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Quantitative Aptitude: Number System

1. Concept Definitions & Explanations

  • Natural Numbers ($\mathbb{N}$): Counting numbers starting from 1. E.g., $1, 2, 3, 4, \dots$
  • Whole Numbers ($\mathbb{W}$): Natural numbers including zero. E.g., $0, 1, 2, 3, \dots$
  • Integers ($\mathbb{Z}$): Complete numbers (both positive, negative, and zero). E.g., $\dots, -3, -2, -1, 0, 1, 2, 3, \dots$
  • Rational Numbers: Numbers that can be expressed in the form $\frac{p}{q}$ where $p$ and $q$ are integers and $q \neq 0$. E.g., $\frac{2}{3}, -5, 0.75$.
  • Irrational Numbers: Numbers that cannot be expressed in the form $\frac{p}{q}$. E.g., $\sqrt{2}, \pi, e$.
  • Real Numbers: The set of all rational and irrational numbers.
  • Prime Numbers: Numbers greater than 1 that have only two factors: 1 and themselves. E.g., $2, 3, 5, 7, 11, 13, \dots$ (Note: 2 is the only even prime number).
  • Composite Numbers: Numbers greater than 1 that are not prime. E.g., $4, 6, 8, 9, \dots$ (Note: 1 is neither prime nor composite).
  • Co-prime Numbers: Two numbers are co-prime if their Highest Common Factor (HCF) is 1. E.g., $(8, 15)$.

2. Key Formulas & Shortcuts

Sum of Series

  1. Sum of first $n$ natural numbers: $$S_n = \frac{n(n+1)}{2}$$
  2. Sum of squares of first $n$ natural numbers: $$S_n^2 = \frac{n(n+1)(2n+1)}{6}$$
  3. Sum of cubes of first $n$ natural numbers: $$S_n^3 = \left[\frac{n(n+1)}{2}\right]^2$$
  4. Sum of first $n$ odd numbers: $$S_{odd} = n^2$$
  5. Sum of first $n$ even numbers: $$S_{even} = n(n+1)$$

Progressions

  • Arithmetic Progression (AP):

    • $n^{\text{th}}$ term: $T_n = a + (n-1)d$
    • Sum of $n$ terms: $S_n = \frac{n}{2}[2a + (n-1)d] = \frac{n}{2}(a + l)$
    • where $a = \text{first term}$, $d = \text{common difference}$, $l = \text{last term}$.

  • Geometric Progression (GP):

    • $n^{\text{th}}$ term: $T_n = a \cdot r^{n-1}$
    • Sum of $n$ terms ($r \neq 1$): $S_n = \frac{a(r^n - 1)}{r - 1}$ for $r > 1$, or $S_n = \frac{a(1 - r^n)}{1 - r}$ for $r < 1$.
    • Sum of infinite terms ($|r| < 1$): $S_{\infty} = \frac{a}{1 - r}$
    • where $a = \text{first term}$, $r = \text{common ratio}$.

Number of Factors

If a number $N$ is written in its prime factorized form: $$N = p^a \cdot q^b \cdot r^c \dots$$ (where $p, q, r$ are prime numbers), then:

  • Total number of factors $= (a+1)(b+1)(c+1)\dots$

3. Practice Problems

Problem 1

Find the unit digit of $(287)^{562} \times (124)^{321}$.

Problem 2

Find the sum of all natural numbers between 100 and 300 which are exactly divisible by 4.

Problem 3

Find the total number of factors of 360 (excluding 1 and the number itself).

Problem 4

A number when divided by 899 leaves a remainder 63. What will be the remainder when the same number is divided by 29?

Problem 5

Find the remainder when $2^{31}$ is divided by 5.

Problem 6

Find the number of zeroes at the end of the product $1 \times 2 \times 3 \times \dots \times 100$ (or $100!$).

Problem 7

Find the sum of all terms of the infinite geometric series: $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$

Problem 8

The sum of a two-digit number and the number obtained by reversing its digits is 121. What is the sum of the digits of the number?

Problem 9

Find the number of positive integers $n$ less than or equal to 100 such that $n^2 - 1$ is divisible by 8.

Problem 10

What is the remainder when $7^{100}$ is divided by 100?

4. Step-by-Step Solutions

Solution 1

  1. Analyze the cycles of unit digits:
    • For base ending in $7$ ($287$): The unit digit cyclicity of $7$ is 4 ($7^1=7$, $7^2=9$, $7^3=3$, $7^4=1$, repeating).
      • Divide the exponent $562$ by $4$: $562 \div 4 = 140$ with a remainder of $2$.
      • Therefore, the unit digit of $(287)^{562}$ is same as $7^2$, which is $9$.
    • For base ending in $4$ ($124$): The unit digit cyclicity of $4$ is 2 ($4^{\text{odd}} = 4$, $4^{\text{even}} = 6$).
      • The exponent $321$ is odd.
      • Therefore, the unit digit of $(124)^{321}$ is $4$.
  2. Combine the results:
    • Multiply the individual unit digits: $9 \times 4 = 36$.
    • The unit digit of the final product is 6.

Solution 2

  1. Identify the terms:
    • The first number after 100 divisible by 4 is $104$. (Since "between" excludes 100 and 300).
    • The last number before 300 divisible by 4 is $296$.
    • This forms an Arithmetic Progression (AP) with first term $a = 104$, last term $l = 296$, and common difference $d = 4$.
  2. Find the number of terms ($n$): $$l = a + (n-1)d$$ $$296 = 104 + (n-1)4$$ $$192 = (n-1)4$$ $$n-1 = 48 \implies n = 49$$
  3. Calculate the sum ($S_n$): $$S_n = \frac{n}{2}(a + l)$$ $$S_{49} = \frac{49}{2}(104 + 296) = \frac{49}{2}(400) = 49 \times 200 = 9800$$
  4. Answer: The sum is 9800.

Solution 3

  1. Prime Factorization of 360: $$360 = 2^3 \times 3^2 \times 5^1$$
  2. Calculate Total Number of Factors: $$\text{Total Factors} = (3 + 1)(2 + 1)(1 + 1) = 4 \times 3 \times 2 = 24$$
  3. Exclude 1 and the number itself: $$\text{Required Factors} = 24 - 2 = 22$$
  4. Answer: The number of factors is 22.

Solution 4

  1. Represent the number:
    • Let the number be $N = 899k + 63$, where $k$ is an integer.
  2. Analyze divisibility by 29:
    • Check if 899 is divisible by 29: $899 \div 29 = 31$.
    • Since 899 is a multiple of 29, the term $899k$ leaves a remainder of 0 when divided by 29.
  3. Find the remainder of the constant term:
    • Divide 63 by 29: $$63 = 29 \times 2 + 5$$
    • The remainder is 5.
  4. Answer: The remainder is 5.

Solution 5

  1. Analyze unit/remainder cycles:
    • We need to find $2^{31} \pmod 5$.
    • Calculate powers of 2 modulo 5:
      • $2^1 \equiv 2 \pmod 5$
      • $2^2 \equiv 4 \pmod 5$
      • $2^3 \equiv 8 \equiv 3 \pmod 5$
      • $2^4 \equiv 16 \equiv 1 \pmod 5$
    • The cycle of remainders is $[2, 4, 3, 1]$ of length 4.
  2. Divide exponent by cycle length: $$31 = 4 \times 7 + 3$$
    • The remainder is 3.
  3. Determine the result:
    • $2^{31} \equiv 2^3 \equiv 3 \pmod 5$.
  4. Answer: The remainder is 3.

Solution 6

  1. Identify the source of zeroes:
    • Zeroes at the end of a factorial product are formed by the factors of 10, which are $2 \times 5$.
    • In any factorial, the prime factor 2 occurs more frequently than 5. Thus, the number of zeroes is determined by the number of times 5 divides $100!$.
  2. Apply Legendre's Formula: $$\text{Number of factors of 5} = \lfloor \frac{100}{5} \rfloor + \lfloor \frac{100}{25} \rfloor + \lfloor \frac{100}{125} \rfloor + \dots$$ $$\text{Number of factors of 5} = 20 + 4 + 0 = 24$$
  3. Answer: The number of trailing zeroes is 24.

Solution 7

  1. Identify type of series:
    • The series is $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \dots$
    • This is an infinite Geometric Progression (GP) where:
      • First term $a = 1$
      • Common ratio $r = \frac{1}{3}$
  2. Check convergence:
    • Since $|r| = \frac{1}{3} < 1$, the series converges.
  3. Apply the sum formula: $$S_{\infty} = \frac{a}{1 - r}$$ $$S_{\infty} = \frac{1}{1 - 1/3} = \frac{1}{2/3} = \frac{3}{2} = 1.5$$
  4. Answer: The sum of the series is 1.5 (or $\frac{3}{2}$).

Solution 8

  1. Represent the two-digit number:
    • Let the tens digit be $x$ and the units digit be $y$.
    • The value of the number is $10x + y$.
  2. Set up the reversed number:
    • Reversing the digits gives the number $10y + x$.
  3. Form the equation:
    • The sum of the numbers is 121: $$(10x + y) + (10y + x) = 121$$ $$11x + 11y = 121$$ $$11(x + y) = 121 \implies x + y = 11$$
  4. Answer: The sum of the digits is 11.

Solution 9

  1. Analyze divisibility of $n^2 - 1$ by 8:
    • $n^2 - 1 = (n-1)(n+1)$.
    • For $(n-1)(n+1)$ to be divisible by 8, $n$ must be odd. If $n$ is even, $n^2$ is even, so $n^2-1$ is odd, which is not divisible by 8.
    • Let $n = 2k+1$ for some integer $k$.
    • Then $n^2 - 1 = (2k+1)^2 - 1 = 4k^2 + 4k = 4k(k+1)$.
    • Since either $k$ or $k+1$ must be even, $k(k+1)$ is always divisible by 2.
    • Thus, $4k(k+1)$ is always divisible by $4 \times 2 = 8$.
    • Hence, $n^2 - 1$ is divisible by 8 for all odd positive integers $n$.
  2. Count the number of odd integers up to 100:
    • The odd positive integers less than or equal to 100 are $1, 3, 5, \dots, 99$.
    • The number of terms is $\frac{99 - 1}{2} + 1 = 50$.
  3. Answer: There are 50 such integers.

Solution 10

  1. Find $7^{100} \pmod{100}$:
    • Compute successive powers of 7 modulo 100:
      • $7^1 \equiv 7 \pmod{100}$
      • $7^2 \equiv 49 \pmod{100}$
      • $7^3 \equiv 49 \times 7 = 343 \equiv 43 \pmod{100}$
      • $7^4 \equiv 43 \times 7 = 301 \equiv 1 \pmod{100}$
    • Since $7^4 \equiv 1 \pmod{100}$, the pattern repeats every 4 powers.
  2. Apply to the exponent:
    • Divide the exponent 100 by the cycle length 4:
      • $100 = 4 \times 25$ (remainder is 0).
    • Therefore, $7^{100} = (7^4)^{25} \equiv 1^{25} \equiv 1 \pmod{100}$.
  3. Answer: The remainder is 1.