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Showing 10 problems, seed 1.

Number Base worksheet

Grades 6–8

Solve each problem and show your work. Round answers to 2 decimal places where needed.

  1. Problem 1.
    • Number: 325
    • From base: Base 23
    • To base: Binary (base 2)
    • Two's complement view: 16-bit

    Find: In binary

  2. Problem 2.
    • Number: 220
    • From base: Base 27
    • To base: Base 15
    • Two's complement view: 64-bit

    Find: In base 15

  3. Problem 3.
    • Number: 240
    • From base: Base 7
    • To base: Base 4
    • Two's complement view: 16-bit

    Find: In base 4

  4. Problem 4.
    • Number: 250
    • From base: Base 25
    • To base: Base 21
    • Two's complement view: 32-bit

    Find: In base 21

  5. Problem 5.
    • Number: 200
    • From base: Base 36
    • To base: Base 7
    • Two's complement view: 32-bit

    Find: In base 7

  6. Problem 6.
    • Number: 250
    • From base: Base 32
    • To base: Base 7
    • Two's complement view: 64-bit

    Find: In base 7

  7. Problem 7.
    • Number: 380
    • From base: Base 35
    • To base: Base 34
    • Two's complement view: 32-bit

    Find: In base 34

  8. Problem 8.
    • Number: 350
    • From base: Octal (base 8)
    • To base: Base 32
    • Two's complement view: 8-bit

    Find: In base 32

  9. Problem 9.
    • Number: 240
    • From base: Base 30
    • To base: Base 31
    • Two's complement view: 8-bit

    Find: In base 31

  10. Problem 10.
    • Number: 230
    • From base: Base 11
    • To base: Base 25
    • Two's complement view: 32-bit

    Find: In base 25

Answer key: Number Base worksheet

  1. 1. In binary: 11001100110

    1. Multiply each digit by its place value, a power of 23, counting powers from 0 at the right

      3×232+2×231+5×230=1,587+46+5
    2. Add the place values to get the decimal value

      1,587+46+5=1,638
    3. Divide by 2: quotient 819, remainder 0

      1,638 ÷ 2 = 819 remainder 0
    4. Divide by 2: quotient 409, remainder 1

      819 ÷ 2 = 409 remainder 1
    5. Divide by 2: quotient 204, remainder 1

      409 ÷ 2 = 204 remainder 1
    6. Divide by 2: quotient 102, remainder 0

      204 ÷ 2 = 102 remainder 0
    7. Keep dividing the quotient by 2 until it reaches 0 (11 divisions in all; the last two are shown)

    8. Divide by 2: quotient 1, remainder 1

      3 ÷ 2 = 1 remainder 1
    9. Divide by 2: quotient 0, remainder 1

      1 ÷ 2 = 0 remainder 1
    10. Read the remainders from the last one to the first to get the digits

      11001100110
    11. Write the pattern in binary, padded with leading zeros to 16 bits

      0000 0110 0110 0110
  2. 2. In base 15: 6AC

    1. Multiply each digit by its place value, a power of 27, counting powers from 0 at the right

      2×272+2×271=1,458+54
    2. Add the place values to get the decimal value

      1,458+54=1,512
    3. Divide by 15: quotient 100, remainder 12 (digit C)

      1,512 ÷ 15 = 100 remainder 12
    4. Divide by 15: quotient 6, remainder 10 (digit A)

      100 ÷ 15 = 6 remainder 10
    5. Divide by 15: quotient 0, remainder 6

      6 ÷ 15 = 0 remainder 6
    6. Read the remainders from the last one to the first to get the digits

      6AC
    7. Write the pattern in binary, padded with leading zeros to 64 bits

      0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0101 1110 1000
  3. 3. In base 4: 1332

    1. Multiply each digit by its place value, a power of 7, counting powers from 0 at the right

      2×72+4×71=98+28
    2. Add the place values to get the decimal value

      98+28=126
    3. Divide by 4: quotient 31, remainder 2

      126 ÷ 4 = 31 remainder 2
    4. Divide by 4: quotient 7, remainder 3

      31 ÷ 4 = 7 remainder 3
    5. Divide by 4: quotient 1, remainder 3

      7 ÷ 4 = 1 remainder 3
    6. Divide by 4: quotient 0, remainder 1

      1 ÷ 4 = 0 remainder 1
    7. Read the remainders from the last one to the first to get the digits

      1332
    8. Write the pattern in binary, padded with leading zeros to 16 bits

      0000 0000 0111 1110
  4. 4. In base 21: 32A

    1. Multiply each digit by its place value, a power of 25, counting powers from 0 at the right

      2×252+5×251=1,250+125
    2. Add the place values to get the decimal value

      1,250+125=1,375
    3. Divide by 21: quotient 65, remainder 10 (digit A)

      1,375 ÷ 21 = 65 remainder 10
    4. Divide by 21: quotient 3, remainder 2

      65 ÷ 21 = 3 remainder 2
    5. Divide by 21: quotient 0, remainder 3

      3 ÷ 21 = 0 remainder 3
    6. Read the remainders from the last one to the first to get the digits

      32A
    7. Write the pattern in binary, padded with leading zeros to 32 bits

      0000 0000 0000 0000 0000 0101 0101 1111
  5. 5. In base 7: 10362

    1. Multiply each digit by its place value, a power of 36, counting powers from 0 at the right

      2×362=2,592
    2. Add the place values to get the decimal value

      2,592=2,592
    3. Divide by 7: quotient 370, remainder 2

      2,592 ÷ 7 = 370 remainder 2
    4. Divide by 7: quotient 52, remainder 6

      370 ÷ 7 = 52 remainder 6
    5. Divide by 7: quotient 7, remainder 3

      52 ÷ 7 = 7 remainder 3
    6. Divide by 7: quotient 1, remainder 0

      7 ÷ 7 = 1 remainder 0
    7. Divide by 7: quotient 0, remainder 1

      1 ÷ 7 = 0 remainder 1
    8. Read the remainders from the last one to the first to get the digits

      10362
    9. Write the pattern in binary, padded with leading zeros to 32 bits

      0000 0000 0000 0000 0000 1010 0010 0000
  6. 6. In base 7: 6303

    1. Multiply each digit by its place value, a power of 32, counting powers from 0 at the right

      2×322+5×321=2,048+160
    2. Add the place values to get the decimal value

      2,048+160=2,208
    3. Divide by 7: quotient 315, remainder 3

      2,208 ÷ 7 = 315 remainder 3
    4. Divide by 7: quotient 45, remainder 0

      315 ÷ 7 = 45 remainder 0
    5. Divide by 7: quotient 6, remainder 3

      45 ÷ 7 = 6 remainder 3
    6. Divide by 7: quotient 0, remainder 6

      6 ÷ 7 = 0 remainder 6
    7. Read the remainders from the last one to the first to get the digits

      6303
    8. Write the pattern in binary, padded with leading zeros to 64 bits

      0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 0000 1000 1010 0000
  7. 7. In base 34: 3EB

    1. Multiply each digit by its place value, a power of 35, counting powers from 0 at the right

      3×352+8×351=3,675+280
    2. Add the place values to get the decimal value

      3,675+280=3,955
    3. Divide by 34: quotient 116, remainder 11 (digit B)

      3,955 ÷ 34 = 116 remainder 11
    4. Divide by 34: quotient 3, remainder 14 (digit E)

      116 ÷ 34 = 3 remainder 14
    5. Divide by 34: quotient 0, remainder 3

      3 ÷ 34 = 0 remainder 3
    6. Read the remainders from the last one to the first to get the digits

      3EB
    7. Write the pattern in binary, padded with leading zeros to 32 bits

      0000 0000 0000 0000 0000 1111 0111 0011
  8. 8. In base 32: 78

    1. Multiply each digit by its place value, a power of 8, counting powers from 0 at the right

      3×82+5×81=192+40
    2. Add the place values to get the decimal value

      192+40=232
    3. Divide by 32: quotient 7, remainder 8

      232 ÷ 32 = 7 remainder 8
    4. Divide by 32: quotient 0, remainder 7

      7 ÷ 32 = 0 remainder 7
    5. Read the remainders from the last one to the first to get the digits

      78
    6. Write the pattern in binary, padded with leading zeros to 8 bits

      1110 1000
  9. 9. In base 31: 1UT

    1. Multiply each digit by its place value, a power of 30, counting powers from 0 at the right

      2×302+4×301=1,800+120
    2. Add the place values to get the decimal value

      1,800+120=1,920
    3. Divide by 31: quotient 61, remainder 29 (digit T)

      1,920 ÷ 31 = 61 remainder 29
    4. Divide by 31: quotient 1, remainder 30 (digit U)

      61 ÷ 31 = 1 remainder 30
    5. Divide by 31: quotient 0, remainder 1

      1 ÷ 31 = 0 remainder 1
    6. Read the remainders from the last one to the first to get the digits

      1UT
  10. 10. In base 25: B0

    1. Multiply each digit by its place value, a power of 11, counting powers from 0 at the right

      2×112+3×111=242+33
    2. Add the place values to get the decimal value

      242+33=275
    3. Divide by 25: quotient 11, remainder 0

      275 ÷ 25 = 11 remainder 0
    4. Divide by 25: quotient 0, remainder 11 (digit B)

      11 ÷ 25 = 0 remainder 11
    5. Read the remainders from the last one to the first to get the digits

      B0
    6. Write the pattern in binary, padded with leading zeros to 32 bits

      0000 0000 0000 0000 0000 0001 0001 0011