Examkin

Number Bases and Conversions

21 soru

Tüm türler21Çoktan seçmeli10Sayısal11
Soru 1Soru →

If (125)x+(32)x=(201)x(125)_x + (32)_x = (201)_x, where xx represents the base of the numerals, what is the value of xx?

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Cevap: 6

Cevap

The base xx is equal to 66.
Expanding the base-xx numbers into base 10 polynomials yields (x2+2x+5)+(3x+2)=2x2+1(x^2 + 2x + 5) + (3x + 2) = 2x^2 + 1. Simplifying this equation gives x2−5x−6=0x^2 - 5x - 6 = 0, which factors as (x−6)(x+1)=0(x - 6)(x + 1) = 0. Since a number base must be positive and greater than the highest digit in the expression (which is 5), the only valid solution is 6.

Adım Adım Çözüm

1
Expand all numbers in base xx into positional power notation (base 10 equivalent)
(125)x=1⋅x2+2⋅x1+5⋅x0=x2+2x+5(125)_x = 1 \cdot x^2 + 2 \cdot x^1 + 5 \cdot x^0 = x^2 + 2x + 5, (32)x=3⋅x1+2⋅x0=3x+2(32)_x = 3 \cdot x^1 + 2 \cdot x^0 = 3x + 2, and (201)x=2⋅x2+0⋅x1+1⋅x0=2x2+1(201)_x = 2 \cdot x^2 + 0 \cdot x^1 + 1 \cdot x^0 = 2x^2 + 1
Converting all terms to base 10 allows setup of a algebraic equation in xx.
2
Substitute the expanded terms back into the original equation and simplify
(x2+2x+5)+(3x+2)=2x2+1  ⟹  x2+5x+7=2x2+1(x^2 + 2x + 5) + (3x + 2) = 2x^2 + 1 \implies x^2 + 5x + 7 = 2x^2 + 1
Combining like terms on the left-hand side prepares the expression for quadratic rearrangement.
3
Rearrange into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0
2x2−x2−5x+1−7=0  ⟹  x2−5x−6=02x^2 - x^2 - 5x + 1 - 7 = 0 \implies x^2 - 5x - 6 = 0
Subtracting (x2+5x+7)(x^2 + 5x + 7) from both sides sets the quadratic equation to zero.
4
Solve the quadratic equation by factoring and evaluate valid base conditions
(x−6)(x+1)=0  ⟹  x=6(x - 6)(x + 1) = 0 \implies x = 6 or x=−1x = -1. Valid base: x=6x = 6
A number base must be a positive integer greater than any single digit present in the given numbers (digits up to 5 appear, so x>5x > 5).

Anahtar Kavram

Conversion of numbers in arbitrary base xx to base 10 via place-value expansion to solve polynomial equations.
Soru 2Soru →

Given that 123x=3135123_x = 313_5, what is the value of the base xx?

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Cevap: 8

Cevap

8
Converting 3135313_5 to base 10 yields 3(25)+1(5)+3(1)=833(25) + 1(5) + 3(1) = 83. Expanding 123x123_x yields x2+2x+3x^2 + 2x + 3. Setting them equal produces x2+2x+3=83x^2 + 2x + 3 = 83, which simplifies to x2+2x−80=0x^2 + 2x - 80 = 0. Solving (x−8)(x+10)=0(x-8)(x+10) = 0 gives x=8x = 8 since a base must be positive.

Adım Adım Çözüm

1
Convert the right-hand side 3135313_5 to base 10.
3(52)+1(51)+3(50)=75+5+3=83103(5^2) + 1(5^1) + 3(5^0) = 75 + 5 + 3 = 83_{10}
Converting all terms to base 10 allows forming a standard algebraic equation.
2
Expand the left-hand side 123x123_x in powers of xx.
1⋅x2+2⋅x1+3⋅x0=x2+2x+31 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3
Expressing the number in terms of its base xx positional values.
3
Set the two base 10 expressions equal and solve the quadratic equation.
x2+2x+3=83  ⟹  x2+2x−80=0  ⟹  (x+10)(x−8)=0  ⟹  x=8x^2 + 2x + 3 = 83 \implies x^2 + 2x - 80 = 0 \implies (x + 10)(x - 8) = 0 \implies x = 8
The base xx must be a positive integer greater than any digit in 123x123_x (so x>3x > 3), which leaves x=8x = 8.

Anahtar Kavram

Solving equations with unknown number bases by expanding into base 10 polynomials.
Soru 3Soru →

Given that 245x−157x=66x245_x - 157_x = 66_x, find the value of the base xx.

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Cevap: 8

Cevap

The value of the base xx is 8.
Expanding all terms in powers of xx gives 2x2+4x+5−(x2+5x+7)=6x+62x^2 + 4x + 5 - (x^2 + 5x + 7) = 6x + 6. Simplifying gives x2−7x−8=0x^2 - 7x - 8 = 0, which factors into (x−8)(x+1)=0(x - 8)(x + 1) = 0. Since a number base must be a positive integer greater than any digit present in the equation (the maximum digit here is 7), the only valid base is x=8x = 8.

Adım Adım Çözüm

1
Convert each positional number into its base 10 polynomial expansion.
245x=2x2+4x+5245_x = 2x^2 + 4x + 5, 157x=x2+5x+7157_x = x^2 + 5x + 7, and 66x=6x+666_x = 6x + 6.
A number d2d1d0d_2 d_1 d_0 in base xx represents d2x2+d1x1+d0x0d_2 x^2 + d_1 x^1 + d_0 x^0 in base 10.
2
Substitute the expanded expressions into the given subtraction equation.
(2x2+4x+5)−(x2+5x+7)=6x+6(2x^2 + 4x + 5) - (x^2 + 5x + 7) = 6x + 6
This translates the base xx relationship into a standard base 10 equation.
3
Simplify and rearrange into standard quadratic form.
x2−7x−8=0x^2 - 7x - 8 = 0
Expanding the subtraction yields x2−x−2=6x+6x^2 - x - 2 = 6x + 6. Subtracting (6x+6)(6x + 6) from both sides produces a quadratic set to zero.
4
Factorize the quadratic expression.
(x−8)(x+1)=0  ⟹  x=8 or x=−1(x - 8)(x + 1) = 0 \implies x = 8 \text{ or } x = -1
Finding the roots provides potential mathematical values for xx.
5
Apply number base constraints to select the valid root.
x=8x = 8
A valid base must be a positive integer strictly greater than any individual digit in the expression. Since 7 appears in 157x157_x, x>7x > 7, ruling out −1-1 and confirming x=8x = 8.

Anahtar Kavram

Converting numbers from an unknown base xx to base 10 polynomials to solve algebraic equations.
Soru 4Soru →

If 214x=3125214_x = 312_5, where xx represents a positive integer base, find the value of xx.

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Cevap: 6

Cevap

The value of the base xx is 6.
Expanding 3125312_5 to base 10 gives 3(25)+1(5)+2(1)=823(25) + 1(5) + 2(1) = 82. Expanding 214x214_x gives 2x2+x+42x^2 + x + 4. Equating both expressions yields 2x2+x−78=02x^2 + x - 78 = 0. Factoring as (2x+13)(x−6)=0(2x + 13)(x - 6) = 0 gives the positive integer root x=6x = 6.

Adım Adım Çözüm

1
Convert the right side of the equation from base 5 to base 10
3125=3×52+1×51+2×50=75+5+2=8210312_5 = 3 \times 5^2 + 1 \times 5^1 + 2 \times 5^0 = 75 + 5 + 2 = 82_{10}
Converting known non-decimal bases to base 10 provides a standard baseline for algebraic manipulation.
2
Expand the left side expression in terms of powers of xx
214x=2x2+x+4214_x = 2x^2 + x + 4
Positional values in base xx correspond to powers of xx (x2,x1,x0x^2, x^1, x^0).
3
Formulate and rearrange the resulting quadratic equation in standard form
2x2+x+4=82  ⟹  2x2+x−78=02x^2 + x + 4 = 82 \implies 2x^2 + x - 78 = 0
Setting the base 10 expansions equal forms a quadratic equation.
4
Solve the quadratic equation for positive integer values of xx
x=6x = 6
Factoring (2x+13)(x−6)=0(2x + 13)(x - 6) = 0 gives x=6x = 6 or x=−6.5x = -6.5. A base must be a positive integer larger than all individual digits appearing in the number (digits are 2, 1, 4, so x>4x > 4).

Anahtar Kavram

Converting non-decimal numbers to base 10 using positional expansion to solve polynomial equations involving unknown bases.
Soru 5Soru →

What is the result of the subtraction 415−23541_5 - 23_5 in base 5?

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Cevap: 13513_5

Cevap

13513_5
The answer 13513_5 is correct because borrowing 1 from the fives place contributes 5 to the units place, making the calculation (1+5)−3=3(1 + 5) - 3 = 3 for the units place and 3−2=13 - 2 = 1 for the fives place.

Adım Adım Çözüm

1
Set up column subtraction for 415−23541_5 - 23_5 in base 5.
In the units position, we evaluate 1−31 - 3.
Since 1 is smaller than 3, a borrow operation from the fives column is required.
2
Borrow 1 from the fives column (reducing 4 to 3) and add the base value 5 to the units digit 1.
Units column value becomes 1+5=61 + 5 = 6. Fives column value becomes 3.
When borrowing in base nn, the borrowed amount is equal to nn, which is 5 for base 5.
3
Subtract digits in each column.
Units column: 6−3=36 - 3 = 3. Fives column: 3−2=13 - 2 = 1. The result is 13513_5.
Completing the column subtraction yields the final digits in base 5.

Anahtar Kavram

Subtraction in Non-Decimal Number Bases
Tahmini Süre:45s
Soru 6Soru →

Convert the fractional binary number 0.110120.1101_2 to its equivalent base 10 (decimal) value. What is the decimal value?

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Cevap: 0.8125

Cevap

The base 10 value of 0.110120.1101_2 is 0.81250.8125.
To convert a fractional binary number to decimal, expand each digit after the radix point using decreasing negative powers of 2 (2−1,2−2,2−3,2−42^{-1}, 2^{-2}, 2^{-3}, 2^{-4}). Evaluating 1(0.5)+1(0.25)+0(0.125)+1(0.0625)1(0.5) + 1(0.25) + 0(0.125) + 1(0.0625) yields 0.81250.8125.

Adım Adım Çözüm

1
Write the given binary fraction in place-value expansion using powers of 2
0.11012=1⋅2−1+1⋅2−2+0⋅2−3+1⋅2−40.1101_2 = 1 \cdot 2^{-1} + 1 \cdot 2^{-2} + 0 \cdot 2^{-3} + 1 \cdot 2^{-4}
Positions after the binary point represent negative powers of 2 starting from 2−12^{-1}.
2
Evaluate each fractional component
2−1=0.52^{-1} = 0.5, 2−2=0.252^{-2} = 0.25, 2−3=0.1252^{-3} = 0.125, 2−4=0.06252^{-4} = 0.0625
Calculating standard decimal values for binary fractional places.
3
Add the non-zero fractional terms together
0.5+0.25+0.0625=0.81250.5 + 0.25 + 0.0625 = 0.8125
Summing the decimal values gives the complete converted decimal representation.

Anahtar Kavram

Conversion of fractional numbers from base 2 to base 10
Soru 7Soru →

What is the result of the subtraction 5238−2678523_8 - 267_8 in base 8?

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Cevap: 2348234_8

Cevap

2348234_8
Performing place-by-place subtraction in base 8 requires borrowing 8 whenever a top digit is smaller than the bottom digit. Borrowing 1 from the tens place adds 8 to the units place (3+8−7=43 + 8 - 7 = 4). The tens place becomes 1, borrowing 1 from the hundreds place adds 8 (1+8−6=31 + 8 - 6 = 3). The hundreds place becomes 4, giving 4−2=24 - 2 = 2. Thus, the answer is 2348234_8.

Adım Adım Çözüm

1
Subtract the units column (3−73 - 7 in base 8)
Borrow 11 from the middle column (which represents 88). The units position becomes 3+8=113 + 8 = 11. Then 11−7=411 - 7 = 4.
Since 3<73 < 7, borrowing from the next higher position (base 8) is required.
2
Subtract the middle column (1−61 - 6 in base 8)
After borrowing, the middle digit 22 becomes 11. Borrow 11 from the hundreds column (representing 88). The middle position becomes 1+8=91 + 8 = 9. Then 9−6=39 - 6 = 3.
The middle digit was reduced by 11 due to the previous borrow, requiring another borrow from the left.
3
Subtract the hundreds column (4−24 - 2 in base 8)
The left digit 55 was reduced to 44. Then 4−2=24 - 2 = 2.
Complete the subtraction for the leading column.
4
Combine the resulting digits
2348234_8
Concatenating the results from left to right gives the final answer in base 8.

Anahtar Kavram

Subtraction in non-decimal number bases
Soru 8Soru →

Convert the decimal number 18.6251018.625_{10} to a number in base 8. What is the equivalent value in base 8?

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Cevap: 22.5822.5_8

Cevap

The equivalent value in base 8 is 22.5822.5_8.
Converting the whole number part 181018_{10} to base 8 gives 22822_8 because 18=2×81+2×8018 = 2 \times 8^1 + 2 \times 8^0. Converting the fractional part 0.625100.625_{10} to base 8 involves multiplying by 8: 0.625×8=5.00.625 \times 8 = 5.0, giving .58.5_8. Thus, 18.62510=22.5818.625_{10} = 22.5_8.

Adım Adım Çözüm

1
Convert the integer part (1818) from base 10 to base 8.
18÷8=218 \div 8 = 2 with a remainder of 22. Reading the digits upwards gives 22822_8.
Successive division by 8 extracts the octal place values for the whole number.
2
Convert the fractional part (0.6250.625) from base 10 to base 8.
0.625×8=5.00.625 \times 8 = 5.0. The whole number part of the product is 55, so 0.62510=0.580.625_{10} = 0.5_8.
Successive multiplication of the fractional part by the target base isolates the negative powers of the base.
3
Combine the converted integer and fractional parts.
228+0.58=22.5822_8 + 0.5_8 = 22.5_8.
The full representation is the sum of the integer and fractional base 8 components.

Anahtar Kavram

Fractional Base Conversion
Soru 9Soru →

If 43x+56x=121x43_x + 56_x = 121_x, where xx represents a positive integer base, find the value of xx.

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Cevap: 8

Cevap

The value of the base xx is 8.
Expanding each base xx number into polynomial form gives (4x+3)+(5x+6)=x2+2x+1(4x + 3) + (5x + 6) = x^2 + 2x + 1. Simplifying yields the quadratic equation x2−7x−8=0x^2 - 7x - 8 = 0, which factors as (x−8)(x+1)=0(x - 8)(x + 1) = 0. Since a number base must be a positive integer greater than any digit present in the problem (x>6x > 6), x=8x = 8.

Adım Adım Çözüm

1
Convert all base xx numbers into base 10 algebraic expressions.
43x=4x+343_x = 4x + 3, 56x=5x+656_x = 5x + 6, and 121x=x2+2x+1121_x = x^2 + 2x + 1.
Place-value expansion expresses numbers in base xx as polynomials in xx.
2
Set up the algebraic equation corresponding to the addition.
(4x+3)+(5x+6)=x2+2x+1  ⟹  9x+9=x2+2x+1(4x + 3) + (5x + 6) = x^2 + 2x + 1 \implies 9x + 9 = x^2 + 2x + 1.
The sum of the left-hand terms equals the right-hand term.
3
Rearrange into standard quadratic form and factor.
x2−7x−8=0  ⟹  (x−8)(x+1)=0x^2 - 7x - 8 = 0 \implies (x - 8)(x + 1) = 0.
Moving all terms to one side allows solving for the roots of the quadratic equation.
4
Determine the valid base value.
x=8x = 8.
Number bases must be positive integers greater than all individual digits present in the expression (x>6x > 6).

Anahtar Kavram

Unknown base equations and expansion
Soru 10Soru →

If 24x×13x=345x24_x \times 13_x = 345_x, where xx represents a positive integer base, find the value of xx.

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Cevap: 7

Cevap

The value of the base xx is 7.
Expanding 24x24_x, 13x13_x, and 345x345_x into base 10 yields (2x+4)(x+3)=3x2+4x+5(2x + 4)(x + 3) = 3x^2 + 4x + 5. Expanding the left side gives 2x2+10x+122x^2 + 10x + 12. Equating and simplifying gives x2−6x−7=0x^2 - 6x - 7 = 0, which factors as (x−7)(x+1)=0(x - 7)(x + 1) = 0. The positive integer solution greater than 5 is x=7x = 7.

Adım Adım Çözüm

1
Convert all base xx numbers to decimal (base 10) expressions.
24x=2x+424_x = 2x + 4, 13x=x+313_x = x + 3, and 345x=3x2+4x+5345_x = 3x^2 + 4x + 5.
Place-value expansion allows algebraic manipulation in standard base 10.
2
Multiply the expanded factors on the left-hand side.
(2x+4)(x+3)=2x2+10x+12(2x + 4)(x + 3) = 2x^2 + 10x + 12.
Applying the distributive property of multiplication.
3
Equate the expanded left-hand side to the right-hand side and rearrange into standard quadratic form.
3x2+4x+5−(2x2+10x+12)=0  ⟹  x2−6x−7=03x^2 + 4x + 5 - (2x^2 + 10x + 12) = 0 \implies x^2 - 6x - 7 = 0.
Setting the quadratic expression equal to zero enables factoring.
4
Factor the quadratic equation and select the valid base.
(x−7)(x+1)=0  ⟹  x=7(x - 7)(x + 1) = 0 \implies x = 7 (rejecting x=−1x = -1).
A number base must be a positive integer strictly greater than any individual digit in the given numbers (max digit is 5).

Anahtar Kavram

Solving polynomial equations derived from number base expansion.
Soru 11Soru →

If 132k+43k=230k132_k + 43_k = 230_k, where kk represents a positive integer base, what is the value of kk?

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Cevap: 5

Cevap

The value of the base kk is 5.
Expanding 132k132_k, 43k43_k, and 230k230_k into base 10 yields (k2+3k+2)+(4k+3)=2k2+3k(k^2 + 3k + 2) + (4k + 3) = 2k^2 + 3k. Simplifying this gives the quadratic equation k2−4k−5=0k^2 - 4k - 5 = 0, which factors as (k−5)(k+1)=0(k - 5)(k + 1) = 0. Since a number base must be a positive integer greater than the maximum digit present (which is 4), the correct base is 5.

Adım Adım Çözüm

1
Expand each term in the equation 132k+43k=230k132_k + 43_k = 230_k into base 10 using place-value notation.
132k=1⋅k2+3⋅k1+2⋅k0=k2+3k+2132_k = 1 \cdot k^2 + 3 \cdot k^1 + 2 \cdot k^0 = k^2 + 3k + 2
43k=4⋅k1+3⋅k0=4k+343_k = 4 \cdot k^1 + 3 \cdot k^0 = 4k + 3
230k=2⋅k2+3⋅k1+0⋅k0=2k2+3k230_k = 2 \cdot k^2 + 3 \cdot k^1 + 0 \cdot k^0 = 2k^2 + 3k
Converting all terms to base 10 allows standard algebraic operations to be performed.
2
Set up the algebraic equation in base 10 and combine like terms.
(k2+3k+2)+(4k+3)=2k2+3k(k^2 + 3k + 2) + (4k + 3) = 2k^2 + 3k
k2+7k+5=2k2+3kk^2 + 7k + 5 = 2k^2 + 3k
Equate the sum of the expanded left-hand side terms to the expanded right-hand side.
3
Rearrange the terms to form a standard quadratic equation ak2+bk+c=0ak^2 + bk + c = 0.
2k2+3k−(k2+7k+5)=02k^2 + 3k - (k^2 + 7k + 5) = 0
k2−4k−5=0k^2 - 4k - 5 = 0
Subtract k2+7k+5k^2 + 7k + 5 from both sides to set the quadratic expression equal to zero.
4
Factor the quadratic equation to find the possible values of kk.
(k−5)(k+1)=0  ⟹  k=5 or k=−1(k - 5)(k + 1) = 0 \implies k = 5 \text{ or } k = -1
Solve for kk using factorization.
5
Select the valid base based on the domain of number bases.
k=5k = 5 (since a number base must be a positive integer greater than the largest digit 4 present in the expression).
Base numbers cannot be negative or smaller than any of their constituent digits.

Anahtar Kavram

Number Bases and Conversions
Tahmini Süre:1m 30s
Soru 12Soru →

If 32x=231032_x = 23_{10}, find the value of the base xx.

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Cevap: 7

Cevap

7
Expanding the number 32x32_x into base 10 gives 3×x+2=3x+23 \times x + 2 = 3x + 2. Setting 3x+2=233x + 2 = 23 leads to 3x=213x = 21, giving x=7x = 7.

Adım Adım Çözüm

1
Expand 32x32_x in terms of powers of xx
3x+23x + 2
In base xx, the rightmost digit has a weight of x0=1x^0 = 1 and the next digit to the left has a weight of x1=xx^1 = x.
2
Equate the expanded form to the given base 10 value and solve for xx
x=7x = 7
Setting 3x+2=233x + 2 = 23 gives 3x=213x = 21, which yields x=7x = 7 upon dividing by 3.

Anahtar Kavram

Expansion and conversion from Base x to Base 10
Tahmini Süre:45s
Soru 13Soru →

If 11012+324=x51101_2 + 32_4 = x_5, what is the value of xx?

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Cevap: 102

Cevap

The value of xx is 102102.
To add numbers given in different bases, convert each term to base 10: 11012=13101101_2 = 13_{10} and 324=141032_4 = 14_{10}. Summing these decimal values gives 13+14=271013 + 14 = 27_{10}. Converting 271027_{10} to base 5 by repeated division by 5 gives remainders 2, 0, 1 (from least significant to most significant digit), resulting in 1025102_5. Therefore, x=102x = 102.

Adım Adım Çözüm

1
Convert 110121101_2 to decimal (base 10)
131013_{10}
Expand in powers of 2: 1(23)+1(22)+0(21)+1(20)=8+4+0+1=131(2^3) + 1(2^2) + 0(2^1) + 1(2^0) = 8 + 4 + 0 + 1 = 13.
2
Convert 32432_4 to decimal (base 10)
141014_{10}
Expand in powers of 4: 3(41)+2(40)=12+2=143(4^1) + 2(4^0) = 12 + 2 = 14.
3
Add the two base-10 numbers together
271027_{10}
Perform standard addition: 13+14=2713 + 14 = 27.
4
Convert 271027_{10} into base 5
1025102_5
Divide 27 repeatedly by 5 and record remainders: 27=5×5+227 = 5 \times 5 + 2, 5=1×5+05 = 1 \times 5 + 0, 1=0×5+11 = 0 \times 5 + 1. Reading remainders upward yields 1025102_5.

Anahtar Kavram

Conversion between different number bases via base 10
Soru 14Soru →

If 1.41x=4625101.41_x = \frac{46}{25}_{10}, where xx is a positive integer base, find the value of xx.

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Cevap: 5

Cevap

The value of the base xx is 5.
Expanding 1.41x1.41_x yields 1+4x+1x21 + \frac{4}{x} + \frac{1}{x^2}. Setting this equal to 4625\frac{46}{25} gives 4x+1x2=2125\frac{4x + 1}{x^2} = \frac{21}{25}. Cross-multiplying results in the quadratic equation 21x2−100x−25=021x^2 - 100x - 25 = 0, which factors as (21x+5)(x−5)=0(21x + 5)(x - 5) = 0. Since a number base must be a positive integer greater than 4, x=5x = 5 is the only valid solution.

Adım Adım Çözüm

1
Expand 1.41x1.41_x using place value powers of xx.
1+4x+1x21 + \frac{4}{x} + \frac{1}{x^2}
Fractional digits to the right of the radix point represent negative powers of the base (x−1,x−2,…x^{-1}, x^{-2}, \dots).
2
Equate the expanded form to 4625\frac{46}{25} and simplify.
4x+1x2=2125\frac{4x + 1}{x^2} = \frac{21}{25}
Subtracting 1 from both sides isolates the fractional place values.
3
Cross-multiply and solve the quadratic equation 21x2−100x−25=021x^2 - 100x - 25 = 0.
(21x+5)(x−5)=0  ⟹  x=5(21x + 5)(x - 5) = 0 \implies x = 5
A base must be a positive integer strictly greater than any individual digit in the number (digits present are 1 and 4).

Anahtar Kavram

Conversion of fractional numbers in non-decimal bases to base 10 and solving polynomial equations in unknown bases.
Soru 15Soru →

If 123x=3810123_x = 38_{10}, what is the value of the base xx?

Cevabı ve açıklamayı göster

Cevap: 5

Cevap

The base xx is 5.
Expanding 123x123_x in terms of powers of xx yields 1⋅x2+2⋅x1+3⋅x0=x2+2x+31 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3. Setting this equal to the decimal value 38 produces the quadratic equation x2+2x+3=38x^2 + 2x + 3 = 38, which simplifies to x2+2x−35=0x^2 + 2x - 35 = 0. Factoring gives (x+7)(x−5)=0(x + 7)(x - 5) = 0, yielding solutions x=−7x = -7 and x=5x = 5. Since a number base must be a positive integer, the correct value for xx is 5.

Adım Adım Çözüm

1
Expand the base xx number into decimal form using place values
123x=1⋅x2+2⋅x1+3⋅x0=x2+2x+3123_x = 1 \cdot x^2 + 2 \cdot x^1 + 3 \cdot x^0 = x^2 + 2x + 3
Each digit position in base xx corresponds to a power of xx, starting from x0x^0 on the right.
2
Set up and rearrange the quadratic equation
x2+2x+3=38  ⟹  x2+2x−35=0x^2 + 2x + 3 = 38 \implies x^2 + 2x - 35 = 0
Subtracting 38 from both sides converts the equation into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
3
Solve the quadratic equation for xx
(x+7)(x−5)=0  ⟹  x=−7 or x=5(x + 7)(x - 5) = 0 \implies x = -7 \text{ or } x = 5
Factoring gives the roots of the quadratic equation.
4
Select the valid positive base
x=5x = 5
A base must be a positive integer greater than the largest digit appearing in the number (which is 3).

Anahtar Kavram

Place value expansion and base conversion to base 10
Soru 16Soru →

If 203n−114n=45n203_n - 114_n = 45_n, where nn represents a positive integer base, what is the value of nn?

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Cevap: 66

Cevap

The correct base is 66.
Expanding the numbers in terms of powers of nn gives (2n2+3)−(n2+n+4)=4n+5(2n^2 + 3) - (n^2 + n + 4) = 4n + 5. Grouping like terms results in the quadratic equation n2−5n−6=0n^2 - 5n - 6 = 0. Factoring yields (n−6)(n+1)=0(n - 6)(n + 1) = 0, giving n=6n = 6 as the only valid positive integer solution greater than 55.

Adım Adım Çözüm

1
Convert each term from base nn to base 10 using positional expansion.
203n=2n2+0n+3=2n2+3203_n = 2n^2 + 0n + 3 = 2n^2 + 3, 114n=1n2+1n+4=n2+n+4114_n = 1n^2 + 1n + 4 = n^2 + n + 4, and 45n=4n+545_n = 4n + 5.
Converting all terms to a common decimal representation allows algebraic manipulation.
2
Substitute the expanded terms back into the original equation and simplify.
(2n2+3)−(n2+n+4)=4n+5  ⟹  n2−n−1=4n+5(2n^2 + 3) - (n^2 + n + 4) = 4n + 5 \implies n^2 - n - 1 = 4n + 5.
Carefully distribute the negative sign across all terms of (n2+n+4)(n^2 + n + 4).
3
Rearrange the expression into standard quadratic form ax2+bx+c=0ax^2 + bx + c = 0.
n2−5n−6=0n^2 - 5n - 6 = 0.
Subtracting 4n+54n + 5 from both sides sets the quadratic equation to zero.
4
Factor the quadratic equation to find the valid base nn.
(n−6)(n+1)=0  ⟹  n=6(n - 6)(n + 1) = 0 \implies n = 6 or n=−1n = -1.
Since a base must be a positive integer strictly greater than any digit present in the equation (maximum digit is 55), n=6n = 6.

Anahtar Kavram

Solving unknown base equations using polynomial expansion in positional notation.
Soru 17Soru →

Convert the base 2 fractional number 110.1012110.101_2 to its equivalent value in base 10 (decimal). What is the decimal value?

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Cevap: 6.625

Cevap

The decimal value of 110.1012110.101_2 is 6.6256.625.
To convert 110.1012110.101_2 to base 10, expand both the integer and fractional parts using powers of 2. The integer component 1102=(1×22)+(1×21)+(0×20)=4+2+0=6110_2 = (1 \times 2^2) + (1 \times 2^1) + (0 \times 2^0) = 4 + 2 + 0 = 6. The fractional component 0.1012=(1×2−1)+(0×2−2)+(1×2−3)=0.5+0+0.125=0.6250.101_2 = (1 \times 2^{-1}) + (0 \times 2^{-2}) + (1 \times 2^{-3}) = 0.5 + 0 + 0.125 = 0.625. Summing these values gives 6+0.625=6.6256 + 0.625 = 6.625.

Adım Adım Çözüm

1
Convert the integer part 1102110_2 to base 10
1×22+1×21+0×20=4+2+0=61 \times 2^2 + 1 \times 2^1 + 0 \times 2^0 = 4 + 2 + 0 = 6
Each position to the left of the binary point corresponds to an increasing non-negative power of 2 (20,21,222^0, 2^1, 2^2).
2
Convert the fractional part 0.10120.101_2 to base 10
1×2−1+0×2−2+1×2−3=12+0+18=0.5+0.125=0.6251 \times 2^{-1} + 0 \times 2^{-2} + 1 \times 2^{-3} = \frac{1}{2} + 0 + \frac{1}{8} = 0.5 + 0.125 = 0.625
Each position to the right of the binary point corresponds to a negative power of 2 (2−1,2−2,2−32^{-1}, 2^{-2}, 2^{-3}).
3
Combine the integer and fractional results
6+0.625=6.6256 + 0.625 = 6.625
The total value in base 10 is the sum of the expanded integer and fractional components.

Anahtar Kavram

Conversion of Fractional Non-Decimal Numbers to Decimal
Soru 18Soru →

Convert the base 5 number 2345234_5 to a base 10 (decimal) number. What is the value in base 10?

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Cevap: 69

Cevap

The decimal (base 10) equivalent of 2345234_5 is 69.
Expanding 2345234_5 gives (2×52)+(3×51)+(4×50)=50+15+4=69(2 \times 5^2) + (3 \times 5^1) + (4 \times 5^0) = 50 + 15 + 4 = 69.

Adım Adım Çözüm

1
Write out the positional expansion of 2345234_5 using powers of 5 starting from position 0 on the right.
2345=(2×52)+(3×51)+(4×50)234_5 = (2 \times 5^2) + (3 \times 5^1) + (4 \times 5^0)
Converting from any non-decimal base to base 10 involves multiplying each digit by its corresponding positional base power.
2
Evaluate the powers of 5 and multiply by the respective digits.
2×25=502 \times 25 = 50, 3×5=153 \times 5 = 15, 4×1=44 \times 1 = 4
Apply basic arithmetic exponents: 52=255^2=25, 51=55^1=5, 50=15^0=1.
3
Sum all calculated values together.
50+15+4=6950 + 15 + 4 = 69
Adding the positional values yields the total value in base 10.

Anahtar Kavram

Converting a non-decimal number to base 10 using expansion by powers of the base.
Soru 19Soru →

What is the result of the subtraction 5248−2678524_8 - 267_8 expressed in base 8?

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Cevap: 2358235_8

Cevap

The correct answer is 2358235_8.
Subtracting 2678267_8 from 5248524_8 using direct base-8 subtraction requires borrowing 8 when a digit is smaller than the digit subtracted from it. In the units column, borrowing 1 (which equals 8) gives (4+8)−7=5(4 + 8) - 7 = 5. In the eights column, the borrowed-from digit 2 becomes 1; borrowing 1 (which equals 8) gives (1+8)−6=3(1 + 8) - 6 = 3. In the sixty-fours column, the borrowed-from digit 5 becomes 4; 4−2=24 - 2 = 2. Thus, the difference is 2358235_8.

Adım Adım Çözüm

1
Convert both octal numbers to base 10.
5248=(5×82)+(2×81)+(4×80)=320+16+4=34010524_8 = (5 \times 8^2) + (2 \times 8^1) + (4 \times 8^0) = 320 + 16 + 4 = 340_{10}, and 2678=(2×82)+(6×81)+(7×80)=128+48+7=18310267_8 = (2 \times 8^2) + (6 \times 8^1) + (7 \times 8^0) = 128 + 48 + 7 = 183_{10}.
Converting to decimal simplifies standard arithmetic operations.
2
Subtract the decimal values.
34010−18310=15710340_{10} - 183_{10} = 157_{10}.
Obtain the difference in base 10.
3
Convert the decimal difference 15710157_{10} back to base 8.
157÷8=19157 \div 8 = 19 remainder 55; 19÷8=219 \div 8 = 2 remainder 33; 2÷8=02 \div 8 = 0 remainder 22. Reading remainders from bottom to top gives 2358235_8.
Successive division by 8 yields the octal representation.

Anahtar Kavram

Non-decimal base arithmetic and borrowing rules
Soru 20Soru →

If 132x−54x=45x132_x - 54_x = 45_x, where xx is a positive integer base, what is the value of xx?

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Cevap: 7

Cevap

The value of the base xx is 7.
Expanding all numbers in terms of powers of xx yields 132x=x2+3x+2132_x = x^2 + 3x + 2, 54x=5x+454_x = 5x + 4, and 45x=4x+545_x = 4x + 5. Substituting these into 132x−54x=45x132_x - 54_x = 45_x gives (x2+3x+2)−(5x+4)=4x+5(x^2 + 3x + 2) - (5x + 4) = 4x + 5. Simplifying this equation results in x2−6x−7=0x^2 - 6x - 7 = 0. Factoring (x−7)(x+1)=0(x - 7)(x + 1) = 0 gives solutions x=7x = 7 and x=−1x = -1. Since a base must be a positive integer greater than 5, the base xx is 7.

Adım Adım Çözüm

1
Convert each term from base xx to base 10 using place-value expansion.
132x=1⋅x2+3⋅x1+2⋅x0=x2+3x+2132_x = 1 \cdot x^2 + 3 \cdot x^1 + 2 \cdot x^0 = x^2 + 3x + 2, 54x=5x+454_x = 5x + 4, and 45x=4x+545_x = 4x + 5.
Converting all terms to a common base (base 10) allows standard algebraic solving.
2
Substitute the expanded expressions into the given equation.
(x2+3x+2)−(5x+4)=4x+5(x^2 + 3x + 2) - (5x + 4) = 4x + 5.
Set up the algebraic equation in terms of xx.
3
Simplify the equation into standard quadratic form.
x2−2x−2=4x+5  ⟹  x2−6x−7=0x^2 - 2x - 2 = 4x + 5 \implies x^2 - 6x - 7 = 0.
Combine like terms and move all terms to one side.
4
Solve the quadratic equation for xx.
(x−7)(x+1)=0  ⟹  x=7(x - 7)(x + 1) = 0 \implies x = 7 or x=−1x = -1.
Factor the quadratic expression.
5
Select the valid base.
x=7x = 7.
A number base must be a positive integer greater than the largest digit present in the equation (which is 5).

Anahtar Kavram

Solving equations involving unknown number bases by expanding in powers of the base.
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Number Bases and Conversions Alıştırma Soruları — JAMB UTME | Examkin