How to Convert Base 4 to Base 6?

Converting numbers between different bases can seem tricky, but with a systematic approach, it becomes manageable. Let’s break down the process of converting a number from base 4 to base 6 step-by-step.

  1. Convert Base 4 to Base 10
    The first step is to convert the base 4 number to a base 10 (decimal) number. Here’s how you can do it:

Example

Let’s convert the base 4 number 123_4 to base 10.

  1. Write down the number and its positional values:
    • $1 times 4^2$
    • $2 times 4^1$
    • $3 times 4^0$
  2. Calculate each term:
    • $1 times 4^2 = 1 times 16 = 16$
    • $2 times 4^1 = 2 times 4 = 8$
    • $3 times 4^0 = 3 times 1 = 3$
  3. Sum these values to get the decimal number:
    • $16 + 8 + 3 = 27$

So, 123_4 is 27_{10} in base 10.

  1. Convert Base 10 to Base 6
    Next, we convert the base 10 number to a base 6 number. This involves repeatedly dividing the number by 6 and keeping track of the remainders.

Example

Let’s convert 27_{10} to base 6.

  1. Divide 27 by 6 and record the quotient and remainder:
    • $27 , div , 6 = 4$ with a remainder of $3$
  2. Divide the quotient (4) by 6:
    • $4 , div , 6 = 0$ with a remainder of $4$
  3. Read the remainders from bottom to top to get the base 6 number:
    • Remainders: $4, 3$
    • So, 27_{10} is 43_6 in base 6.

Conclusion

To summarize, converting a base 4 number to base 6 involves two main steps: first, convert the base 4 number to a base 10 number, and second, convert that base 10 number to a base 6 number. By following these steps, you can accurately convert between any number bases.

Practice Problem

Try converting the base 4 number 302_4 to base 6.

  1. Convert 302_4 to base 10:
    • $3 times 4^2 + 0 times 4^1 + 2 times 4^0 = 3 times 16 + 0 times 4 + 2 times 1 = 48 + 0 + 2 = 50$
  2. Convert 50_{10} to base 6:
    • $50 , div , 6 = 8$ with a remainder of $2$
    • $8 , div , 6 = 1$ with a remainder of $2$
    • $1 , div , 6 = 0$ with a remainder of $1$
    • Read remainders from bottom to top: 122_6

So, 302_4 is 122_6 in base 6.

Citations

  1. 1. Khan Academy – Number Systems
  2. 2. Math Is Fun – Number Bases
  3. 3. Purplemath – Converting Between Bases

Related

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H + HO2 → O2 + H2 k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O2 k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) H + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-5 s^-1) φ

Table 1 Reactions, rate constants and activation energies used in the model* No. Reaction kopt (M⁻¹ s⁻¹) 1 OH + H₂ → H + H₂O 3.74 x 10⁷ 2 OH + HO₂ → HO₂ + OH⁻ 5 x 10⁹ 3 OH + H₂O₂ → HO₂ + H₂O 3.8 x 10⁷ 4 OH + O₂ → O₂ + OH 9.96 x 10⁹ 5 OH + HO₂ → O₂ + H₂O 7.1 x 10⁹ 6 OH + OH → H₂O₂ 5.3 x 10⁹ 7 OH + e⁻aq → OH⁻ 3 x 10¹⁰ 8 H + O₂ → HO₂ 2.0 x 10¹⁰ 9 H + HO₂ → H₂O₂ 2.0 x 10¹⁰ 10 H + H₂O₂ → OH + H₂O 3.44 x 10⁷ 11 H + OH → H₂O 1.4 x 10¹⁰ 12 H + H → H₂ 1.94 x 10¹⁰ 13 e⁻aq + O₂ → O₂⁻ 1.9 x 10¹⁰ 14 e⁻aq + O₂ → HO₂⁻ + OH⁻ 1.3 x 10¹⁰ 15 e⁻aq + HO₂ 2.0 x 10¹⁰ 16 e⁻aq + H₂O₂ 1.1 x 10¹⁰ 17 e⁻aq + HO₂ → OH + OH⁻ 1.3 x 10¹⁰ 18 e⁻aq + H⁺ → H 2.3 x 10¹⁰ 19 e⁻aq + e⁻aq → H₂ + OH⁻ + OH⁻ 2.5 x 10⁹ 20 HO₂ + O₂ → O₂ + HO₂ 1.3 x 10⁹ 21 HO₂ + HO₂ → O₂ + H₂O₂ 8.3 x 10⁵ 22 HO₂ + HO₂ → O₂ + OH + H₂O 3.7 23 HO₂ + HO₂ → O₂ + O₂ + OH + H₂O 7 x 10⁵ s⁻¹ 24 H⁺ + O₂⁻ → HO₂ 4.5 x 10¹⁰ 25 H⁺ + O₂⁻ → O₂ 2.0 x 10¹⁰ 26 H⁺ + OH⁻ 1.4 x 10¹¹ 27 H⁺ + HO₂⁻ 2 x 10¹⁰ 28 H₂O₂ → HO₂ + H⁺ + OH⁻ 2.5 x 10⁻⁵ s⁻¹ 29 H₂O₂ → H⁺ + OH⁻ 1.4 x 10⁻⁷ s⁻¹ 30 O₂ + O₂ → O₂ + HO₂ + OH⁻ 0.3 31 O₂ + H₂O₂ → O₂ + OH + OH 16 32

(2) O3 + H → O2 + OH k2 = 1.78×10^-11 cm^3 s^-1 (3) O + OH → O2 + H k3 = 4.40×10^-11 cm^3 s^-1 (5) O + HO2 → O2 + OH k5 = 3.50×10^-11 cm^3 s^-1 (6) H2O + O → 2 OH k6 = 5.40×10^-12 cm^3 s^-1 (9) OH + HO2 → O2 + H2O k9 = 4.00×10^-11 cm^3 s^-1 (10) HO2 + HO2 → O2 + H2O2 k10 = 2.50×10^-12 cm s^-1 (11) O + O2 + M → O3 + M k11 = 1.05×10^-34 cm^6 s^-1 (14) H + O2 + M → HO2 + M k14 = 8.08×10^-32 cm^6 s^-1 (15) OH + H + M → H2O + M k15 = 3.31×10^-27 cm^6 s^-1 (16) O2 + hv → 2 O k16 = (1.26×10^-8 s^-1) φ (17) H2O + hv → H + OH k17 = (3.4×10^-6 s^-1) φ (18) O3 + hv → O2 + O k18 = (7.10×10^-8 s^-1) φ