OBJ (Objective Questions)

1. If \(23_x = 32_5\), find the value of x. A. 7. B. 6. C. 5. D. 4. [2014/2]

• Topic: Converting between different number bases and solving for an unknown base.
• Correct Option: A. 7
• Detailed Explanation: Convert both numbers to base 10:

  \(23_x = 2 \times x^1 + 3 \times x^0 = 2x + 3\)

  \(32_5 = 3 \times 5^1 + 2 \times 5^0 = 15 + 2 = 17\)

  Equate the base 10 expressions:

  \[2x + 3 = 17\] \[2x = 17 - 3\] \[2x = 14\] \[x = 7\]

  For \(x=7\), the number \(23_7\) is valid as \(7 > 3\).

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2. The sum of \(11011_2\), \(11111_2\), and \(10000_2\) is \(10m10n0_2\). Find the values of m and n. A. m=0,n=0. B. m=1,n=0. C. m=0,n=1. D. m=1,n=1. [2015/2]

• Topic: Binary addition and identifying digits in a binary number.
• Correct Option: C. m=0, n=1
• Detailed Explanation: Perform binary addition:

        11011₂
        11111₂
      + 10000₂
      ---------
       101010₂
      

  Comparing the sum \(101010_2\) with the given format \(10m10n0_2\):

  • The second digit from the left is \(0\), so \(m = 0\).
  • The fifth digit from the left (second from the right) is \(1\), so \(n = 1\).

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3. The subtraction below is in base seven. Find the missing number.

  5162₇
- 2644₇
-------
  2x15₇

A. 2. B. 3. C. 4. D. 5. [2010/43]

• Topic: Subtraction in base seven.
• Correct Option: A. 2
• Detailed Explanation: Perform subtraction in base seven, column by column from right to left, borrowing 7 when necessary.

        5162₇
      - 2644₇
      -------
      

  • Units column (\(7^0\)): \(2 - 4\). Borrow from 6 (\(6 \to 5\)). \((7+2) - 4 = 9 - 4 = 5\). (Matches)
  • \(7^1\) column: \(5 - 4 = 1\). (Matches)
  • \(7^2\) column: \(1 - 6\). Borrow from 5 (\(5 \to 4\)). \((7+1) - 6 = 8 - 6 = 2\). This is \(x\).
  • \(7^3\) column: \(4 - 2 = 2\). (Matches)

  So, the missing digit \(x\) is 2.

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4. Convert \(42_5\) to a base three numeral. A. 201₃. B. 210₃. C. 211₃.D. 222₃. [2008/6]

• Topic: Converting from one non-decimal base to another non-decimal base.
• Correct Option: C. 211₃
• Detailed Explanation: First, convert \(42_5\) to base 10: \[42_5 = 4 \times 5^1 + 2 \times 5^0 = 20 + 2 = 22_{10}\]

  Next, convert \(22_{10}\) to base 3 using repeated division:

  \[22 \div 3 = 7 \text{ remainder } 1\] \[7 \div 3 = 2 \text{ remainder } 1\] \[2 \div 3 = 0 \text{ remainder } 2\]

  Reading the remainders from bottom to top gives \(211_3\).

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5. If \(23_x + 101_x = 130_x\), find the value of x. A. 7. B. 6. C. 5. D. 4. [2016/1]

• Topic: Addition in an unknown base and solving for the base.
• Correct Option: D. 4
• Detailed Explanation: Convert all numbers to base 10:

  \(23_x = 2x + 3\)

  \(101_x = 1 \times x^2 + 0 \times x^1 + 1 \times x^0 = x^2 + 1\)

  \(130_x = 1 \times x^2 + 3 \times x^1 + 0 \times x^0 = x^2 + 3x\)

  Substitute into the equation:

  \[(2x + 3) + (x^2 + 1) = x^2 + 3x\] \[x^2 + 2x + 4 = x^2 + 3x\]

  Subtract \(x^2\) from both sides:

  \[2x + 4 = 3x\]

  Subtract \(2x\) from both sides:

  \[4 = x\]

  The base \(x\) must be greater than the largest digit used in the numbers (which is 3). Since \(x=4\) is greater than 3, it is a valid base.

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6. Given that \(124_x = 7(14_x)\), find the value of x. A. 12. B. 11.C. 9. D. 8. [2011/38]

• Topic: Equations involving unknown bases and solving for the base.
• Correct Option: D. 8
• Detailed Explanation: Convert all numbers to base 10:

  \(124_x = 1 \times x^2 + 2 \times x^1 + 4 \times x^0 = x^2 + 2x + 4\)

  \(14_x = 1 \times x^1 + 4 \times x^0 = x + 4\)

  \(7\) is already in base 10.

  Substitute into the equation:

  \[x^2 + 2x + 4 = 7(x + 4)\] \[x^2 + 2x + 4 = 7x + 28\] \[x^2 + 2x - 7x + 4 - 28 = 0\] \[x^2 - 5x - 24 = 0\]

  We need two numbers that multiply to -24 and add to -5. These numbers are -8 and 3.

  \[(x - 8)(x + 3) = 0\]

  This gives two possible values for \(x\):

  \[x = 8 \quad \text{or} \quad x = -3\]

  Since a number base cannot be negative, \(x = 8\). The base \(x\) must also be greater than the largest digit (4) in \(124_x\) and \(14_x\). Since \(8 > 4\), the answer is valid.

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7. Evaluate \((111_2)^2\) and leave your answer in base 2. A. \(111001_2\). B. \(110001_2\). C. \(101001_2\). D. \(10010_2\). [2007/1]

• Topic: Binary multiplication.
• Correct Option: B. \(110001_2\)
• Detailed Explanation: First, convert \(111_2\) to base 10: \[111_2 = 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 4 + 2 + 1 = 7_{10}\]

  Calculate the square in base 10:

  \[7^2 = 49_{10}\]

  Now convert \(49_{10}\) to base 2 using repeated division:

  \[49 \div 2 = 24 \text{ remainder } 1\] \[24 \div 2 = 12 \text{ remainder } 0\] \[12 \div 2 = 6 \text{ remainder } 0\] \[6 \div 2 = 3 \text{ remainder } 0\] \[3 \div 2 = 1 \text{ remainder } 1\] \[1 \div 2 = 0 \text{ remainder } 1\]

  Reading the remainders from bottom to top gives \(110001_2\).

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8. Convert \(35_{10}\) to a number in base 2. A. 1011. B. 10011. C. 100011.D. 11001. [2012/4]

• Topic: Converting from base 10 to base 2.
• Correct Option: C. 100011₂
• Detailed Explanation: Use repeated division by 2: \[35 \div 2 = 17 \text{ remainder } 1\] \[17 \div 2 = 8 \text{ remainder } 1\] \[8 \div 2 = 4 \text{ remainder } 0\] \[4 \div 2 = 2 \text{ remainder } 0\] \[2 \div 2 = 1 \text{ remainder } 0\] \[1 \div 2 = 0 \text{ remainder } 1\]

  Reading the remainders from bottom to top gives \(100011_2\).

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9. Arrange the following numbers in descending order of magnitude: \(22_3\), \(34_5\), \(21_6\). A. \(21_6\) \(22_3\) \(34_5\). B. \(22_6\) \(34_5\) \(22_3\). C. \(22_3\) \(34_5\) \(21_6\). D. \(34_5\) \(21_6\) \(22_3\). [2009/4]

• Topic: Comparing numbers in different bases.
• Correct Option: D. \(34_5\) \(21_6\) \(22_3\)
• Detailed Explanation: Convert all numbers to base 10:

  \(22_3 = 2 \times 3^1 + 2 \times 3^0 = 6 + 2 = 8_{10}\)

  \(34_5 = 3 \times 5^1 + 4 \times 5^0 = 15 + 4 = 19_{10}\)

  \(21_6 = 2 \times 6^1 + 1 \times 6^0 = 12 + 1 = 13_{10}\)

  In base 10, the numbers are: 8, 19, 13.

  Arranging these in descending order (largest to smallest): 19, 13, 8.

  Converting back to their original base forms:

  \(19_{10} = 34_5\)

  \(13_{10} = 21_6\)

  \(8_{10} = 22_3\)

  So, the descending order is \(34_5, 21_6, 22_3\).

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10. In what number base is the addition \(465 + 24 + 225 = 1050\)? A. Ten. B. Nine. C. Eight. D. Seven. [2013/3]

• Topic: Finding the base of an addition equation.
• Correct Option: D. Seven
• Detailed Explanation: Let the unknown base be \(x\). Convert all numbers to base 10:

  \(465_x = 4x^2 + 6x + 5\)

  \(24_x = 2x + 4\)

  \(225_x = 2x^2 + 2x + 5\)

  \(1050_x = 1x^3 + 0x^2 + 5x + 0 = x^3 + 5x\)

  The equation is:

  \[(4x^2 + 6x + 5) + (2x + 4) + (2x^2 + 2x + 5) = x^3 + 5x\]

  Combine like terms on the left side:

  \[6x^2 + 10x + 14 = x^3 + 5x\]

  Rearrange into a cubic equation:

  \[x^3 - 6x^2 - 5x - 14 = 0\]

  The base \(x\) must be greater than the largest digit used in the numbers (which is 6). So, \(x\) must be at least 7. Test \(x = 7\):

  \[7^3 - 6(7^2) - 5(7) - 14\] \[343 - 6(49) - 35 - 14\] \[343 - 294 - 35 - 14\] \[49 - 35 - 14 = 14 - 14 = 0\]

  Since the equation evaluates to 0 when \(x = 7\), the base is 7.

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11. In what number base was the addition \(1 + nn = 100\), where \(n > 0\), done? A. n - 1. B. n. C. n + 1. D. n + 2. [2017/33]

• Topic: Finding the base of an addition equation with variables as digits.
• Correct Option: C. n + 1
• Detailed Explanation: Let the base be \(b\). Convert the numbers to base 10:

  \(1_{10} = 1\)

  \(nn_b = n \times b^1 + n \times b^0 = nb + n\)

  \(100_b = 1 \times b^2 + 0 \times b^1 + 0 \times b^0 = b^2\)

  Substitute into the equation \(1 + nn = 100\):

  \[1 + (nb + n) = b^2\]

  Rearrange to solve for \(b\), or test the options. The base \(b\) must be greater than the digit \(n\).

  Let's test option C, \(b = n + 1\):

  \[1 + n(n+1) + n = (n+1)^2\] \[1 + n^2 + n + n = n^2 + 2n + 1\] \[n^2 + 2n + 1 = n^2 + 2n + 1\]

  This identity holds true, so the base is \(n+1\).

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12. Find the value of x for which \(32_x = 22_x\). A. three. B. five. C. six. D. seven. [2018/4]

• Topic: Solving equations involving numbers in an unknown base.
• Correct Option: No valid option based on the equation.
• Detailed Explanation: Convert both numbers to base 10:

  \(32_x = 3x + 2\)

  \(22_x = 2x + 2\)

  Equate them:

  \[3x + 2 = 2x + 2\] \[3x = 2x\] \[x = 0\]

  A number base cannot be 0. Additionally, the base \(x\) must be greater than the largest digit used (which is 3 in \(32_x\)). Thus, there is no valid solution for \(x\). The question is ill-posed.

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13. Convert \(101101_2\) to a number in base ten. A. 61. B. 46.C. 45. D. 44. [2005/11]

• Topic: Converting from base 2 to base 10.
• Correct Option: C. 45
• Detailed Explanation: \[101101_2 = 1 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0\] \[= 1 \times 32 + 0 \times 16 + 1 \times 8 + 1 \times 4 + 0 \times 2 + 1 \times 1\] \[= 32 + 0 + 8 + 4 + 0 + 1 = 45_{10}\]

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14. Two numbers \(24_x\) and \(31_y\) are equal in value when converted to base ten. Find the equation connecting x and y. A. \(2x = 3 (y - 1)\). B. \(4x - y = 1\). C. \(3y + 2x = 3\). D. \(3y = 2 (x + 3)\). [1999/35]

• Topic: Forming an algebraic equation from number base conversions.
• Correct Option: A. \(2x = 3 (y - 1)\)
• Detailed Explanation: Convert both numbers to base 10:

  \(24_x = 2 \times x^1 + 4 \times x^0 = 2x + 4\)

  \(31_y = 3 \times y^1 + 1 \times y^0 = 3y + 1\)

  Since they are equal in base 10:

  \[2x + 4 = 3y + 1\]

  Rearrange the equation to match the options:

  \[2x + 3 = 3y\]

  From option A: \(2x = 3(y - 1) \Rightarrow 2x = 3y - 3 \Rightarrow 3y = 2x + 3\).

  This matches our derived equation.

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15. Find \((101_2)^2\), expressing the answer in base 2. A. \(10101_2\). B. \(11001_2\). C. \(10010_2\). D. \(11101_2\). [1995/1]

• Topic: Binary multiplication.
• Correct Option: B. \(11001_2\)
• Detailed Explanation: First, convert \(101_2\) to base 10: \[101_2 = 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 4 + 0 + 1 = 5_{10}\]

  Calculate the square in base 10:

  \[5^2 = 25_{10}\]

  Now convert \(25_{10}\) to base 2 using repeated division:

  \[25 \div 2 = 12 \text{ remainder } 1\] \[12 \div 2 = 6 \text{ remainder } 0\] \[6 \div 2 = 3 \text{ remainder } 0\] \[3 \div 2 = 1 \text{ remainder } 1\] \[1 \div 2 = 0 \text{ remainder } 1\]

  Reading the remainders from bottom to top gives \(11001_2\).

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16. Simplify: \(11011_2 – 1101_2\) A. \(101000_2\). B. \(1100_2\). C. \(1110_2\). D. \(1011_2\). [2006/2]

• Topic: Binary subtraction.
• Correct Option: C. \(1110_2\)
• Detailed Explanation: Convert to base 10, perform subtraction, then convert back to base 2.

  \(11011_2 = 1 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 16 + 8 + 0 + 2 + 1 = 27_{10}\)

  \(1101_2 = 1 \times 2^3 + 1 \times 2^2 + 0 \times 2^1 + 1 \times 2^0 = 8 + 4 + 0 + 1 = 13_{10}\)

  Perform the subtraction in base 10:

  \[27_{10} - 13_{10} = 14_{10}\]

  Now convert \(14_{10}\) to base 2:

  \[14 \div 2 = 7 \text{ remainder } 0\] \[7 \div 2 = 3 \text{ remainder } 1\] \[3 \div 2 = 1 \text{ remainder } 1\] \[1 \div 2 = 0 \text{ remainder } 1\]

  Reading the remainders from bottom to top gives \(1110_2\).

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17. Find the missing number in the addition of the following numbers, in base seven.

  4321₇
  1234₇
+ ____
-------
  12341₇

A. 3453. B. 5556. C. 6016. D. 13453. [2000/4]

• Topic: Addition in base seven and finding a missing addend.
• Correct Option: A. \(3453_7\)
• Detailed Explanation: Let the missing number be \(X\). The equation is \((4321_7 + 1234_7) + X = 12341_7\).

  First, add \(4321_7\) and \(1234_7\):

        4321₇
      + 1234₇
      -------
        5555₇
      

  Now, subtract \(5555_7\) from \(12341_7\) to find \(X\):

        12341₇
      -  5555₇
      --------
         3453₇
      

  Step-by-step subtraction:

  • \(7^0\): \(1 - 5\). Borrow 1 from 4 (\(4 \to 3\)). \((1+7) - 5 = 8 - 5 = 3\).
  • \(7^1\): \(3 - 5\). Borrow 1 from 3 (\(3 \to 2\)). \((3+7) - 5 = 10 - 5 = 5\).
  • \(7^2\): \(2 - 5\). Borrow 1 from 2 (\(2 \to 1\)). \((2+7) - 5 = 9 - 5 = 4\).
  • \(7^3\): \(1 - 5\). Borrow 1 from 1 (\(1 \to 0\), effectively \(7^4\) becomes \(0 \times 7^4\)). \((1+7) - 5 = 8 - 5 = 3\).

  So, \(X = 3453_7\).

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18. If \(104_x = 68\), find the value of x. A. 5. B. 7. C. 8. D. 9. [2000/8]

• Topic: Converting from an unknown base to base 10 and solving for the base.
• Correct Option: C. 8
• Detailed Explanation: Convert \(104_x\) to base 10: \[1 \times x^2 + 0 \times x^1 + 4 \times x^0 = x^2 + 4\]

  Equate this to 68:

  \[x^2 + 4 = 68\] \[x^2 = 68 - 4\] \[x^2 = 64\] \[x = \sqrt{64}\] \[x = 8\]

  Since a base must be positive, \(x=8\). The base \(x\) must be greater than the largest digit in \(104_x\) (which is 4). Since \(8 > 4\), the answer is valid.

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19. Arrange in ascending order of magnitude: \(26_8\), \(36_7\) and \(25_9\). A. \(25_9\) \(26_8\) \(36_7\). B. \(26_8\) \(25_9\) \(36_7\). C. \(36_7\) \(26_8\) \(25_9\). D. \(36_7\) \(25_9\) \(26_8\). [1999/7]

• Topic: Comparing numbers in different bases.
• Correct Option: B. \(26_8\) \(25_9\) \(36_7\)
• Detailed Explanation: Convert all numbers to base 10:

  \(26_8 = 2 \times 8^1 + 6 \times 8^0 = 16 + 6 = 22_{10}\)

  \(36_7 = 3 \times 7^1 + 6 \times 7^0 = 21 + 6 = 27_{10}\)

  \(25_9 = 2 \times 9^1 + 5 \times 9^0 = 18 + 5 = 23_{10}\)

  In base 10, the numbers are: 22, 27, 23.

  Arrange these in ascending order (smallest to largest): 22, 23, 27.

  Convert them back to their original base forms:

  \(22_{10} = 26_8\)

  \(23_{10} = 25_9\)

  \(27_{10} = 36_7\)

  So, the ascending order is \(26_8, 25_9, 36_7\).

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20. Evaluate \((20_3)^2 - (11_3)^2\) in base three. A. \(101_3\). B. \(121_3\).C. \(202_3\). D. \(2020_3\). [2001/2]

• Topic: Operations (squaring and subtraction) in a given base.
• Correct Option: C. \(202_3\)
• Detailed Explanation: First, convert the numbers to base 10, perform the operations, then convert back to base 3.

  \(20_3 = 2 \times 3^1 + 0 \times 3^0 = 6 + 0 = 6_{10}\)

  \(11_3 = 1 \times 3^1 + 1 \times 3^0 = 3 + 1 = 4_{10}\)

  Now, evaluate the expression in base 10:

  \[(6)^2 - (4)^2 = 36 - 16 = 20_{10}\]

  Next, convert \(20_{10}\) to base 3:

  \[20 \div 3 = 6 \text{ remainder } 2\] \[6 \div 3 = 2 \text{ remainder } 0\] \[2 \div 3 = 0 \text{ remainder } 2\]

  Reading the remainders from bottom to top gives \(202_3\).

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21. Convert 77 to a number in base two. A. 1001101. B. 111001.C. 100110. D. 10101. [1992/4]

• Topic: Converting from base 10 to base 2.
• Correct Option: A. \(1001101_2\)
• Detailed Explanation: Use repeated division by 2: \[77 \div 2 = 38 \text{ remainder } 1\] \[38 \div 2 = 19 \text{ remainder } 0\] \[19 \div 2 = 9 \text{ remainder } 1\] \[9 \div 2 = 4 \text{ remainder } 1\] \[4 \div 2 = 2 \text{ remainder } 0\] \[2 \div 2 = 1 \text{ remainder } 0\] \[1 \div 2 = 0 \text{ remainder } 1\]

  Reading the remainders from bottom to top gives \(1001101_2\).

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22. If \(M5_{10} = 1001011_2\), find the value of M. A. 5. B. 6. C. 7. D. 8. [2002/7]

• Topic: Converting between base 2 and base 10, and interpreting notation.
• Correct Option: C. 7
• Detailed Explanation: Assume \(M5_{10}\) represents a two-digit number \(10M + 5\).

  First, convert \(1001011_2\) to base 10:

  \[1 \times 2^6 + 0 \times 2^5 + 0 \times 2^4 + 1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0\] \[= 64 + 0 + 0 + 8 + 0 + 2 + 1 = 75_{10}\]

  Now, equate \(10M + 5\) to \(75\):

  \[10M + 5 = 75\] \[10M = 70\] \[M = 7\]

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23. Given, that \(4P4_5 = 119_{10}\) find the value of P. A. 1. B. 2.C. 3. D. 4. [2003/19]

• Topic: Converting from an unknown digit in a given base to base 10, and solving for the digit.
• Correct Option: C. 3
• Detailed Explanation: Convert \(4P4_5\) to base 10: \[4 \times 5^2 + P \times 5^1 + 4 \times 5^0\] \[4 \times 25 + 5P + 4 \times 1\] \[100 + 5P + 4 = 104 + 5P\]

  Equate this to \(119_{10}\):

  \[104 + 5P = 119\] \[5P = 119 - 104\] \[5P = 15\] \[P = \frac{15}{5}\] \[P = 3\]

  The digit \(P\) must be less than the base 5. Since \(3 < 5\), the answer is valid.

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24. Evaluate \((111_2)^2 - (101_2)^2\) A. \(10_2\). B. \(100_2\). C. \(1100_2\). D. \(11000_2\). [2003/27]

• Topic: Operations (squaring and subtraction) with binary numbers.
• Correct Option: D. \(11000_2\)
• Detailed Explanation: Convert to base 10 first, perform calculations, then convert back to base 2.

  \(111_2 = 1 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 = 4 + 2