Number System Concepts

1. Unit Digit & Trailing Zeros

• Formula: For unit digits, identify the repeating cycle of the base. Common cycles:

  • (2:\ 2,4,8,6)
  • (3:\ 3,9,7,1)
  • (7:\ 7,9,3,1)
  • (8:\ 8,4,2,6)
  • (4:\ 4,6), (9:\ 9,1)
  • (0,1,5,6): unit digit remains unchanged.

  For trailing zeros in (n!):

$$Z=\left\lfloor\frac n5\right\rfloor+\left\lfloor\frac n{25}\right\rfloor+\left\lfloor\frac n{125}\right\rfloor+\cdots$$

• Example: Find the unit digit of (7^{103}). Solution: Cycle of 7:

$$7,9,3,1$$

Cycle length (=4).

$$103\div4\Rightarrow \text{remainder}=3$$

The 3rd digit in the cycle is 3.

Therefore, unit digit (=\boxed{3}).

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2. Trailing Zeros in Factorials

• Formula: The number of trailing zeros in (n!) equals the number of factors of 5:

$$Z=\left\lfloor\frac n5\right\rfloor+\left\lfloor\frac n{25}\right\rfloor+\left\lfloor\frac n{125}\right\rfloor+\cdots$$

• Example: Find the number of trailing zeros in (100!). Solution:

$$\left\lfloor\frac{100}{5}\right\rfloor=20$$ $$\left\lfloor\frac{100}{25}\right\rfloor=4$$ $$\left\lfloor\frac{100}{125}\right\rfloor=0$$

Therefore:

$$Z=20+4=\boxed{24}$$

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3. Number of Factors / Divisors

• Formula: If

$$N=p_1^{a}p_2^{b}p_3^{c}\cdots$$

then the number of positive factors is:

$$(a+1)(b+1)(c+1)\cdots$$

• Example: Find the number of factors of 360. Solution:

$$360=2^3\times3^2\times5$$

Therefore:

$$(3+1)(2+1)(1+1)$$ $$=4\times3\times2=\boxed{24}$$

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4. Sum of Factors

• Formula: If

$$N=p_1^{a}p_2^{b}\cdots$$

then the sum of all positive factors is:

$$\sigma(N)= \left(\frac{p_1^{a+1}-1}{p_1-1}\right) \left(\frac{p_2^{b+1}-1}{p_2-1}\right)\cdots$$

• Example: Find the sum of all factors of 12. Solution:

$$12=2^2\times3$$

Therefore:

$$\sigma(12)= \left(\frac{2^3-1}{2-1}\right) \left(\frac{3^2-1}{3-1}\right)$$ $$=7\times4=\boxed{28}$$

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5. Number of Odd and Even Factors

• Formula: If

$$N=2^a p_1^{b}p_2^{c}\cdots$$

then:

$$\text{Total factors}=(a+1)(b+1)(c+1)\cdots$$ $$\text{Odd factors}=(b+1)(c+1)\cdots$$ $$\text{Even factors}=\text{Total factors}-\text{Odd factors}$$

• Example: Find the number of odd factors of 360. Solution:

$$360=2^3\times3^2\times5$$

For an odd factor, the power of 2 must be zero.

Thus:

$$(2+1)(1+1)=3\times2=\boxed6$$

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6. Prime Factors and Distinct Prime Factors

• Formula: Prime factorization expresses a number as:

$$N=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$$

Number of distinct prime factors (=k). Total prime factors counted with multiplicity:

$$a_1+a_2+\cdots+a_k$$

• Example: Find the number of distinct prime factors and total prime factors of 360. Solution:

$$360=2^3\times3^2\times5$$

Distinct prime factors:

$$\boxed{3}$$

Total prime factors:

$$3+2+1=\boxed6$$

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7. Modular Arithmetic & Remainders

• Formula: If

$$a\equiv r\pmod n$$

then powers of (a) can be reduced using the repeating remainder cycle:

$$a^k\bmod n$$

For (\gcd(a,n)=1), Euler’s theorem gives:

$$a^{\phi(n)}\equiv1\pmod n$$

• Example: Find the remainder when (2^{256}) is divided by 7. Solution:

$$2^3=8\equiv1\pmod7$$

Now:

$$256=3(85)+1$$

Therefore:

$$2^{256}=(2^3)^{85}\times2$$ $$\equiv1^{85}\times2\equiv\boxed2\pmod7$$

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8. Remainder of Large Powers

• Formula: Find the cycle of (a^n\pmod m), then reduce the exponent using the cycle length.
• Example: Find the remainder when (3^{50}) is divided by 7. Solution:

$$3^1\equiv3,\quad3^2\equiv2,\quad3^3\equiv6,\quad3^4\equiv4,$$ $$3^5\equiv5,\quad3^6\equiv1\pmod7$$

Cycle length (=6).

$$50\div6\Rightarrow \text{remainder}=2$$

Hence:

$$3^{50}\equiv3^2\equiv\boxed2\pmod7$$

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9. Euler’s Totient Function

• Formula: If

$$n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k},$$

then:

$$\phi(n)=n\left(1-\frac1{p_1}\right) \left(1-\frac1{p_2}\right)\cdots \left(1-\frac1{p_k}\right)$$

For a prime (p):

$$\phi(p)=p-1$$

• Example: Find (\phi(12)). Solution:

$$12=2^2\times3$$ $$\phi(12)=12\left(1-\frac12\right)\left(1-\frac13\right)$$ $$=12\times\frac12\times\frac23=\boxed4$$

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10. Divisibility Rules

• Formula: Important rules:

  • Divisible by 2 → last digit even.
  • Divisible by 3 → sum of digits divisible by 3.
  • Divisible by 4 → last two digits divisible by 4.
  • Divisible by 5 → last digit 0 or 5.
  • Divisible by 8 → last three digits divisible by 8.
  • Divisible by 9 → sum of digits divisible by 9.
  • Divisible by 11 → difference between alternating digit sums is divisible by 11.

• Example: Is 7,236 divisible by 9? Solution: Sum of digits:

$$7+2+3+6=18$$

Since 18 is divisible by 9, 7,236 is divisible by 9.

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11. HCF and LCM Using Prime Factorization

• Formula: For prime factorizations:

  • HCF → take the minimum power of every common prime.
  • LCM → take the maximum power of every prime.

  For two positive integers:

$$\text{HCF}\times\text{LCM}=\text{Product of the numbers}$$

• Example: Find the HCF and LCM of 36 and 48. Solution:

$$36=2^2\times3^2$$ $$48=2^4\times3$$

HCF:

$$2^2\times3=\boxed{12}$$

LCM:

$$2^4\times3^2=\boxed{144}$$

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12. Number of Divisors from a Given Condition

• Formula: If

$$N=p^aq^b,$$

the number of divisors is:

$$(a+1)(b+1)$$

A perfect square has an odd number of divisors.

• Example: How many divisors does (2^4\times3^3) have? Solution:

$$(4+1)(3+1)=5\times4=\boxed{20}$$

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13. Perfect Square & Perfect Cube

• Formula: In prime factorization:

  • Perfect square → every prime has an even exponent.
  • Perfect cube → every prime has an exponent divisible by 3.

• Example: Is 3600 a perfect square? Solution:

$$3600=2^4\times3^2\times5^2$$

All exponents are even.

Therefore, 3600 is a perfect square.

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14. Base Conversion: Any Base to Decimal

• Formula:

$$(d_nd_{n-1}\ldots d_1d_0)*b =d_nb^n+d*{n-1}b^{n-1}+\cdots+d_1b+d_0$$

• Example: Convert ((101101)_2) to decimal. Solution:

$$1(2^5)+0(2^4)+1(2^3)+1(2^2)+0(2)+1$$ $$=32+8+4+1=\boxed{45}$$

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15. Decimal to Binary

• Formula: Repeatedly divide the decimal number by 2 and read the remainders from bottom to top.
• Example: Convert (45) to binary. Solution:

$$45\div2=22\text{ remainder }1$$ $$22\div2=11\text{ remainder }0$$ $$11\div2=5\text{ remainder }1$$ $$5\div2=2\text{ remainder }1$$ $$2\div2=1\text{ remainder }0$$ $$1\div2=0\text{ remainder }1$$

Reading upward:

$$\boxed{101101_2}$$

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16. Binary, Octal & Hexadecimal Conversion

• Formula: Group binary digits:

  • Binary (\leftrightarrow) Octal: groups of 3 bits.
  • Binary (\leftrightarrow) Hexadecimal: groups of 4 bits.

• Example: Convert ((101101)_2) to octal. Solution: Group from right:

$$101\quad101$$ $$101_2=5,\qquad101_2=5$$

Therefore:

$$(101101)_2=\boxed{55_8}$$

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Advanced Variants

17. Unit Digit of a Product of Large Powers

• Formula: Find the unit digit of each factor separately using its cycle, then multiply the resulting unit digits.
• Example: Find the unit digit of (7^{25}\times3^{14}). Solution: For (7^{25}):

$$25\mod4=1\Rightarrow\text{unit digit}=7$$

For (3^{14}):

$$14\mod4=2\Rightarrow\text{unit digit}=9$$

Therefore:

$$7\times9=63$$

Unit digit (=\boxed3).

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18. Last Two Digits of a Large Power

• Formula: Work modulo (100):

$$a^n\pmod{100}$$

Use the repeating cycle or modular reduction.

• Example: Find the last two digits of (3^{20}). Solution:

$$3^{10}=59049\Rightarrow3^{20}=59049^2$$

Alternatively, modulo 100:

$$3^{20}\equiv1\pmod{100}$$

Therefore, the last two digits are:

$$\boxed{01}$$

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19. Trailing Zeros in a Product

• Formula: Trailing zeros are determined by:

$$\min(\text{number of factors of 2},\text{number of factors of 5})$$

• Example: Find the number of trailing zeros in (25\times40\times16). Solution:

$$25=5^2,\quad40=2^3\times5,\quad16=2^4$$

Total factors of 2:

$$3+4=7$$

Total factors of 5:

$$2+1=3$$

Therefore:

$$\text{Trailing zeros}=\min(7,3)=\boxed3$$

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20. Highest Power of a Prime in (n!)

• Formula: Highest power of prime (p) dividing (n!):

$$v_p(n!)= \left\lfloor\frac np\right\rfloor+ \left\lfloor\frac n{p^2}\right\rfloor+ \left\lfloor\frac n{p^3}\right\rfloor+\cdots$$

• Example: Find the highest power of 3 dividing (50!). Solution:

$$\left\lfloor\frac{50}{3}\right\rfloor=16$$ $$\left\lfloor\frac{50}{9}\right\rfloor=5$$ $$\left\lfloor\frac{50}{27}\right\rfloor=1$$

Therefore:

$$v_3(50!)=16+5+1=\boxed{22}$$

Hence (3^{22}) divides (50!).

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21. Number of Factors of a Factorial

• Formula: First find the prime factorization of (n!):

$$n!=2^{a}3^{b}5^{c}\cdots$$

Then:

$$d(n!)=(a+1)(b+1)(c+1)\cdots$$

• Example: Find the number of factors of (10!). Solution:

$$v_2(10!)=5+2+1=8$$ $$v_3(10!)=3+1=4$$ $$v_5(10!)=2$$ $$v_7(10!)=1$$

Therefore:

$$d(10!)=(8+1)(4+1)(2+1)(1+1)$$ $$=9\times5\times3\times2=\boxed{270}$$

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22. Perfect Numbers

• Formula: A perfect number equals the sum of its positive proper divisors.

$$\sigma(n)=2n$$

Common examples:

$$6,\ 28,\ 496,\ 8128$$

• Example: Check whether 28 is a perfect number. Solution: Proper divisors of 28:

$$1,2,4,7,14$$

Their sum:

$$1+2+4+7+14=28$$

Therefore, 28 is a perfect number.