Basic Numeracy

295 soru

Soru 201Soru
Calculate the exact numerical value of the given mathematical expression using the VBODMAS rule:
35 of 250[42+{24÷(7.54.2+0.3)}]\frac{3}{5} \text{ of } 250 - \left[ 42 + \left\{ 24 \div \left( 7.5 - \overline{4.2 + 0.3} \right) \right\} \right]
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Cevap: 100

Cevap

100
Evaluating strictly according to VBODMAS order of operations yields 100: Vinculum gives 4.5, round brackets give 3, division in curly brackets gives 8, addition in square brackets gives 50, 'of' operation gives 150, and 150 - 50 = 100.

Adım Adım Çözüm

1
Evaluate the expression under the vinculum (bar line)
4.2 + 0.3 = 4.5
According to the VBODMAS rule, the vinculum takes precedence over standard brackets and operations.
2
Simplify the innermost round brackets
7.5 - 4.5 = 3
Operations inside round parentheses ( ) are resolved next.
3
Evaluate the expression inside the curly brackets
24 / 3 = 8
Perform the division operation inside curly braces { }.
4
Evaluate the expression inside the square brackets
42 + 8 = 50
Perform addition within the outer square brackets [ ].
5
Calculate the 'of' operation
(3/5) * 250 = 150
The 'of' operation takes priority over standard addition and subtraction outside brackets.
6
Perform final subtraction
150 - 50 = 100
Subtract the total evaluated bracketed quantity from the 'of' calculation result.

Anahtar Kavram

Hierarchical priority of operations in VBODMAS involving vinculum, nested brackets, and fractional 'of' operations
Soru 202Soru

A positive integer NN has exactly 1515 positive factors and is divisible by 66. If NN has exactly two distinct prime factors, what is the sum of all possible values of NN that are less than 500500?

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Cevap: 468

Cevap

The sum of all possible values of N less than 500 is 468.
For a number N=pa×qbN = p^a \times q^b to have 15 factors, (a+1)(b+1)=15(a+1)(b+1) = 15. The factor pairs of 15 for two prime factors are (5,3)(5, 3) and (3,5)(3, 5), which correspond to exponent pairs (4,2)(4, 2) and (2,4)(2, 4). Since NN is divisible by 6, its prime factors are 2 and 3. Computing both cases gives 24×32=1442^4 \times 3^2 = 144 and 22×34=3242^2 \times 3^4 = 324. Both are less than 500, making their sum 144+324=468144 + 324 = 468.

Adım Adım Çözüm

1
Determine the prime factors of N
N has prime factors 2 and 3
Since N is divisible by 6, it must have at least 2 and 3 as prime factors. The problem states N has exactly two distinct prime factors, so its prime factorization is of the form N=2a×3bN = 2^a \times 3^b.
2
Apply the total factors formula
(a+1)(b+1)=15(a + 1)(b + 1) = 15
The total number of positive factors of N=pa×qbN = p^a \times q^b is given by (a+1)(b+1)=15(a + 1)(b + 1) = 15.
3
Find valid non-negative integer pairs (a, b)
Either (a=4,b=2)(a=4, b=2) or (a=2,b=4)(a=2, b=4)
Since 15 factors into 5×35 \times 3 or 3×53 \times 5 (note that 15×115 \times 1 would imply only one prime factor, which contradicts having two distinct prime factors), the possible exponent pairs are (4,2)(4, 2) and (2,4)(2, 4).
4
Calculate the values of N and check bounds
N1=24×32=144N_1 = 2^4 \times 3^2 = 144 and N2=22×34=324N_2 = 2^2 \times 3^4 = 324
Both 144<500144 < 500 and 324<500324 < 500 meet all conditions.
5
Sum the valid values of N
144+324=468144 + 324 = 468
Adding the two valid integers yields the total sum.

Anahtar Kavram

Factors and Prime Factorization Exponent Rule
Soru 203Soru

Three church bells toll together at intervals of 99 minutes, 1212 minutes, and 1515 minutes respectively. If they all toll together at 8:00 AM8:00\text{ AM}, at what time will they next toll together?

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Cevap: 11:00 AM

Cevap

11:00 AM
The bells will toll together after a duration equal to the Least Common Multiple (LCM) of their individual tolling intervals. The prime factorizations are 9=329 = 3^2, 12=22×312 = 2^2 \times 3, and 15=3×515 = 3 \times 5. Taking the highest power of each prime factor yields LCM(9,12,15)=22×32×5=180\text{LCM}(9, 12, 15) = 2^2 \times 3^2 \times 5 = 180 minutes, which is equal to 33 hours. Adding 33 hours to 8:00 AM8:00\text{ AM} gives 11:00 AM11:00\text{ AM}.

Adım Adım Çözüm

1
Find prime factorizations of each interval
9=329 = 3^2, 12=22×312 = 2^2 \times 3, and 15=3×515 = 3 \times 5
To compute the Least Common Multiple (LCM), express each number in prime factor form.
2
Calculate the LCM
LCM(9,12,15)=22×32×5=4×9×5=180 minutes\text{LCM}(9, 12, 15) = 2^2 \times 3^2 \times 5 = 4 \times 9 \times 5 = 180\text{ minutes}
Take the highest power of each prime factor involved.
3
Convert minutes to hours and add to the initial time
180 minutes=3 hours180\text{ minutes} = 3\text{ hours}; 8:00 AM+3 hours=11:00 AM8:00\text{ AM} + 3\text{ hours} = 11:00\text{ AM}
Determine the exact time the bells will next chime together.

Anahtar Kavram

Application of Least Common Multiple (LCM) in Simultaneous Periodic Events
Soru 204Soru
What is the simplified numerical value of the following mathematical expression when evaluated strictly according to the VBODMAS rule?
40% of 150[16+{20÷(8.53.1+1.4)}]40\% \text{ of } 150 - \left[ 16 + \left\{ 20 \div \left( 8.5 - \overline{3.1 + 1.4} \right) \right\} \right]
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Cevap: 39

Cevap

39
Evaluating strictly by VBODMAS priority: first the vinculum yields 4.5, then the round bracket gives 8.5 - 4.5 = 4, then division yields 20 ÷ 4 = 5, followed by square bracket addition 16 + 5 = 21. Finally, subtracting 21 from 40% of 150 (which is 60) gives 60 - 21 = 39.

Adım Adım Çözüm

1
Evaluate the expression under the vinculum (bar)
\overline{3.1 + 1.4} = 4.5
The vinculum has the highest priority and must be evaluated first.
2
Evaluate the terms inside the round brackets
8.5 - 4.5 = 4
The next priority is resolving the innermost parentheses.
3
Perform division inside the curly brackets
20÷4=520 \div 4 = 5
Inside curly brackets, division takes precedence over addition.
4
Evaluate the terms inside the square brackets
16 + 5 = 21
Resolve the outer square bracket by adding the terms.
5
Calculate the percentage term
40\% \text{ of } 150 = \frac{40}{100} \times 150 = 60
'Of' represents multiplication in percentage operations.
6
Perform final subtraction
60 - 21 = 39
Subtract the total bracket value from the percentage calculation.

Anahtar Kavram

VBODMAS Rule (Vinculum, Brackets, Orders, Division, Multiplication, Addition, Subtraction)
Soru 205Soru

Consider the positive integer N=720N = 720, which has a prime factorization of 24×32×512^4 \times 3^2 \times 5^1. Which of the following statements regarding the positive factors of NN are correct?

Geçerli olan tümünü seçin

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Cevap: The number of positive factors of NN that are perfect squares is 66.; The sum of all positive even factors of NN is 23402340.; The product of all positive factors of NN is equal to 72015720^{15}.

Cevap

The correct statements are those asserting that NN has 6 perfect square factors, that the sum of all positive even factors of NN is 2340, and that the product of all positive factors of NN is 72015720^{15}.
The statements confirming 6 perfect square factors, a sum of 2340 for even factors, and a factor product of 72015720^{15} are all mathematically correct applications of prime factorization principles.

Adım Adım Çözüm

1
Determine perfect square factors of N=24×32×51N = 2^4 \times 3^2 \times 5^1
For a factor 2a×3b×5c2^a \times 3^b \times 5^c to be a square, exponents a,b,ca, b, c must be even. Possible values: a{0,2,4}a \in \{0, 2, 4\} (3 options), b{0,2}b \in \{0, 2\} (2 options), c{0}c \in \{0\} (1 option). Number of square factors =3×2×1=6= 3 \times 2 \times 1 = 6.
Perfect squares require all prime factors to have even exponents.
2
Calculate the sum of all positive even factors
Sum of all factors =(20+21+22+23+24)(30+31+32)(50+51)=31×13×6=2418= (2^0+2^1+2^2+2^3+2^4)(3^0+3^1+3^2)(5^0+5^1) = 31 \times 13 \times 6 = 2418. Sum of odd factors (only 202^0) =1×13×6=78= 1 \times 13 \times 6 = 78. Sum of even factors =241878=2340= 2418 - 78 = 2340.
Even factors are obtained by subtracting odd factor sum from total factor sum.
3
Count factors divisible by 15
Since 15=31×5115 = 3^1 \times 5^1, factors of 720 divisible by 15 require a{0,1,2,3,4}a \in \{0, 1, 2, 3, 4\} (5 choices), b{1,2}b \in \{1, 2\} (2 choices), and c{1}c \in \{1\} (1 choice). Total factors =5×2×1=10= 5 \times 2 \times 1 = 10.
Divisibility by 15 requires at least power 1 for both prime factors 3 and 5.
4
Compute the product of all positive factors
Total factors T=(4+1)(2+1)(1+1)=30T = (4+1)(2+1)(1+1) = 30. Product of factors =NT/2=72030/2=72015= N^{T/2} = 720^{30/2} = 720^{15}.
Factors pair up such that fi×fT+1i=Nf_i \times f_{T+1-i} = N, giving T/2T/2 pairs.

Anahtar Kavram

Properties of positive factors derived from prime factorization: square factor counts, even factor sums, constrained divisibility factor counts, and total factor product formula.
Soru 206Soru
What is the simplified numerical value of the following mathematical expression when evaluated strictly using the BODMAS rule?
50[8+{16÷(873)}]50 - \left[ 8 + \left\{ 16 \div \left( 8 - \overline{7 - 3} \right) \right\} \right]
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Cevap: 38

Cevap

The simplified numerical value of the expression is 38.
Evaluating step-by-step according to VBODMAS order gives: first vinculum 7 - 3 = 4, then round brackets 8 - 4 = 4, followed by curly bracket division 16 ÷ 4 = 4, then square bracket addition 8 + 4 = 12, and finally 50 - 12 = 38. The value 38 correctly demonstrates strict adherence to order of operations.

Adım Adım Çözüm

1
Evaluate the expression under the vinculum bar
\overline{7 - 3} = 4
According to VBODMAS rules, operations under a vinculum bar have the highest priority.
2
Evaluate the expression inside the round brackets
8 - 4 = 4
Substitute the vinculum result into the round bracket.
3
Perform division inside the curly brackets
16÷4=416 \div 4 = 4
Curly brackets are evaluated next after simplifying the inner round brackets.
4
Perform addition inside the square brackets
8 + 4 = 12
Resolve the remaining operations inside the outermost square brackets.
5
Perform final subtraction outside the brackets
50 - 12 = 38
Complete the evaluation by subtracting the bracket value from 50.

Anahtar Kavram

VBODMAS Rule (Vinculum, Brackets, Orders, Division, Multiplication, Addition, Subtraction)
Soru 207Soru

A positive integer NN has a prime factorization of the form paqbp^a \cdot q^b, where pp and qq are distinct prime numbers, and a,b1a, b \ge 1. If the total number of positive factors of N2N^2 is 4545 and the total number of positive factors of N3N^3 is 9191, what is the total number of positive factors of NN?

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Cevap: 1515

Cevap

The total number of positive factors of NN is 1515.
For a composite integer N=paqbN = p^a q^b, the number of positive factors is (a+1)(b+1)(a+1)(b+1). Given that N2=p2aq2bN^2 = p^{2a} q^{2b} has 4545 positive factors, we have (2a+1)(2b+1)=45(2a+1)(2b+1) = 45. Factoring 4545 into odd pairs yields two possibilities for (a,b)(a, b): (1,7)(1, 7) or (2,4)(2, 4). Using the second condition that N3=p3aq3bN^3 = p^{3a} q^{3b} has 9191 factors, we check (3a+1)(3b+1)=91(3a+1)(3b+1) = 91. For (1,7)(1, 7), (4)(22)=8891(4)(22) = 88 \neq 91. For (2,4)(2, 4), (7)(13)=91(7)(13) = 91, which is correct. Therefore, a=2a=2 and b=4b=4, so the number of positive factors of NN is (2+1)(4+1)=15(2+1)(4+1) = 15.

Adım Adım Çözüm

1
Set up the factor count equation for N2N^2
Since N=paqbN = p^a q^b, N2=p2aq2bN^2 = p^{2a} q^{2b}. The number of positive factors of N2N^2 is (2a+1)(2b+1)=45(2a+1)(2b+1) = 45.
For a prime factorization pxqyp^x q^y, the total number of positive factors is (x+1)(y+1)(x+1)(y+1).
2
Find possible integer solutions for (a,b)(a, b)
The odd factor pairs of 4545 are (3,15)(3, 15) and (5,9)(5, 9). This gives two possible sets of exponents: Case 1: 2a+1=3    a=12a+1=3 \implies a=1 and 2b+1=15    b=72b+1=15 \implies b=7. Case 2: 2a+1=5    a=22a+1=5 \implies a=2 and 2b+1=9    b=42b+1=9 \implies b=4.
Since a,b1a, b \ge 1, both factors 2a+12a+1 and 2b+12b+1 must be odd integers greater than 11.
3
Evaluate candidate solutions using the second condition for N3N^3
For N3=p3aq3bN^3 = p^{3a} q^{3b}, the number of factors is (3a+1)(3b+1)=91(3a+1)(3b+1) = 91. Testing Case 1 (a=1,b=7a=1, b=7): (3(1)+1)(3(7)+1)=4×22=8891(3(1)+1)(3(7)+1) = 4 \times 22 = 88 \neq 91. Testing Case 2 (a=2,b=4a=2, b=4): (3(2)+1)(3(4)+1)=7×13=91(3(2)+1)(3(4)+1) = 7 \times 13 = 91, which matches.
The true exponent values must simultaneously satisfy both given conditions.
4
Calculate the total number of positive factors of NN
With a=2a=2 and b=4b=4, N=p2q4N = p^2 q^4. The total number of positive factors of NN is (2+1)(4+1)=3×5=15(2+1)(4+1) = 3 \times 5 = 15.
Apply the total factor formula (a+1)(b+1)(a+1)(b+1) to NN.

Anahtar Kavram

Calculating factor counts of a prime-factored number from equations involving factor counts of its powers
Tahmini Süre:2m 0s
Soru 208Soru

A positive integer NN is divisible by 1010. The sum of all distinct prime factors of NN is 1010, and NN has exactly 1212 positive factors. What is the smallest possible value of NN?

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Cevap: 60

Cevap

The smallest possible value of NN is 6060.
Because NN is divisible by 1010, its prime factorization must contain 22 and 55. The sum of all distinct prime factors is 1010, which requires 2+5+p=102 + 5 + p = 10, giving p=3p = 3. Hence, N=2a×3b×5cN = 2^a \times 3^b \times 5^c with a,b,c1a, b, c \ge 1. The number of factors is (a+1)(b+1)(c+1)=12(a+1)(b+1)(c+1) = 12. The unique product representation of 1212 using three integers 2\ge 2 is 3×2×23 \times 2 \times 2, meaning the exponents are one 22 and two 11 s. Minimizing NN requires assigning the highest exponent 22 to the smallest prime 22, giving N=22×31×51=60N = 2^2 \times 3^1 \times 5^1 = 60.

Adım Adım Çözüm

1
Identify the distinct prime factors of NN
The distinct prime factors are 2,3,2, 3, and 55.
Divisibility by 1010 implies 22 and 55 are prime factors of NN. The sum of distinct prime factors is 1010, so the remaining prime factor is 10(2+5)=310 - (2 + 5) = 3.
2
Determine the exponents in the prime factorization of NN
The set of exponents {a,b,c}\{a, b, c\} must be {2,1,1}\{2, 1, 1\}.
The total number of factors is (a+1)(b+1)(c+1)=12(a+1)(b+1)(c+1) = 12. Since a,b,c1a, b, c \ge 1, the only integer partition of 1212 into three factors 2\ge 2 is 3×2×23 \times 2 \times 2.
3
Assign exponents to minimize the value of NN
N=22×31×51=60N = 2^2 \times 3^1 \times 5^1 = 60.
To minimize NN, the largest exponent 22 must be assigned to the smallest prime base 22.

Anahtar Kavram

Prime factorization, distinct prime factor properties, and total factor count formula
Soru 209Soru
What is the simplified value of the following mathematical expression when evaluated strictly using the VBODMAS rule?
15+2.5×[16{10÷(3.52.10.6)}]15 + 2.5 \times \left[ 16 - \left\{ 10 \div \left( 3.5 - \overline{2.1 - 0.6} \right) \right\} \right]
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Cevap: 42.5

Cevap

The simplified value of the expression is 42.5.
Evaluating strictly according to VBODMAS rules, we first simplify the expression under the vinculum bar (2.1 - 0.6 = 1.5), then the round brackets (3.5 - 1.5 = 2), followed by the curly brackets (10 ÷ 2 = 5), and then the square brackets (16 - 5 = 11). Finally, executing multiplication before addition yields 15 + (2.5 × 11) = 15 + 27.5 = 42.5.

Adım Adım Çözüm

1
Evaluate the expression under the vinculum (bar)
\overline{2.1 - 0.6} = 1.5
According to VBODMAS, the vinculum takes highest priority over standard brackets.
2
Evaluate the terms inside the innermost parentheses (round brackets)
3.5 - 1.5 = 2.0
Parentheses are evaluated next after resolving the vinculum.
3
Evaluate the expression inside the curly brackets
10÷2.0=510 \div 2.0 = 5
Division within the curly brackets takes priority before moving outward.
4
Evaluate the expression inside the square brackets
16 - 5 = 11
Resolve the remaining operations within the square brackets.
5
Perform multiplication before addition
15 + (2.5 \times 11) = 15 + 27.5 = 42.5
Multiplication takes precedence over addition according to BODMAS rules.

Anahtar Kavram

Order of Operations (VBODMAS Rule)
Soru 210Soru

If 5x+1+5x1=6505^{x+1} + 5^{x-1} = 650, what is the value of 22x12^{2x - 1}?

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Cevap: 3232

Cevap

32
Factoring out 5x15^{x-1} yields 265x1=65026 \cdot 5^{x-1} = 650, which simplifies to 5x1=255^{x-1} = 25, giving x=3x = 3. Substituting x=3x = 3 into 22x12^{2x - 1} gives 25=322^5 = 32.

Adım Adım Çözüm

1
Factor out common exponential terms from the given equation.
5x1(52+1)=650    5x1(25+1)=650    265x1=6505^{x-1}(5^2 + 1) = 650 \implies 5^{x-1}(25 + 1) = 650 \implies 26 \cdot 5^{x-1} = 650
Use the product law of indices 5x+1=5x1525^{x+1} = 5^{x-1} \cdot 5^2 to rewrite the expression.
2
Solve for the unknown variable xx.
5x1=65026=25=52    x1=2    x=35^{x-1} = \frac{650}{26} = 25 = 5^2 \implies x - 1 = 2 \implies x = 3
Since the bases are identical, equate the exponents.
3
Substitute x=3x = 3 into the expression 22x12^{2x - 1}.
22(3)1=261=25=322^{2(3) - 1} = 2^{6 - 1} = 2^5 = 32
Evaluate the power after computing the exponent value.

Anahtar Kavram

Solving Exponential Equations using Laws of Indices
Soru 211Soru

If x=7+575x = \frac{\sqrt{7} + \sqrt{5}}{\sqrt{7} - \sqrt{5}} and y=757+5y = \frac{\sqrt{7} - \sqrt{5}}{\sqrt{7} + \sqrt{5}}, what is the value of x2+y2+xyx^2 + y^2 + xy?

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Cevap: 143143

Cevap

The value of x2+y2+xyx^2 + y^2 + xy is 143143.
Rationalizing both surds gives x=6+35x = 6 + \sqrt{35} and y=635y = 6 - \sqrt{35}. The sum x+y=12x + y = 12 and product xy=1xy = 1. Substituting into the algebraic identity x2+y2+xy=(x+y)2xyx^2 + y^2 + xy = (x + y)^2 - xy gives 1221=14312^2 - 1 = 143.

Adım Adım Çözüm

1
Rationalize the denominator for xx and yy
x=(7+5)2(7)2(5)2=7+5+2352=6+35x = \frac{(\sqrt{7} + \sqrt{5})^2}{(\sqrt{7})^2 - (\sqrt{5})^2} = \frac{7 + 5 + 2\sqrt{35}}{2} = 6 + \sqrt{35}, and similarly y=635y = 6 - \sqrt{35}.
Eliminating radicals from denominators simplifies calculations.
2
Compute the sum (x+y)(x + y) and product xyxy
x+y=(6+35)+(635)=12x + y = (6 + \sqrt{35}) + (6 - \sqrt{35}) = 12, and xy=(6+35)(635)=3635=1xy = (6 + \sqrt{35})(6 - \sqrt{35}) = 36 - 35 = 1.
Using symmetric expressions simplifies evaluating degree 2 polynomials.
3
Express x2+y2+xyx^2 + y^2 + xy in terms of (x+y)(x + y) and xyxy
x2+y2+xy=(x+y)2xy=1221=1441=143x^2 + y^2 + xy = (x + y)^2 - xy = 12^2 - 1 = 144 - 1 = 143.
Applying the identity x2+y2=(x+y)22xyx^2 + y^2 = (x + y)^2 - 2xy gives x2+y2+xy=(x+y)2xyx^2 + y^2 + xy = (x + y)^2 - xy.

Anahtar Kavram

Rationalization of surds and application of algebraic identities
Soru 212Soru

If 32m+19m227m1=81\frac{3^{2m + 1} \cdot 9^{m - 2}}{27^{m - 1}} = 81, what is the value of mm?

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Cevap: 4

Cevap

The value of mm is 4.
By converting all terms in the equation to base 3 and applying index rules (axay=ax+ya^x \cdot a^y = a^{x+y} and axay=axy\frac{a^x}{a^y} = a^{x-y}), the expression simplifies directly to 3m=343^m = 3^4, leading to m=4m = 4.

Adım Adım Çözüm

1
Express all terms with a common base of 3
9m2=(32)m2=32m49^{m-2} = (3^2)^{m-2} = 3^{2m-4} and 27m1=(33)m1=33m327^{m-1} = (3^3)^{m-1} = 3^{3m-3}
Rewriting powers with a single prime base allows for simplification using the laws of indices.
2
Simplify the numerator using the product rule of exponents
32m+132m4=3(2m+1)+(2m4)=34m33^{2m+1} \cdot 3^{2m-4} = 3^{(2m+1) + (2m-4)} = 3^{4m-3}
According to the product rule axay=ax+ya^x \cdot a^y = a^{x+y}, exponents with the same base are added during multiplication.
3
Apply the quotient rule of exponents to simplify the left-hand side
34m333m3=3(4m3)(3m3)=3m\frac{3^{4m-3}}{3^{3m-3}} = 3^{(4m-3) - (3m-3)} = 3^m
According to the quotient rule axay=axy\frac{a^x}{a^y} = a^{x-y}, exponents are subtracted during division.
4
Express the right-hand side in base 3 and solve for mm
3m=81=34    m=43^m = 81 = 3^4 \implies m = 4
Since the bases on both sides of the equation are equal, the exponents must also be equal.

Anahtar Kavram

Exponential Equations and Laws of Indices
Soru 213Soru

If (35)x+1=(12527)x1\left(\sqrt{\frac{3}{5}}\right)^{x + 1} = \left(\frac{125}{27}\right)^{x - 1}, what is the value of xx?

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Cevap: 57\frac{5}{7}

Cevap

57\frac{5}{7}
By converting both sides of the equation to the common base 35\frac{3}{5}, the left side becomes (35)x+12\left(\frac{3}{5}\right)^{\frac{x+1}{2}} and the right side becomes (35)3(x1)\left(\frac{3}{5}\right)^{-3(x-1)}. Equating exponents gives x+12=33x\frac{x+1}{2} = 3 - 3x, which simplifies directly to x=57x = \frac{5}{7}.

Adım Adım Çözüm

1
Express the left side of the equation using fractional exponents
(35)x+1=((35)12)x+1=(35)x+12\left(\sqrt{\frac{3}{5}}\right)^{x+1} = \left(\left(\frac{3}{5}\right)^{\frac{1}{2}}\right)^{x+1} = \left(\frac{3}{5}\right)^{\frac{x+1}{2}}
The square root of a quantity corresponds to an exponent of 12\frac{1}{2}.
2
Express the right side with base 35\frac{3}{5}
(12527)x1=((53)3)x1=((35)3)x1=(35)3(x1)\left(\frac{125}{27}\right)^{x-1} = \left(\left(\frac{5}{3}\right)^3\right)^{x-1} = \left(\left(\frac{3}{5}\right)^{-3}\right)^{x-1} = \left(\frac{3}{5}\right)^{-3(x-1)}
Since 125=53125 = 5^3 and 27=3327 = 3^3, 12527=(53)3=(35)3\frac{125}{27} = \left(\frac{5}{3}\right)^3 = \left(\frac{3}{5}\right)^{-3}.
3
Equate the exponents since the bases are identical
x+12=3(x1)    x+12=33x\frac{x+1}{2} = -3(x-1) \implies \frac{x+1}{2} = 3 - 3x
If am=ana^m = a^n for a>0a > 0 and a1a \neq 1, then m=nm = n.
4
Solve the linear equation for xx
x+1=2(33x)    x+1=66x    7x=5    x=57x + 1 = 2(3 - 3x) \implies x + 1 = 6 - 6x \implies 7x = 5 \implies x = \frac{5}{7}
Standard algebraic simplification to isolate xx.

Anahtar Kavram

Laws of Indices and Rational Base Equivalence
Soru 214Soru

If x=743x = 7 - 4\sqrt{3}, what is the value of x+1x\sqrt{x} + \frac{1}{\sqrt{x}}?

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Cevap: 4

Cevap

The value of x+1x\sqrt{x} + \frac{1}{\sqrt{x}} is 4.
Expressing 7437 - 4\sqrt{3} as (23)2(2 - \sqrt{3})^2 allows taking the square root to get x=23\sqrt{x} = 2 - \sqrt{3}. Rationalizing its reciprocal gives 1x=2+3\frac{1}{\sqrt{x}} = 2 + \sqrt{3}. Adding these two values cancels the irrational component 3\sqrt{3}, leaving 2+2=42 + 2 = 4.

Adım Adım Çözüm

1
Simplify the nested surd x=743\sqrt{x} = \sqrt{7 - 4\sqrt{3}}
x=23\sqrt{x} = 2 - \sqrt{3}
Rewrite 7437 - 4\sqrt{3} as 22+(3)22(2)(3)=(23)22^2 + (\sqrt{3})^2 - 2(2)(\sqrt{3}) = (2 - \sqrt{3})^2 and take the principal square root.
2
Calculate the reciprocal 1x\frac{1}{\sqrt{x}} by rationalizing the denominator
1x=2+3\frac{1}{\sqrt{x}} = 2 + \sqrt{3}
Multiply numerator and denominator of 123\frac{1}{2 - \sqrt{3}} by its conjugate (2+3)(2 + \sqrt{3}).
3
Add x\sqrt{x} and 1x\frac{1}{\sqrt{x}}
4
Sum (23)+(2+3)(2 - \sqrt{3}) + (2 + \sqrt{3}) so that the radical terms cancel out.

Anahtar Kavram

Simplification of surds of the form a±b\sqrt{a \pm \sqrt{b}} and rationalization using conjugates
Soru 215Soru

What is the remainder when the value of the expression 2405×32^{40} - 5 \times 3 is divided by 1010?

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Cevap: 1

Cevap

1
The remainder when dividing by 10 is determined by the unit digit of the expression. First, evaluate the multiplication: 5×3=155 \times 3 = 15. Next, find the unit digit of 2402^{40}. The powers of 2 follow a cyclicity of 4 (2, 4, 8, 6). Since 40 is perfectly divisible by 4, the unit digit corresponds to the 4th value in the cycle, which is 6. Substituting these into the expression gives 615=96 - 15 = -9. To find the valid positive remainder modulo 10, the divisor 10 must be added to the negative result: 9+10=1-9 + 10 = 1.

Adım Adım Çözüm

1
Evaluate the multiplication part of the expression.
The term 5×35 \times 3 equals 1515. The expression simplifies to 240152^{40} - 15.
The order of operations (BODMAS) requires multiplication to be performed before subtraction.
2
Determine the unit digit (remainder modulo 10) of 2402^{40} by identifying the cyclicity of base 2.
The unit digits of powers of 2 repeat in a cycle of 4: 2, 4, 8, 6.
Since dividing by 10 is equivalent to finding the unit digit, we use power cyclicity to simplify large exponents.
3
Divide the exponent 40 by the cyclicity length 4.
40 divided by 4 leaves a remainder of 0. A remainder of 0 corresponds to the 4th position in the cycle.
The remainder determines the position in the cyclicity sequence. A remainder of 0 means the sequence has completed a full cycle.
4
Substitute the determined unit digit back into the simplified expression.
The 4th power in the cycle ends in 6. Substituting this into the expression yields 615=96 - 15 = -9.
Replacing the large power with its unit digit equivalent allows us to calculate the raw remainder.
5
Convert the negative remainder into a valid positive remainder modulo 10.
9+10=1-9 + 10 = 1. The final remainder is 1.
Remainders must be non-negative integers smaller than the divisor. Adding the divisor to a negative remainder provides the correct positive equivalent.

Anahtar Kavram

Applying cyclicity rules, the remainder theorem, and the order of operations to evaluate complex numerical expressions.
Soru 216Soru

If 2x+1+2x1=3202^{x+1} + 2^{x-1} = 320, what is the value of xx?

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Cevap: 7

Cevap

The value of xx is 7.
Factoring out 2x12^{x-1} from the expression gives 2x1(22+1)=3202^{x-1}(2^2 + 1) = 320, which simplifies to 52x1=3205 \cdot 2^{x-1} = 320. Dividing by 5 yields 2x1=64=262^{x-1} = 64 = 2^6. Equating exponents gives x1=6x - 1 = 6, so x=7x = 7.

Adım Adım Çözüm

1
Rewrite terms using exponent rules to factor out the common power 2x12^{x-1}
2x1(22+1)=3202^{x-1}(2^2 + 1) = 320
Using the product rule am+n=amana^{m+n} = a^m \cdot a^n, we have 2x+1=2x1222^{x+1} = 2^{x-1} \cdot 2^2.
2
Simplify the numerical factor inside the parentheses
52x1=3205 \cdot 2^{x-1} = 320
22+1=4+1=52^2 + 1 = 4 + 1 = 5.
3
Isolate the exponential term by dividing by 5
2x1=642^{x-1} = 64
Dividing 320320 by 55 gives 6464.
4
Write 64 as a power of 2 and solve for xx
2x1=26    x1=6    x=72^{x-1} = 2^6 \implies x - 1 = 6 \implies x = 7
Equating exponents when bases are equal.

Anahtar Kavram

Solving Exponential Equations by Factoring Common Powers
Soru 217Soru
What is the remainder when the sum 222026+392026+562026+732026+90202622^{2026} + 39^{2026} + 56^{2026} + 73^{2026} + 90^{2026} is divided by 1717?
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Cevap: 11

Cevap

11
By reducing all bases modulo 17, the sum becomes five identical terms of 520265^{2026}, which simplifies to 520275^{2027}. Using Fermat's Little Theorem (5161(mod17)5^{16} \equiv 1 \pmod{17}), the exponent 2027 is reduced modulo 16 to 11. Finally, calculating 511(mod17)5^{11} \pmod{17} yields 40-40, which corresponds to a positive remainder of 11.

Adım Adım Çözüm

1
Reduce each base in the expression modulo 17.
The expression simplifies to 52026+52026+52026+52026+52026(mod17)5^{2026} + 5^{2026} + 5^{2026} + 5^{2026} + 5^{2026} \pmod{17}.
Using the property (A(modM))k(modM)=Ak(modM)(A \pmod M)^k \pmod M = A^k \pmod M, we find that 22, 39, 56, 73, and 90 all leave a remainder of 5 when divided by 17.
2
Combine the identical terms into a single expression.
The sum becomes 5×52026=52027(mod17)5 \times 5^{2026} = 5^{2027} \pmod{17}.
Since there are exactly 5 identical terms, adding them together is equivalent to multiplying the term by 5, which conveniently increases the exponent by 1.
3
Apply Fermat's Little Theorem to reduce the large exponent.
Fermat's Little Theorem states 5161(mod17)5^{16} \equiv 1 \pmod{17}. Dividing the exponent 2027 by 16 yields 2027=16×126+112027 = 16 \times 126 + 11.
Because 17 is a prime number and does not divide 5, we can use ap11(modp)a^{p-1} \equiv 1 \pmod p to eliminate full cycles of 16 in the exponent.
4
Simplify the remaining expression 511(mod17)5^{11} \pmod{17}.
The expression reduces to 6(mod17)-6 \pmod{17}.
We break down the power: 52=2585^2 = 25 \equiv 8, 5482=6445^4 \equiv 8^2 = 64 \equiv -4, and 58(4)2=1615^8 \equiv (-4)^2 = 16 \equiv -1. Thus, 511=58×52×51(1)×8×5=405^{11} = 5^8 \times 5^2 \times 5^1 \equiv (-1) \times 8 \times 5 = -40. Since 40=3×17+11-40 = -3 \times 17 + 11, the remainder is 11 (or 6-6).
5
Convert any negative remainder to a valid positive remainder.
6+17=11-6 + 17 = 11. The final remainder is 11.
A standard remainder must be a positive integer strictly less than the divisor.

Anahtar Kavram

Fermat's Little Theorem and Modulo Arithmetic Reductions
Soru 218Soru

What is the smallest positive integer that has exactly 2121 positive divisors?

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Cevap: 576

Cevap

The smallest positive integer with exactly 21 positive divisors is 576.
The number 576 factors completely into 26×322^6 \times 3^2. According to the divisor formula, we take the exponents, add 1 to each, and multiply them: (6+1)×(2+1)=7×3=21(6+1) \times (2+1) = 7 \times 3 = 21. Because 2 is the smallest prime and is paired with the largest exponent, 576 is the absolute minimum positive integer that can produce exactly 21 divisors.

Adım Adım Çözüm

1
Use the divisor counting formula.
The number of divisors of N=pa×qbN = p^a \times q^b is (a+1)(b+1)(a+1)(b+1). We set this equal to 2121.
This establishes the relationship between prime exponents and the given divisor count.
2
Find the factor pairs of 21.
The ways to factor 21 are 2121 (as 21×121 \times 1) or 7×37 \times 3.
This determines the possible combinations of (a+1)(a+1) and (b+1)(b+1).
3
Test the first case (single prime factor).
a+1=21    a=20a+1 = 21 \implies a = 20. The smallest number is 220=1,048,5762^{20} = 1,048,576.
We must evaluate all cases to ensure we find the absolute minimum.
4
Test the second case (two prime factors).
(a+1)(b+1)=7×3    a=6,b=2(a+1)(b+1) = 7 \times 3 \implies a = 6, b = 2. The number form is p6×q2p^6 \times q^2.
This evaluates the only other valid prime exponent combination.
5
Minimize the two-prime case.
Assign p=2p=2 and q=3q=3 to get 26×32=64×9=5762^6 \times 3^2 = 64 \times 9 = 576.
Assigning the smallest available prime to the largest exponent minimizes the product.
6
Compare the results of both cases.
576<1,048,576576 < 1,048,576.
To conclusively identify the smallest positive integer.

Anahtar Kavram

Prime Factorization and the Divisor Counting Formula
Soru 219Soru

What is the positive remainder when the expression 1277267712^{77} - 26^{77} is divided by 1919?

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Cevap: 16

Cevap

The correct positive remainder is 16.
By reducing the bases modulo 19, we get 12712 \equiv -7 and 26726 \equiv 7. The expression transforms to (7)77777(-7)^{77} - 7^{77}, which simplifies to 2×777-2 \times 7^{77} since the exponent is odd. Finding the cycle of 7(mod19)7 \pmod{19} reveals that 731(mod19)7^3 \equiv 1 \pmod{19}. Because 77=3×25+277 = 3 \times 25 + 2, we have 7777211(mod19)7^{77} \equiv 7^2 \equiv 11 \pmod{19}. Finally, 2×11=22-2 \times 11 = -22, and adjusting this to a positive remainder by adding 38 (a multiple of 19) gives 16.

Adım Adım Çözüm

1
Express the bases of the exponents in terms of their remainders when divided by 19.
127(mod19)12 \equiv -7 \pmod{19} and 267(mod19)26 \equiv 7 \pmod{19}.
Converting large bases to smaller equivalent modular values (especially symmetric ones) simplifies the exponentiation.
2
Substitute the simplified bases back into the original expression.
12772677(7)77777(mod19)12^{77} - 26^{77} \equiv (-7)^{77} - 7^{77} \pmod{19}.
This reduces the problem to calculating powers of 7.
3
Simplify the expression by handling the negative sign.
Since 77 is an odd number, (7)77=(777)(-7)^{77} = -(7^{77}). The expression becomes (777)777=2×777(mod19)-(7^{77}) - 7^{77} = -2 \times 7^{77} \pmod{19}.
Factoring out 7777^{77} isolates the exponentiation part of the problem.
4
Determine the remainder of 7777^{77} divided by 19.
Using modulo operations, 71=77^1 = 7, 72=49117^2 = 49 \equiv 11, 73=7×11=771(mod19)7^3 = 7 \times 11 = 77 \equiv 1 \pmod{19}. Since 7317^3 \equiv 1, the cyclicity is 3. We divide the exponent by 3: 77=3×25+277 = 3 \times 25 + 2. Thus, 7777211(mod19)7^{77} \equiv 7^2 \equiv 11 \pmod{19}.
Finding the cycle of powers (where the remainder is 1) drastically reduces large exponents.
5
Multiply by the coefficient and convert to a positive remainder.
2×11=22-2 \times 11 = -22. To find the positive remainder, add multiples of 19: 22+19×2=22+38=16-22 + 19 \times 2 = -22 + 38 = 16.
Remainders must be positive integers strictly less than the divisor.

Anahtar Kavram

Modular Arithmetic and Fermat's Little Theorem

Alternatif Yöntem

Instead of finding the specific cyclicity of 3, Fermat's Little Theorem states that a181(mod19)a^{18} \equiv 1 \pmod{19}. Dividing 77 by 18 gives a remainder of 5. Calculating 75(mod19)7^5 \pmod{19} will also yield 11, leading to the same final answer.
Tahmini Süre:1m 30s
Soru 220Soru

If the integer generated by the mathematical expression 15+35×368715 + 35 \times 36^{87} is divided by 3737, which of the following represents the correct positive remainder?

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Cevap: 17

Cevap

17
Applying modular arithmetic properties, we can evaluate each part of the expression relative to the divisor 3737. First, the base 3636 is equivalent to 1(mod37)-1 \pmod{37}. Raising 1-1 to the odd power of 8787 results in 1-1. Substituting this back into the equation yields 15+35×(1)15 + 35 \times (-1). Following the standard order of operations, we multiply first to get 35-35, and then add 1515, resulting in an intermediate value of 20-20. Since remainders must be positive, adding the divisor (3737) to 20-20 produces the final valid remainder of 1717.

Adım Adım Çözüm

1
Simplify the base 3636 modulo 3737.
361(mod37)36 \equiv -1 \pmod{37}
Reducing the base to a smaller equivalent modulus makes the large exponentiation manageable.
2
Evaluate the exponentiated term 3687(mod37)36^{87} \pmod{37}.
(1)87=1(-1)^{87} = -1
An odd power of a negative base retains the negative sign.
3
Substitute the reduced term back into the expression and apply the correct order of operations (BODMAS/PEMDAS).
15+35×(1)=1535=2015 + 35 \times (-1) = 15 - 35 = -20
Multiplication must be resolved prior to addition.
4
Convert the intermediate negative remainder to a standard positive remainder.
20+37=17-20 + 37 = 17
A final remainder must always be a positive integer strictly less than the divisor.

Anahtar Kavram

Modular Arithmetic with Negative Bases and Operational Order
Tahmini Süre:1m 15s
ÖncekiSayfa 11 / 15Sonraki
Basic Numeracy Alıştırma Soruları — State PSC Exam — Sayfa 11 | Examkin