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Write the unit digit in the cube of the following numbers: 19, 201, 164, 1307, 510, 35, 716.

Mathematics

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Here are the solutions to the problems.

1. Write the unit digit in the cube of the following numbers:
To find the unit digit of a cube, we only need to cube the unit digit of the original number and find its unit digit.

  • a) 19
  1. The unit digit of 19 is 9.

  2. Cube the unit digit

    93=7299^3 = 729.

  3. The unit digit of 729 is 9.

    The unit digit is $\boxed{9}$.
    
    • b) 201
  4. The unit digit of 201 is 1.

  5. Cube the unit digit

    13=11^3 = 1.

  6. The unit digit of 1 is 1.

    The unit digit is $\boxed{1}$.
    
    • c) 164
  7. The unit digit of 164 is 4.

  8. Cube the unit digit

    43=644^3 = 64.

  9. The unit digit of 64 is 4.

    The unit digit is $\boxed{4}$.
    
    • d) 1307
  10. The unit digit of 1307 is 7.

  11. Cube the unit digit

    73=3437^3 = 343.

  12. The unit digit of 343 is 3.

    The unit digit is $\boxed{3}$.
    
    • e) 510
  13. The unit digit of 510 is 0.

  14. Cube the unit digit

    03=00^3 = 0.

  15. The unit digit of 0 is 0.

    The unit digit is $\boxed{0}$.
    
    • f) 35
  16. The unit digit of 35 is 5.

  17. Cube the unit digit

    53=1255^3 = 125.

  18. The unit digit of 125 is 5.

    The unit digit is $\boxed{5}$.
    
    • g) 716
  19. The unit digit of 716 is 6.

  20. Cube the unit digit

    63=2166^3 = 216.

  21. The unit digit of 216 is 6.

    The unit digit is $\boxed{6}$.
    
    • h) 1718
  22. The unit digit of 1718 is 8.

  23. Cube the unit digit

    83=5128^3 = 512.

  24. The unit digit of 512 is 2.

    The unit digit is $\boxed{2}$.
    

    2. State which of the following numbers are perfect cubes:
    A number is a perfect cube if it can be expressed as n3n^3 for some integer nn.

    • a) 1000
  25. Find the cube root of 1000.

    $$
    \sqrt[3]{1000} = 10
    $$
    
  26. Since 1000 is the cube of an integer (10), it is a perfect cube.

    $\boxed{\text{Perfect cube}}$
    
    • b) 53×735^3 \times 7^3
  27. Use the property (a×b)n=an×bn(a \times b)^n = a^n \times b^n.

    $$
    5^3 \times 7^3 = (5 \times 7)^3 = 35^3
    $$
    
  28. Since the number can be expressed as the cube of an integer (35), it is a perfect cube.

    $\boxed{\text{Perfect cube}}$
    
    • c) 100000
  29. Count the number of zeros.

    There are 5 zeros.

  30. For a number ending in zeros to be a perfect cube, the number of zeros must be a multiple of 3.

    Since 5 is not a multiple of 3, 100000 is not a perfect cube.
    Not a perfect cube\boxed{\text{Not a perfect cube}}

    • d) (100)2×10(100)^2 \times 10
  31. Simplify the expression.

    $$
    (100)^2 \times 10 = 10000 \times 10 = 100000
    $$
    
  32. As determined in part (c), 100000 is not a perfect cube.

    $\boxed{\text{Not a perfect cube}}$
    
    • e) (1000)2(1000)^2
  33. Simplify the expression.

    $$
    (1000)^2 = 1000 \times 1000 = 1000000
    $$
    
  34. Find the cube root of 1000000.

    $$
    \sqrt[3]{1000000} = 100
    $$
    
  35. Since 1000000 is the cube of an integer (100), it is a perfect cube.

    $\boxed{\text{Perfect cube}}$
    

    3. Find the cube of the following numbers:

    • a) 21
  36. Calculate 21321^3.

    $$
    21^3 = 21 \times 21 \times 21 = 441 \times 21 = 9261
    $$
    The cube of 21 is $\boxed{9261}$.
    
    • b) 100
  37. Calculate 1003100^3.

    $$
    100^3 = 100 \times 100 \times 100 = 1000000
    $$
    The cube of 100 is $\boxed{1000000}$.
    
    • c) 17
  38. Calculate 17317^3.

    $$
    17^3 = 17 \times 17 \times 17 = 289 \times 17 = 4913
    $$
    The cube of 17 is $\boxed{4913}$.
    
    • d) 11311^3
  39. Calculate 11311^3.

    $$
    11^3 = 11 \times 11 \times 11 = 121 \times 11 = 1331
    $$
    The cube of 11 is $\boxed{1331}$.
    

    4. Find the sum of consecutive odd numbers to obtain the following cubes:
    The sum of nn consecutive odd numbers starting from n2−n+1n^2 - n + 1 equals n3n^3.

    • a) 434^3
  40. Identify nn.

    Here n=4n=4.

  41. Find the first odd number in the sequence

    n2−n+1=42−4+1=16−4+1=13n^2 - n + 1 = 4^2 - 4 + 1 = 16 - 4 + 1 = 13.

  42. Write the sum of 4 consecutive odd numbers starting from 13.

    $$
    4^3 = 13 + 15 + 17 + 19 = 64
    $$
    The sum is $\boxed{13 + 15 + 17 + 19}$.
    
    • b) 737^3
  43. Identify nn.

    Here n=7n=7.

  44. Find the first odd number in the sequence

    n2−n+1=72−7+1=49−7+1=43n^2 - n + 1 = 7^2 - 7 + 1 = 49 - 7 + 1 = 43.

  45. Write the sum of 7 consecutive odd numbers starting from 43.

    $$
    7^3 = 43 + 45 + 47 + 49 + 51 + 53 + 55 = 343
    $$
    The sum is $\boxed{43 + 45 + 47 + 49 + 51 + 53 + 55}$.
    
    • c) 838^3
  46. Identify nn.

    Here n=8n=8.

  47. Find the first odd number in the sequence

    n2−n+1=82−8+1=64−8+1=57n^2 - n + 1 = 8^2 - 8 + 1 = 64 - 8 + 1 = 57.

  48. Write the sum of 8 consecutive odd numbers starting from 57.

    $$
    8^3 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512
    $$
    The sum is $\boxed{57 + 59 + 61 + 63 + 65 + 67 + 69 + 71}$.
    
    • d) 11311^3
  49. Identify nn.

    Here n=11n=11.

  50. Find the first odd number in the sequence

    n2−n+1=112−11+1=121−11+1=111n^2 - n + 1 = 11^2 - 11 + 1 = 121 - 11 + 1 = 111.

  51. Write the sum of 11 consecutive odd numbers starting from 111.

    $$
    11^3 = 111 + 113 + 115 + 117 + 119 + 121 + 123 + 125 + 127 + 129 + 131 = 1331
    $$
    The sum is $\boxed{111 + 113 + 115 + 117 + 119 + 121 + 123 + 125 + 127 + 129 + 131}$.
    
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