Tüm alıştırma soruları

2195 soru

Soru 221Soru

Set SS consists of nn consecutive integers. The sum of the smallest three integers in set SS is 33-33, and the sum of the largest three integers in set SS is 6969. What is the value of nn?

Cevabı ve açıklamayı göster

Cevap: 37

Cevap

The total number of consecutive integers in set SS is 37.
Solving 3x+3=333x + 3 = -33 yields a first term of 12-12, and solving 3y3=693y - 3 = 69 yields a last term of 2424. The total count of consecutive integers in an inclusive range is lastfirst+1\text{last} - \text{first} + 1, giving 24(12)+1=3724 - (-12) + 1 = 37.

Adım Adım Çözüm

1
Find the smallest integer in set S
The smallest integer is -12
Let the smallest integer be xx. The sum of the smallest three consecutive integers is x+(x+1)+(x+2)=3x+3x + (x + 1) + (x + 2) = 3x + 3. Setting 3x+3=333x + 3 = -33 yields 3x=363x = -36, so x=12x = -12.
2
Find the largest integer in set S
The largest integer is 24
Let the largest integer be yy. The sum of the largest three consecutive integers is (y2)+(y1)+y=3y3(y - 2) + (y - 1) + y = 3y - 3. Setting 3y3=693y - 3 = 69 yields 3y=723y = 72, so y=24y = 24.
3
Calculate the total number of elements n in set S
n = 37
For an inclusive set of consecutive integers bounded by first term aa and last term bb, the total number of terms is n=ba+1n = b - a + 1. Here, n=24(12)+1=37n = 24 - (-12) + 1 = 37.

Anahtar Kavram

Counting inclusive terms in a sequence of consecutive integers
Soru 222Soru

A set SS consists of nn consecutive integers. The sum of all the integers in set SS is 675675, and the product of the smallest integer and the largest integer in set SS is 19761{}976. What is the value of nn?

Cevabı ve açıklamayı göster

Cevap: 15

Cevap

15
For any set of nn consecutive integers, the median mm equals the arithmetic mean 675n\frac{675}{n}. The smallest and largest elements can be written as mn12m - \frac{n-1}{2} and m+n12m + \frac{n-1}{2}, respectively. Their product is m2(n12)2=1976m^2 - \left(\frac{n-1}{2}\right)^2 = 1976. Substituting n=15n = 15 gives m=45m = 45, leading to 45272=202549=197645^2 - 7^2 = 2025 - 49 = 1976, which satisfies all conditions.

Adım Adım Çözüm

1
Relate the sum of the set to its mean and number of terms
The mean (arithmetic average) of nn consecutive integers is equal to the median mm, so Sum=nm=675\text{Sum} = n \cdot m = 675, which implies m=675nm = \frac{675}{n}.
For any evenly spaced set, the sum equals the number of terms times the mean.
2
Express the smallest and largest elements in terms of the median mm and number of terms nn
The smallest element is a=mn12a = m - \frac{n-1}{2} and the largest element is b=m+n12b = m + \frac{n-1}{2}.
In a set of nn consecutive integers, the distance from the median to either endpoint is n12\frac{n-1}{2}.
3
Formulate the product equation using the difference of squares
The product of the smallest and largest elements is ab=(mn12)(m+n12)=m2(n12)2=1976a \cdot b = \left(m - \frac{n-1}{2}\right)\left(m + \frac{n-1}{2}\right) = m^2 - \left(\frac{n-1}{2}\right)^2 = 1976.
Applying the difference of squares identity (xy)(x+y)=x2y2(x-y)(x+y) = x^2 - y^2 simplifies the expression.
4
Substitute m=675nm = \frac{675}{n} and solve for nn
Substituting mm yields (675n)2(n12)2=1976\left(\frac{675}{n}\right)^2 - \left(\frac{n-1}{2}\right)^2 = 1976. Testing odd integer factors nn of 675675:
- If n=9n = 9: m=75m = 75, product =75242=562516=56091976= 75^2 - 4^2 = 5625 - 16 = 5609 \neq 1976.
- If n=15n = 15: m=45m = 45, product =45272=202549=1976= 45^2 - 7^2 = 2025 - 49 = 1976.
Hence, n=15n = 15.
Since nn must be a positive integer factor of 675675, checking candidate factors narrows down the unique solution.

Anahtar Kavram

Properties of consecutive integer sets: mean-median equivalence and difference of squares decomposition for endpoints.

Alternatif Yöntem

Let the set be {a,a+1,,a+n1}\{a, a+1, \dots, a+n-1\}. The sum is n(2a+n1)2=675    n(2a+n1)=1350\frac{n(2a + n - 1)}{2} = 675 \implies n(2a + n - 1) = 1350. The product of endpoints is a(a+n1)=1976a(a + n - 1) = 1976. Solving the system of equations for integer values of aa and nn yields a=38a = 38 and n=15n = 15.
Tahmini Süre:2m 30s
Soru 223Soru

Let kk be a positive integer with the prime factorization k=2x3y5zk = 2^x \cdot 3^y \cdot 5^z, where xx, yy, and zz are positive integers. If kk is divisible by both 1818 and 7575, and kk has exactly 3636 positive integer divisors, what is the maximum possible value of x+y+zx + y + z?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The maximum possible value of x+y+zx + y + z is 8.
The correct answer is 8. The prime factorization of kk requires x1x \ge 1, y2y \ge 2, and z2z \ge 2 because kk is a multiple of 18=213218 = 2^1 \cdot 3^2 and 75=315275 = 3^1 \cdot 5^2. The number of positive divisors is given by (x+1)(y+1)(z+1)=36(x+1)(y+1)(z+1) = 36. Under the constraints x+12x+1 \ge 2, y+13y+1 \ge 3, and z+13z+1 \ge 3, the factorizations of 36 into three factors yield the sums x+y+z=8x+y+z = 8 (from factors 2,3,62, 3, 6) and x+y+z=7x+y+z = 7 (from factors 4,3,34, 3, 3). Therefore, the maximum possible value is 8.

Adım Adım Çözüm

1
Determine the lower bounds for the exponents xx, yy, and zz based on divisibility conditions.
Since kk is divisible by 18=213218 = 2^1 \cdot 3^2, we must have x1x \ge 1 and y2y \ge 2. Since kk is divisible by 75=315275 = 3^1 \cdot 5^2, we must have y2y \ge 2 and z2z \ge 2.
For a prime factorization to be divisible by another number, each prime factor's exponent in kk must be at least as large as its corresponding exponent in the divisor.
2
Set up the equation for the total number of positive integer divisors of kk.
(x+1)(y+1)(z+1)=36(x + 1)(y + 1)(z + 1) = 36, with constraints x+12x + 1 \ge 2, y+13y + 1 \ge 3, and z+13z + 1 \ge 3.
The total number of positive divisors of 2x3y5z2^x \cdot 3^y \cdot 5^z is given by (x+1)(y+1)(z+1)(x+1)(y+1)(z+1).
3
Find all valid integer factor triples (A,B,C)=(x+1,y+1,z+1)(A, B, C) = (x+1, y+1, z+1) multiplying to 36 under the given constraints.
The valid triples (A,B,C)(A, B, C) with A2,B3,C3A \ge 2, B \ge 3, C \ge 3 are (2,3,6)(2, 3, 6), (2,6,3)(2, 6, 3), and (4,3,3)(4, 3, 3).
Testing factorizations of 3636: 236=362 \cdot 3 \cdot 6 = 36, 263=362 \cdot 6 \cdot 3 = 36, and 433=364 \cdot 3 \cdot 3 = 36 all meet the inequality bounds for each term.
4
Calculate the sum x+y+z=(A+B+C)3x + y + z = (A + B + C) - 3 for each valid triple and identify the maximum.
For (2,3,6)(2, 3, 6): x+y+z=(21)+(31)+(61)=1+2+5=8x+y+z = (2-1) + (3-1) + (6-1) = 1 + 2 + 5 = 8.
For (2,6,3)(2, 6, 3): x+y+z=(21)+(61)+(31)=1+5+2=8x+y+z = (2-1) + (6-1) + (3-1) = 1 + 5 + 2 = 8.
For (4,3,3)(4, 3, 3): x+y+z=(41)+(31)+(31)=3+2+2=7x+y+z = (4-1) + (3-1) + (3-1) = 3 + 2 + 2 = 7.
The maximum possible value is 88.
Comparing all valid scenarios yields 88 as the maximum sum of exponents.

Anahtar Kavram

Calculating total positive integer divisors from prime factorizations and analyzing exponent constraints derived from divisibility conditions.
Tahmini Süre:2m 0s
Soru 224Soru

What is the smallest positive integer nn such that nn is a multiple of 4545, nn is not divisible by 88, and nn has exactly 1818 positive integer divisors?

Cevabı ve açıklamayı göster

Cevap: 180

Cevap

180
The smallest positive integer satisfying all conditions is 180 because 180 = 2^2 * 3^2 * 5^1, which has (2+1)(2+1)(1+1) = 18 positive divisors, is a multiple of 45, and is not divisible by 8.

Adım Adım Çözüm

1
Determine the prime factorization constraints for n
n must be of the form 2^a * 3^b * 5^c * ..., where b >= 2, c >= 1, and a <= 2.
n is a multiple of 45 = 3^2 * 5^1, requiring at least 3^2 and 5^1. Since n is not divisible by 8 = 2^3, the exponent of 2 cannot exceed 2.
2
Analyze the total number of divisors constraint
(a+1)(b+1)(c+1)... = 18
The total number of positive divisors is calculated by adding 1 to each exponent in the prime factorization and multiplying them together.
3
Find the optimal prime exponent configuration to minimize n
Factor 18 as 3 * 3 * 2, corresponding to exponents 2, 2, and 1.
Using three prime factors (2, 3, and 5) with smaller exponents yields a smaller total integer than using fewer prime factors with larger exponents.
4
Assign exponents to prime bases to yield the minimum integer value
n = 2^2 * 3^2 * 5^1 = 180
Assigning exponent 2 to base 2, exponent 2 to base 3, and exponent 1 to base 5 satisfies all constraints (b = 2 >= 2, c = 1 >= 1, a = 2 < 3) and minimizes the result.

Anahtar Kavram

Divisor count formula and prime factorization constraints
Soru 225Soru

If nn is a positive integer such that the units digit of 7n7^n is 33, what is the remainder when 3n+2+8n+13^{n+2} + 8^{n+1} is divided by 55?

Cevabı ve açıklamayı göster

Cevap: 44

Cevap

The correct remainder is 44.
The cyclicity pattern of the units digit of 7n7^n repeats every 44 terms: 7,9,3,17, 9, 3, 1. For the units digit to be 33, nn must leave a remainder of 33 when divided by 44 (n3(mod4)n \equiv 3 \pmod 4). Using this form, n+251(mod4)n+2 \equiv 5 \equiv 1 \pmod 4, which means 3n+2313(mod5)3^{n+2} \equiv 3^1 \equiv 3 \pmod 5. Similarly, n+140(mod4)n+1 \equiv 4 \equiv 0 \pmod 4, and since 83(mod5)8 \equiv 3 \pmod 5, we have 8n+1301(mod5)8^{n+1} \equiv 3^0 \equiv 1 \pmod 5. Adding these values gives 3+1=43 + 1 = 4, so the remainder when divided by 55 is 44.

Adım Adım Çözüm

1
Determine the remainder of nn when divided by 44 using the units digit cyclicity of 7n7^n.
n3(mod4)n \equiv 3 \pmod 4.
The units digits of powers of 77 follow a repeating pattern of period 44: 7177^1 \rightarrow 7, 7297^2 \rightarrow 9, 7337^3 \rightarrow 3, 7417^4 \rightarrow 1. Since the units digit of 7n7^n is 33, nn must be of the form 4k+34k + 3.
2
Evaluate 3n+2(mod5)3^{n+2} \pmod 5.
3n+23(mod5)3^{n+2} \equiv 3 \pmod 5.
Substitute n=4k+3n = 4k + 3 into the exponent: n+2=4k+5=4(k+1)+1n + 2 = 4k + 5 = 4(k+1) + 1. The powers of 3(mod5)3 \pmod 5 repeat every 44 powers (313,324,332,3413^1 \equiv 3, 3^2 \equiv 4, 3^3 \equiv 2, 3^4 \equiv 1). Thus, 34k+5313(mod5)3^{4k+5} \equiv 3^1 \equiv 3 \pmod 5.
3
Evaluate 8n+1(mod5)8^{n+1} \pmod 5.
8n+11(mod5)8^{n+1} \equiv 1 \pmod 5.
First simplify the base: 83(mod5)8 \equiv 3 \pmod 5, so 8n+13n+1(mod5)8^{n+1} \equiv 3^{n+1} \pmod 5. Substitute n=4k+3n = 4k + 3: n+1=4k+4=4(k+1)n + 1 = 4k + 4 = 4(k+1). Since the exponent is a multiple of 44, 34(k+1)341(mod5)3^{4(k+1)} \equiv 3^4 \equiv 1 \pmod 5.
4
Sum the modular results to find the final remainder modulo 55.
(3+1)(mod5)=4(3 + 1) \pmod 5 = 4.
By properties of modular addition, (3n+2+8n+1)(mod5)(3+1)(mod5)=4(3^{n+2} + 8^{n+1}) \pmod 5 \equiv (3 + 1) \pmod 5 = 4.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic Exponent Rules
Tahmini Süre:2m 0s
Soru 226Soru

Let S=2202632026+4202672026S = 2^{2026} - 3^{2026} + 4^{2026} - 7^{2026}. What is the remainder when SS is divided by 1010?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The remainder when SS is divided by 1010 is 22.
Finding the remainder when an expression is divided by 10 is equivalent to finding the units digit of that expression. The units digits of powers of 2, 3, 4, and 7 repeat in periodic cycles of length 4, 4, 2, and 4, respectively. Since 20262(mod4)2026 \equiv 2 \pmod 4, the units digits correspond to the 2nd term of each cycle: 2242^2 \rightarrow 4, 3293^2 \rightarrow 9, 4264^2 \rightarrow 6, and 7297^2 \rightarrow 9. Evaluating the expression yields 49+69=84 - 9 + 6 - 9 = -8. In modular arithmetic, a negative remainder 8(mod10)-8 \pmod{10} is equivalent to 8+10=2-8 + 10 = 2. Therefore, the value representing 2 is correct.

Adım Adım Çözüm

1
Determine the remainder of each term divided by 10 by finding the units digit cyclicity.
Powers of 2 cycle with period 4 (2, 4, 8, 6). Since 2026=4×506+22026 = 4 \times 506 + 2, 22026224(mod10)2^{2026} \equiv 2^2 \equiv 4 \pmod{10}.
Dividing an integer by 10 yields a remainder equal to its units digit.
2
Evaluate the units digits for the remaining terms 320263^{2026}, 420264^{2026}, and 720267^{2026}.
Powers of 3 cycle with period 4 (3, 9, 7, 1); 32026329(mod10)3^{2026} \equiv 3^2 \equiv 9 \pmod{10}. Powers of 4 cycle with period 2 (4, 6); 42026426(mod10)4^{2026} \equiv 4^2 \equiv 6 \pmod{10}. Powers of 7 cycle with period 4 (7, 9, 3, 1); 72026729(mod10)7^{2026} \equiv 7^2 \equiv 9 \pmod{10}.
Each base follows a repeating pattern of units digits when raised to successive positive integer powers.
3
Substitute the congruent remainder values back into the expression for SS.
S49+69=8(mod10)S \equiv 4 - 9 + 6 - 9 = -8 \pmod{10}.
Modular arithmetic operations preserve addition and subtraction equivalences.
4
Convert the negative result to a standard non-negative remainder.
8+10=2-8 + 10 = 2. Thus, the remainder is 22.
By definition, the remainder rr when an integer is divided by dd must satisfy 0r<d0 \leq r < d.

Anahtar Kavram

Units Digit Cyclicity and Negative Remainder Rules
Tahmini Süre:2m 30s
Soru 227Soru

A bakery sells blueberry muffins for 3eachandchocolatechipmuffinsfor3 each and chocolate chip muffins for 4 each. On Saturday, the bakery sold a total of 50 muffins for $170. How many chocolate chip muffins were sold on Saturday?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The bakery sold 20 chocolate chip muffins on Saturday.
Let bb represent the number of blueberry muffins and cc represent the number of chocolate chip muffins sold. Based on the problem text, we construct the system of equations: b+c=50b + c = 50 and 3b+4c=1703b + 4c = 170. From the first equation, b=50cb = 50 - c. Substituting this into the second equation yields 3(50c)+4c=1703(50 - c) + 4c = 170, which expands to 1503c+4c=170150 - 3c + 4c = 170. Simplifying gives 150+c=170150 + c = 170, so c=20c = 20.

Adım Adım Çözüm

1
Define variables and set up the system of linear equations.
b+c=50b + c = 50 and 3b+4c=1703b + 4c = 170
The total number of muffins sold is 50, and the total revenue from selling blueberry muffins at 3eachandchocolatechipmuffinsat3 each and chocolate chip muffins at 4 each is $170.
2
Substitute b=50cb = 50 - c into the cost equation.
3(50c)+4c=170    150+c=1703(50 - c) + 4c = 170 \implies 150 + c = 170
Substituting bb in terms of cc creates a single-variable linear equation for cc.
3
Solve for cc.
c=20c = 20
Subtracting 150 from both sides gives the exact number of chocolate chip muffins sold.

Anahtar Kavram

Solving Systems of Linear Equations by Substitution or Elimination
Soru 228Soru

What is the sum of all real solutions to the absolute value equation 2x3+x+5=12x|2x - 3| + |x + 5| = 12 - x?

Cevabı ve açıklamayı göster

Cevap: 92-\frac{9}{2}

Cevap

The sum of all real solutions to the equation is 92-\frac{9}{2}.
By splitting the real number line into three intervals based on the critical points x=5x = -5 and x=32x = \frac{3}{2}, we find two valid solutions: x=52x = \frac{5}{2} (from x32x \ge \frac{3}{2}) and x=7x = -7 (from x<5x < -5). Their sum is 52+(7)=92\frac{5}{2} + (-7) = -\frac{9}{2}.

Adım Adım Çözüm

1
Identify critical points for the absolute value expressions
The critical points are x=32x = \frac{3}{2} and x=5x = -5.
The expressions inside the absolute values, 2x32x - 3 and x+5x + 5, change signs at x=32x = \frac{3}{2} and x=5x = -5 respectively.
2
Evaluate Region 1 (x32x \ge \frac{3}{2})
(2x3)+(x+5)=12x    3x+2=12x    4x=10    x=52(2x - 3) + (x + 5) = 12 - x \implies 3x + 2 = 12 - x \implies 4x = 10 \implies x = \frac{5}{2}.
In this region, both 2x302x - 3 \ge 0 and x+5>0x + 5 > 0, so absolute value bars can be removed directly. Since 5232\frac{5}{2} \ge \frac{3}{2}, x=52x = \frac{5}{2} is a valid solution.
3
Evaluate Region 2 (5x<32-5 \le x < \frac{3}{2})
(2x3)+(x+5)=12x    x+8=12x    8=12-(2x - 3) + (x + 5) = 12 - x \implies -x + 8 = 12 - x \implies 8 = 12 (No solution).
In this interval, 2x3<02x - 3 < 0 while x+50x + 5 \ge 0. The resulting equation produces a contradiction, so there are no solutions in this interval.
4
Evaluate Region 3 (x<5x < -5)
(2x3)(x+5)=12x    3x2=12x    2x=14    x=7-(2x - 3) - (x + 5) = 12 - x \implies -3x - 2 = 12 - x \implies -2x = 14 \implies x = -7.
In this region, both 2x3<02x - 3 < 0 and x+5<0x + 5 < 0. Since 7<5-7 < -5, x=7x = -7 is a valid solution.
5
Calculate the sum of all valid solutions
52+(7)=52142=92\frac{5}{2} + (-7) = \frac{5}{2} - \frac{14}{2} = -\frac{9}{2}.
Combining the valid roots from Region 1 and Region 3 gives the final requested sum.

Anahtar Kavram

Solving piecewise linear equations involving multiple absolute value terms.
Tahmini Süre:2m 30s
Soru 229Soru

Which of the following is equivalent to 810+41584+46\sqrt{\frac{8^{10} + 4^{15}}{8^4 + 4^6}}?

Cevabı ve açıklamayı göster

Cevap: 292^9

Cevap

The expression is equivalent to 292^9.
By converting all terms to base 2, 810+415=(23)10+(22)15=230+230=2318^{10} + 4^{15} = (2^3)^{10} + (2^2)^{15} = 2^{30} + 2^{30} = 2^{31} in the numerator, and 84+46=(23)4+(22)6=212+212=2138^4 + 4^6 = (2^3)^4 + (2^2)^6 = 2^{12} + 2^{12} = 2^{13} in the denominator. The expression simplifies to 231213=218=29\sqrt{\frac{2^{31}}{2^{13}}} = \sqrt{2^{18}} = 2^9.

Adım Adım Çözüm

1
Convert all terms in the numerator and denominator to a common base of 2.
Numerator: 810+415=(23)10+(22)15=230+2308^{10} + 4^{15} = (2^3)^{10} + (2^2)^{15} = 2^{30} + 2^{30}. Denominator: 84+46=(23)4+(22)6=212+2128^4 + 4^6 = (2^3)^4 + (2^2)^6 = 2^{12} + 2^{12}.
Expressing terms with prime bases enables simplification using exponent rules.
2
Factor out common terms to simplify addition in numerator and denominator.
Numerator: 230+230=2230=2312^{30} + 2^{30} = 2 \cdot 2^{30} = 2^{31}. Denominator: 212+212=2212=2132^{12} + 2^{12} = 2 \cdot 2^{12} = 2^{13}.
Adding two equal quantities x+xx + x equals 2x2x, increasing the power of 2 by 1.
3
Simplify the fraction inside the square root.
\frac{2^{31}}{2^{13}} = 2^{31 - 13} = 2^{18}.
Apply the quotient rule of exponents: aman=amn\frac{a^m}{a^n} = a^{m-n}.
4
Evaluate the radical expression.
\sqrt{2^{18}} = (2^{18})^{1/2} = 2^{18/2} = 2^9.
Taking the square root of a power halves its exponent.

Anahtar Kavram

Exponents, Roots, and Powers of Integers
Tahmini Süre:2m 0s
Soru 230Soru

An event planner ordered a total of 3030 gift baskets for a corporate conference. Standard gift baskets cost $25\$25 each, and Deluxe gift baskets cost $40\$40 each. If the total cost of all 3030 gift baskets was $900\$900, how many Deluxe gift baskets were ordered?

Cevabı ve açıklamayı göster

Cevap: 1010

Cevap

The correct number of Deluxe gift baskets ordered is 1010.
By setting up the linear system x+y=30x + y = 30 and 25x+40y=90025x + 40y = 900, substituting x=30yx = 30 - y into the cost equation yields 750+15y=900750 + 15y = 900, which solves directly to y=10y = 10. Therefore, 1010 Deluxe baskets were ordered.

Adım Adım Çözüm

1
Define variables and set up the system of linear equations
Let xx be the number of Standard baskets and yy be the number of Deluxe baskets. Equation 1: x+y=30x + y = 30. Equation 2: 25x+40y=90025x + 40y = 900.
The total number of baskets establishes a quantity relation, while the individual prices establish a cost relation.
2
Express xx in terms of yy using Equation 1
x=30yx = 30 - y
Isolating xx allows direct substitution into the total cost equation to solve for yy.
3
Substitute x=30yx = 30 - y into Equation 2 and solve for yy
25(30y)+40y=900    75025y+40y=900    15y=150    y=1025(30 - y) + 40y = 900 \implies 750 - 25y + 40y = 900 \implies 15y = 150 \implies y = 10.
Simplifying the single-variable equation yields the exact number of Deluxe gift baskets.

Anahtar Kavram

Solving Systems of Linear Equations using Substitution or Elimination

Alternatif Yöntem

Use the elimination method: multiply the total quantity equation by 2525 (25x+25y=75025x + 25y = 750) and subtract it from the total cost equation (25x+40y=90025x + 40y = 900) to get 15y=15015y = 150, leading directly to y=10y = 10.
Tahmini Süre:45s
Soru 231Soru

An integer sequence is defined by Tn=7n(2)nT_n = 7^n - (-2)^n for all positive integers nn. What is the remainder when T40T_{40} is divided by 55?

Cevabı ve açıklamayı göster

Cevap: 0

Cevap

The remainder when T40T_{40} is divided by 55 is 00.
Modulo 55, 727 \equiv 2, so 7402407^{40} \equiv 2^{40}. Since 4040 is an even exponent, (2)40=240(-2)^{40} = 2^{40}. Substituting these into the formula yields T40240240=0(mod5)T_{40} \equiv 2^{40} - 2^{40} = 0 \pmod 5. Thus, the remainder is 00.

Adım Adım Çözüm

1
Reduce the base 77 modulo 55
72(mod5)7 \equiv 2 \pmod 5, so 740240(mod5)7^{40} \equiv 2^{40} \pmod 5.
Simplifying the base makes modular exponentiation straightforward.
2
Evaluate the negative base term (2)40(-2)^{40}
Since 4040 is an even integer, (2)40=240(-2)^{40} = 2^{40}.
An even power of a negative number yields a positive result.
3
Compute T40T_{40} modulo 55
T40=740(2)40240240=0(mod5)T_{40} = 7^{40} - (-2)^{40} \equiv 2^{40} - 2^{40} = 0 \pmod 5.
Subtracting identical values yields 00.

Anahtar Kavram

Modular Arithmetic and Exponent Parity
Tahmini Süre:1m 30s
Soru 232Soru

When the positive integer nn is divided by 1212, the remainder is 77. What is the units digit of 9n+4n+17n+29^n + 4^{n+1} - 7^{n+2}?

Cevabı ve açıklamayı göster

Cevap: 8

Cevap

The units digit of the expression is 8.
The correct answer is 8 because evaluating each component using unit digit cyclicity gives 9n9(mod10)9^n ≡ 9 \pmod{10} (since nn is odd), 4n+16(mod10)4^{n+1} ≡ 6 \pmod{10} (since n+1n+1 is even), and 7n+27(mod10)7^{n+2} ≡ 7 \pmod{10} (since n+21(mod4)n+2 ≡ 1 \pmod 4). Combining these yields (9+67)=8(9 + 6 - 7) = 8.

Adım Adım Çözüm

1
Express nn using division algorithm and determine its properties.
Since n=12k+7n = 12k + 7 for some non-negative integer kk, nn is odd, n+1n+1 is even, and n+2=12k+9n+2 = 12k + 9.
Establishing the form of nn determines the exponents for cyclicity calculations.
2
Find the units digit of 9n9^n.
Units digit of 9n9^n is 9.
Powers of 9 alternate units digits: 91=9,92=1,93=9...9^1 = 9, 9^2 = 1, 9^3 = 9... Any odd power of 9 ends in 9. Since n=12k+7n = 12k+7 is odd, 9n9^n ends in 9.
3
Find the units digit of 4n+14^{n+1}.
Units digit of 4n+14^{n+1} is 6.
Powers of 4 alternate units digits: 41=4,42=6,43=4...4^1 = 4, 4^2 = 6, 4^3 = 4... Any even power of 4 ends in 6. Since nn is odd, n+1n+1 is even, so 4n+14^{n+1} ends in 6.
4
Find the units digit of 7n+27^{n+2}.
Units digit of 7n+27^{n+2} is 7.
Powers of 7 follow a 4-step cyclicity pattern: 7, 9, 3, 1. The exponent n+2=12k+9=4(3k+2)+11(mod4)n+2 = 12k + 9 = 4(3k+2) + 1 ≡ 1 \pmod 4. Thus, 7n+27^{n+2} has the same units digit as 717^1, which is 7.
5
Combine the units digits.
Units digit = 9+67=89 + 6 - 7 = 8.
Adding and subtracting the respective units digits gives 157=815 - 7 = 8.

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic
Tahmini Süre:2m 0s
Soru 233Soru

A courier service calculates its total delivery charge using a fixed base fee plus a constant per-mile rate. On Monday, a delivery of 15 miles received a 20% discount on the fixed base fee and a 25% surcharge on the per-mile rate, resulting in a total charge of 30.50.OnTuesday,adeliveryof20milesincurreda4030.50. On Tuesday, a delivery of 20 miles incurred a 40% increase on the fixed base fee and received a 15% discount on the per-mile rate, resulting in a total charge of 34.40. What is the standard total delivery charge, in dollars, for a 25-mile delivery with no fee adjustments or rate changes?

Cevabı ve açıklamayı göster

Cevap: 40

Cevap

The standard total delivery charge for a 25-mile delivery is 40 dollars.
Translating the scenario into linear equations gives 0.80B+18.75r=30.500.80B + 18.75r = 30.50 and 1.40B+17.00r=34.401.40B + 17.00r = 34.40. Solving this linear system yields a standard base fee of B=10.00B = 10.00 dollars and a standard per-mile rate of r=1.20r = 1.20 dollars. Substituting these into the standard 25-mile cost expression B+25rB + 25r yields 10.00+25(1.20)=40.0010.00 + 25(1.20) = 40.00 dollars.

Adım Adım Çözüm

1
Formulate linear equations from the word problem context.
System of equations: 0.80B+18.75r=30.500.80B + 18.75r = 30.50 and 1.40B+17.00r=34.401.40B + 17.00r = 34.40.
Applying the percentage adjustments to the fixed base fee BB and the rate per mile rr for the given distances yields exact linear expressions.
2
Eliminate variable BB to solve for rr.
Multiply equations to equate coefficients of BB: 5.60B+131.25r=213.505.60B + 131.25r = 213.50 and 5.60B+68.00r=137.605.60B + 68.00r = 137.60. Subtracting gives 63.25r=75.9063.25r = 75.90, so r=1.20r = 1.20.
Finding the per-mile rate rr allows determination of the standard mileage component.
3
Solve for base fee BB using r=1.20r = 1.20.
1.40B+17(1.20)=34.40    1.40B=14.00    B=10.001.40B + 17(1.20) = 34.40 \implies 1.40B = 14.00 \implies B = 10.00.
Substituting rr into either linear equation gives the fixed base fee.
4
Calculate the target standard cost for 25 miles.
B+25r=10.00+25(1.20)=40.00B + 25r = 10.00 + 25(1.20) = 40.00.
Evaluating the standard pricing expression B+25rB + 25r with B=10B = 10 and r=1.20r = 1.20 gives the total cost.

Anahtar Kavram

Solving Systems of Two-Variable Linear Equations from Word Problems
Soru 234Soru

What is the value of the positive integer nn if nn is a multiple of 18 and nn has exactly 9 positive integer divisors?

Cevabı ve açıklamayı göster

Cevap: 36

Cevap

36
The prime factorization of 18 is 21322^1 \cdot 3^2. Any multiple nn of 18 must take the form n=2a3bn = 2^a \cdot 3^b \dots where a1a \ge 1 and b2b \ge 2. The total number of positive divisors of nn is given by (a+1)(b+1)=9(a+1)(b+1)\dots = 9. Given that a+12a+1 \ge 2 and b+13b+1 \ge 3, the only product of integers equal to 9 is 3×33 \times 3. This requires a+1=3    a=2a+1 = 3 \implies a = 2 and b+1=3    b=2b+1 = 3 \implies b = 2, with no additional prime factors present. Therefore, n=2232=36n = 2^2 \cdot 3^2 = 36.

Adım Adım Çözüm

1
Determine the prime factorization constraints imposed by 18.
18=213218 = 2^1 \cdot 3^2, so n=2a3bn = 2^a \cdot 3^b \dots with a1a \ge 1 and b2b \ge 2.
Any multiple of 18 must contain at least one factor of 2 and two factors of 3.
2
Apply the divisor count formula to set up an equation.
(a+1)(b+1)=9(a+1)(b+1) = 9
The number of positive divisors of 2a3b2^a \cdot 3^b is given by (a+1)(b+1)(a+1)(b+1).
3
Solve for the exponents aa and bb.
a=2a = 2 and b=2b = 2
Because 9 can only be factored as 3×33 \times 3 for integer components where a+12a+1 \ge 2 and b+13b+1 \ge 3, both a+1a+1 and b+1b+1 must equal 3.
4
Compute the value of nn.
n=2232=36n = 2^2 \cdot 3^2 = 36
Multiply the prime power factors together to find the value of nn.

Anahtar Kavram

The total number of positive divisors of a positive integer n=p1e1p2e2pkekn = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k} is given by (e1+1)(e2+1)(ek+1)(e_1 + 1)(e_2 + 1) \dots (e_k + 1).
Soru 235Soru

If mm and nn are positive integers such that 2m2n=19202^m - 2^n = 1920, what is the value of m+nm + n?

Cevabı ve açıklamayı göster

Cevap: 18

Cevap

18
To solve 2m2n=19202^m - 2^n = 1920, factor out 2n2^n to express the left side as 2n(2mn1)2^n(2^{m-n} - 1). Prime factorizing 1920 gives 27×152^7 \times 15. Since mm and nn are positive integers with m>nm > n, the term (2mn1)(2^{m-n} - 1) is an odd integer. Therefore, the power-of-2 term 2n2^n must equal 272^7, which implies n=7n = 7. The odd term (2mn1)(2^{m-n} - 1) must equal 1515, leading to 2mn=16=242^{m-n} = 16 = 2^4, so mn=4m - n = 4. Solving for mm gives m=11m = 11. Finally, m+n=11+7=18m + n = 11 + 7 = 18.

Adım Adım Çözüm

1
Factor the exponential expression
2n(2mn1)=19202^n(2^{m-n} - 1) = 1920
Factoring out the smaller power of 2 separates the expression into a power of 2 and an odd integer multiplier.
2
Determine the prime factorization of 1920
1920=27×151920 = 2^7 \times 15
Prime factorization isolates the highest power of 2 (272^7) from the remaining odd factor (1515).
3
Equate corresponding power-of-2 and odd factors
n=7n = 7 and mn=4m - n = 4
The even component 2n2^n must equal 272^7, giving n=7n = 7. The odd component 2mn12^{m-n} - 1 must equal 1515, so 2mn=16=242^{m-n} = 16 = 2^4, giving mn=4m - n = 4.
4
Solve for mm and compute m+nm + n
m=11m = 11 and m+n=18m + n = 18
Adding n=7n = 7 to mn=4m - n = 4 yields m=11m = 11. The requested sum is m+n=11+7=18m + n = 11 + 7 = 18.

Anahtar Kavram

Factoring difference of powers using fundamental exponent rules and equating even/odd prime components.
Soru 236Soru

What is the smallest positive integer nn such that n!n! is divisible by 101010^{10}?

Cevabı ve açıklamayı göster

Cevap: 45

Cevap

The smallest positive integer nn such that n!n! is divisible by 101010^{10} is 45.
To find the smallest integer nn such that n!n! is divisible by 101010^{10}, we need n!n! to contain at least 10 prime factors of 5 (since 2s are abundant). Using Legendre's formula, E5(40!)=40/5+40/25=8+1=9E_5(40!) = \lfloor 40/5 \rfloor + \lfloor 40/25 \rfloor = 8 + 1 = 9, which is insufficient. For n=45n = 45, E5(45!)=45/5+45/25=9+1=10E_5(45!) = \lfloor 45/5 \rfloor + \lfloor 45/25 \rfloor = 9 + 1 = 10, satisfying the requirement. Thus, 45 is the smallest positive integer.

Adım Adım Çözüm

1
Determine the prime factorization requirement for divisibility by 101010^{10}.
1010=210×51010^{10} = 2^{10} \times 5^{10}. Thus, n!n! must contain at least 10 factors of 5.
The power of 2 in any factorial n!n! (where n5n \ge 5) is always strictly greater than the power of 5, making 5 the limiting prime factor.
2
Apply Legendre's formula for the exponent of prime p=5p = 5 in n!n!.
E5(n!)=n5+n25+10E_5(n!) = \lfloor \frac{n}{5} \rfloor + \lfloor \frac{n}{25} \rfloor + \dots \ge 10.
Legendre's formula accounts for single multiples of 5, double multiples of 5 (25), etc.
3
Evaluate candidate values for nn.
For n=40n = 40, E5(40!)=8+1=9E_5(40!) = 8 + 1 = 9 factors. For n=45n = 45, E5(45!)=9+1=10E_5(45!) = 9 + 1 = 10 factors.
Testing multiples of 5 systematically pinpoints the exact boundary where the total count of prime factor 5 reaches 10.

Anahtar Kavram

Finding the exponent of a prime factor in a factorial using Legendre's formula
Tahmini Süre:2m 0s
Soru 237Soru

A set SS consists of nn consecutive positive integers, where n>1n > 1 is an odd integer. If the sum of all elements in set SS is equal to 3103^{10}, what is the minimum possible value of the median of set SS?

Cevabı ve açıklamayı göster

Cevap: 243

Cevap

The minimum possible value of the median of set SS is 243.
For an odd number nn of consecutive integers, the sum of the set equals n×mn \times m, where mm is the median. Given that n×m=310n \times m = 3^{10}, both nn and mm must be powers of 3, so n=3kn = 3^k and m=310km = 3^{10-k}. To ensure all terms in the set are positive, the smallest term mn12m - \frac{n-1}{2} must be at least 1, which requires 2m>n2m > n. Substituting the powers of 3 yields 2310k>3k    32k<2310=118,0982 \cdot 3^{10-k} > 3^k \implies 3^{2k} < 2 \cdot 3^{10} = 118,098. The largest integer kk satisfying this condition is k=5k = 5 (since 310=59,049<118,0983^{10} = 59,049 < 118,098 while 312=531,441>118,0983^{12} = 531,441 > 118,098). Maximizing kk minimizes the median m=3105=35=243m = 3^{10-5} = 3^5 = 243.

Adım Adım Çözüm

1
Relate the sum of an evenly spaced set to its number of terms and median.
n×m=310n \times m = 3^{10}, where nn is the number of terms and mm is the median.
For any set of nn consecutive integers where nn is odd, the sum of the set equals the number of terms times the middle term (median).
2
Express nn and mm as powers of 3.
n=3kn = 3^k and m=310km = 3^{10-k} where k1k \ge 1.
Because 3103^{10} has only 3 as a prime factor, any integer factors nn and mm must be powers of 3.
3
Establish the positivity constraint for the terms in set SS.
2m>n2m > n
The smallest term in the set is mn12m - \frac{n-1}{2}. Requiring mn121m - \frac{n-1}{2} \ge 1 gives 2mn+12m \ge n + 1, or strictly 2m>n2m > n.
4
Solve the inequality 2m>n2m > n in terms of kk.
32k<2310=118,0983^{2k} < 2 \cdot 3^{10} = 118,098
Substituting n=3kn = 3^k and m=310km = 3^{10-k} yields 2310k>3k2 \cdot 3^{10-k} > 3^k, which rearranges to 32k<23103^{2k} < 2 \cdot 3^{10}.
5
Find the maximum valid integer value of kk and calculate the corresponding minimum median mm.
Maximum k=5k = 5, giving minimum median m=35=243m = 3^{5} = 243.
For k=5k = 5, 310=59,049<118,0983^{10} = 59,049 < 118,098. For k=6k = 6, 312=531,441>118,0983^{12} = 531,441 > 118,098. Thus k=5k=5 is the maximum integer kk, which yields the minimum median m=3105=243m = 3^{10-5} = 243.

Anahtar Kavram

Sum of consecutive integers set formula and positivity constraints
Soru 238Soru

When a positive integer nn is divided by 1212, the remainder is 77. What is the remainder when n2+5n+11n^2 + 5n + 11 is divided by 1212?

Cevabı ve açıklamayı göster

Cevap: 11

Cevap

11
Since nn leaves a remainder of 77 when divided by 1212, we can substitute n7(mod12)n \equiv 7 \pmod{12} directly into n2+5n+11n^2 + 5n + 11. Evaluating 72+5(7)+117^2 + 5(7) + 11 gives 49+35+11=9549 + 35 + 11 = 95. Dividing 9595 by 1212 gives 77 with a remainder of 1111.

Adım Adım Çözüm

1
Express nn using modular arithmetic
n7(mod12)n \equiv 7 \pmod{12}
A positive integer nn that leaves a remainder of 77 when divided by 1212 can be expressed as n=12k+7n = 12k + 7 for some non-negative integer kk.
2
Substitute n7(mod12)n \equiv 7 \pmod{12} into the target expression
n2+5n+1172+5(7)+11(mod12)n^2 + 5n + 11 \equiv 7^2 + 5(7) + 11 \pmod{12}
By the algebraic properties of remainders (modular arithmetic), substituting the remainder 77 for nn yields an equivalent remainder modulo 1212.
3
Evaluate the arithmetic sum
49+35+11=9549 + 35 + 11 = 95
Computing 72=497^2 = 49, 5×7=355 \times 7 = 35, and adding 1111 gives 9595.
4
Find the remainder of 9595 modulo 1212
95=12×7+11    9511(mod12)95 = 12 \times 7 + 11 \implies 95 \equiv 11 \pmod{12}
Dividing 9595 by 1212 yields a quotient of 77 and a remainder of 1111.

Anahtar Kavram

Modular Arithmetic and Polynomial Remainders
Soru 239Soru

If 3x+2y=183x + 2y = 18 and x+4y=16x + 4y = 16, what is the value of x+yx + y?

Cevabı ve açıklamayı göster

Cevap: 7

Cevap

7
Solving the system of linear equations 3x+2y=183x + 2y = 18 and x+4y=16x + 4y = 16 yields x=4x = 4 and y=3y = 3. Adding these values together produces 4+3=74 + 3 = 7.

Adım Adım Çözüm

1
Express xx in terms of yy using the second equation
x=164yx = 16 - 4y
Isolating one variable allows straightforward substitution into the other equation.
2
Substitute x=164yx = 16 - 4y into the first equation
3(164y)+2y=18    4812y+2y=18    4810y=183(16 - 4y) + 2y = 18 \implies 48 - 12y + 2y = 18 \implies 48 - 10y = 18
This creates an equation containing only the variable yy.
3
Solve for yy
10y=1848=30    y=3-10y = 18 - 48 = -30 \implies y = 3
Isolating the numerical term isolates the value of yy.
4
Substitute y=3y = 3 back to find xx
x=164(3)=1612=4x = 16 - 4(3) = 16 - 12 = 4
Determines the specific value of xx.
5
Calculate the requested sum x+yx + y
x+y=4+3=7x + y = 4 + 3 = 7
Provides the final requested expression value.

Anahtar Kavram

Solving a system of linear equations by substitution or elimination to evaluate a combined linear expression.
Tahmini Süre:45s
Soru 240Soru

If nn is a positive integer greater than 2020 such that 228+220+2n\sqrt{2^{28} + 2^{20} + 2^n} is an integer, what is the least possible value of nn?

Cevabı ve açıklamayı göster

Cevap: 25

Cevap

The least possible value of nn is 25.
The expression inside the radical must be a perfect square. Matching 228+220+2n2^{28} + 2^{20} + 2^n to (214+210)2=228+2214210+220=228+225+220(2^{14} + 2^{10})^2 = 2^{28} + 2 \cdot 2^{14} \cdot 2^{10} + 2^{20} = 2^{28} + 2^{25} + 2^{20} reveals that 2n=2252^n = 2^{25}, so n=25n = 25. Since 25 is greater than 20 and smaller than the other valid solution n=34n = 34, it is the least possible value.

Adım Adım Çözüm

1
Set up the perfect square structure for the radical expression.
For 228+220+2n\sqrt{2^{28} + 2^{20} + 2^n} to be an integer, the expression 228+220+2n2^{28} + 2^{20} + 2^n must equal (2a+2b)2=22a+2a+b+1+22b(2^a + 2^b)^2 = 2^{2a} + 2^{a+b+1} + 2^{2b} for some positive integers a>ba > b.
Expanding a binomial power of 2 generates three power-of-2 terms matching the three terms in the expression.
2
Analyze Case 1 where 2282^{28} corresponds to the leading term 22a2^{2a}.
2a=28    a=142a = 28 \implies a = 14. The remaining terms 2202^{20} and 2n2^n must correspond to 22b2^{2b} and 2a+b+12^{a+b+1}.
Matching highest powers establishes the value of the parameter aa.
3
Evaluate sub-cases for matching 2202^{20}.
Subcase 1: If 2b=20    b=102b = 20 \implies b = 10, then the middle term exponent is a+b+1=14+10+1=25a + b + 1 = 14 + 10 + 1 = 25. Thus n=25n = 25.
Subcase 2: If a+b+1=20    14+b+1=20    b=5a + b + 1 = 20 \implies 14 + b + 1 = 20 \implies b = 5, then 2b=102b = 10. Thus n=10n = 10.
Checking both term assignments determines possible integer values for nn.
4
Analyze Case 2 where 2202^{20} corresponds to 22b2^{2b} and 2282^{28} is the middle term.
If 2b=20    b=102b = 20 \implies b = 10 and a+b+1=28    a+11=28    a=17a + b + 1 = 28 \implies a + 11 = 28 \implies a = 17, then 2a=342a = 34. Thus n=34n = 34.
Checking alternative middle term assignments finds all valid values for nn.
5
Filter by constraint n>20n > 20 and select the minimum.
The valid values satisfying n>20n > 20 are n=25n = 25 and n=34n = 34. The least possible value is 25.
The question asks specifically for the least possible value greater than 20.

Anahtar Kavram

Perfect Square Trinomial Expansion with Exponent Rules
Tahmini Süre:2m 0s
ÖncekiSayfa 12 / 110Sonraki
Tüm alıştırma soruları — GMAT | Examkin