 4 remainder 5   ✓
1029 / 60 = 17 remainder 9  ✓
1813 / 60 = 30 remainder 13 ✓

Problem 21: Last Two Digits

Q: Find the last two digits of 7^200.

Solution:

Method: Using Euler's theorem
phi(100) = 40

Since GCD(7, 100) = 1:
7^40 mod 100 = 1

7^200 = (7^40)^5

7^200 mod 100 = 1^5 = 1

Last two digits = 01

Answer: 01

Problem 22: How Many Numbers Divisible by Exactly One of Two Numbers

Q: How many numbers from 1 to 100 are divisible by 3 but NOT by 5?

Solution:

Numbers divisible by 3 (from 1 to 100) = floor(100/3) = 33
Numbers divisible by 15 (both 3 and 5) = floor(100/15) = 6

Numbers divisible by 3 but NOT 5 = 33 - 6 = 27

Answer: 27

Problem 23: Finding a Missing Digit Using Divisibility

Q: The 6-digit number 5x2y6z is divisible by 7, 11, and 13. Find x + y + z.

Solution:

A number divisible by 7, 11, and 13 is divisible by 7 x 11 x 13 = 1001.

Property of 1001: A 6-digit number is divisible by 1001 if and only if
the first 3 digits and last 3 digits form pairs like:
  abcabc = abc x 1001

So 5x2y6z must be of the form: 5x2 repeated as 5x25x2? No, let's reconsider.

Actually, for the form abcabc:
  abc x 1001 = abcabc

So 5x2y6z = 5x2 5x2 --> y=5, 6=x, z=2

Wait, let's match: 5x2y6z = 5x25x2
Position matching: y=5, 6=x, z=2
So x=6, y=5, z=2

x + y + z = 6 + 5 + 2 = 13

Verification: 562562 / 1001 = 562  ✓
562562 / 7 = 80366 ✓  (562 x 143 = 80366, wait let me verify differently)
562 x 1001 = 562562, and 1001 = 7 x 11 x 13, so 562562 is divisible by 7, 11, and 13.  ✓

Answer: x + y + z = 13

Problem 24: Euler's Totient Application

Q: Find the remainder when 7^222 is divided by 10.

Solution:

phi(10) = 10 x (1 - 1/2) x (1 - 1/5) = 10 x 1/2 x 4/5 = 4

Since GCD(7, 10) = 1:
7^4 mod 10 = 1

222 = 4 x 55 + 2

7^222 = (7^4)^55 x 7^2 = 1^55 x 49

49 mod 10 = 9

Answer: Remainder = 9

Alternative (cyclicity check): Unit digit of 7 cycle = {7,9,3,1}
222 mod 4 = 2, second in cycle = 9  ✓

Problem 25: Expressing as Product of Two Factors

Q: In how many ways can 120 be expressed as a product of two factors?

Solution:

120 = 2^3 x 3^1 x 5^1

Number of factors = (3+1)(1+1)(1+1) = 16

Since 120 is NOT a perfect square:
Number of ways to express as product of two factors = 16/2 = 8

The 8 pairs: (1, 120), (2, 60), (3, 40), (4, 30), (5, 24), (6, 20), (8, 15), (10, 12)

Answer: 8 ways.

Problem 26: Highest Power of Composite in Factorial

Q: Find the highest power of 12 that divides 48!

Solution:

12 = 2^2 x 3

The highest power of 12 in 48! is determined by:
  Power of 2 in 48!: floor(48/2) + floor(48/4) + floor(48/8) + floor(48/16) + floor(48/32)
                    = 24 + 12 + 6 + 3 + 1 = 46
  Power of 3 in 48!: floor(48/3) + floor(48/9) + floor(48/27)
                    = 16 + 5 + 1 = 22

For 12 = 2^2 x 3:
  From 2: 46 powers available, each 12 needs 2 --> floor(46/2) = 23
  From 3: 22 powers available, each 12 needs 1 --> 22

Highest power of 12 = min(23, 22) = 22

Answer: 12^22 is the highest power of 12 dividing 48!

Problem 27: Sum of All Numbers Formed

Q: Find the sum of all 3-digit numbers formed using digits 1, 2, 3 (with repetition allowed).

Solution:

With repetition, total 3-digit numbers = 3^3 = 27

Each digit (1, 2, 3) appears at each place (hundreds, tens, units) equally.
Each digit appears at each position = 27/3 = 9 times.

Sum of digits = 1 + 2 + 3 = 6

Sum at units place    = 9 x 6 x 1   = 54
Sum at tens place     = 9 x 6 x 10  = 540
Sum at hundreds place = 9 x 6 x 100 = 5400

Total sum = 54 + 540 + 5400 = 5994

Answer: 5994

Problem 28: Remainder When Dividing Factorial

Q: Find the remainder when 1! + 2! + 3! + ... + 100! is divided by 10.

Solution:

For n >= 5, n! contains both 2 and 5 as factors, so n! is divisible by 10.
This means 5!, 6!, 7!, ..., 100! all give remainder 0 when divided by 10.

We only need: 1! + 2! + 3! + 4!
= 1 + 2 + 6 + 24
= 33

33 mod 10 = 3

Answer: Remainder = 3

Problem 29: Divisibility and Missing Digits

Q: Find the value of * in the number 34*56 so that it is divisible by 9.

Solution:

For divisibility by 9, the sum of digits must be divisible by 9.

Sum = 3 + 4 + * + 5 + 6 = 18 + *

For (18 + *) to be divisible by 9:
18 is already divisible by 9, so * must be 0 or 9.

Since * is a single digit: * = 0 or * = 9.

Answer: * = 0 or * = 9

Problem 30: Complex Remainder

Q: Find the remainder when 3^444 + 4^333 is divided by 5.

Solution:

Part 1: 3^444 mod 5
Cycle of 3 mod 5: 3, 9->4, 27->2, 81->1 --> cycle {3, 4, 2, 1}, length 4
444 mod 4 = 0 --> last in cycle = 1
3^444 mod 5 = 1

Part 2: 4^333 mod 5
4 = 5 - 1, so 4 mod 5 = -1
(-1)^333 = -1
4^333 mod 5 = -1 mod 5 = 4

Sum: (1 + 4) mod 5 = 5 mod 5 = 0

Answer: Remainder = 0

Summary of Problem Types

Problem TypeKey ConceptProblems
Divisibility checkDivisibility rules1, 9
Number of factors(a+1)(b+1)(c+1)2, 7
Unit digitCyclicity3, 11
Place value/face valuePosition-based value4
Sum of factors[(p^(a+1)-1)/(p-1)] product5
Trailing zerosfloor(n/5) + floor(n/25) + ...6, 19
Remainder problemsFermat, Euler, patterns8, 16, 18, 24, 28, 30
Range countingInclusion-exclusion13, 22
Co-prime countEuler's totient14
Perfect square factorsEven exponents15
Product of factorsN^(d/2)17
HCF applicationSubtract remainders, find HCF20
Last two digitsEuler mod 10021
Missing digitDivisibility rules23, 29
Factors as productsd(N)/225
Highest power in factorialLegendre's formula10, 26
Digit-based sumsCombinatorial counting27