Arithmetic

306 soru

Soru 41Soru

Let mm and nn be negative integers such that m<nm < n and mnm - n is an odd integer. Which of the following expressions must be a positive even integer?

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Cevap: (mn)2+1(m - n)^2 + 1

Cevap

(mn)2+1(m - n)^2 + 1 must be a positive even integer.
Because mnm - n is given as an odd integer and m<nm < n, mnm - n is a negative odd integer. Squaring any negative odd integer gives a positive odd integer. Adding 11 to a positive odd integer yields a positive even integer, which guarantees the result is always positive and even.

Adım Adım Çözüm

1
Analyze the parity and sign of mnm - n
mnm - n is a negative odd integer.
Since m<nm < n, mn<0m - n < 0. The problem specifies that mnm - n is odd.
2
Evaluate the expression (mn)2(m - n)^2
(mn)2(m - n)^2 is a positive odd integer.
Squaring any non-zero real number yields a positive result. Squaring an odd integer always yields an odd integer.
3
Add 1 to (mn)2(m - n)^2
(mn)2+1(m - n)^2 + 1 is an even integer greater than or equal to 2.
Adding 1 to a positive odd integer converts it into a positive even integer.

Anahtar Kavram

Parity rules under arithmetic operations and exponents with signed integers
Tahmini Süre:1m 15s
Soru 42Soru

Let xx and yy be non-zero integers that satisfy all of the following conditions:
I. (x)y+1<0(-x)^{y + 1} < 0
II. x2y+xx^2 y + x is odd
III. x<yx < y

Which of the following statements MUST be true? Select all that apply.

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Cevap: yxy^x is an even integer; xyyx^y - y is an odd integer; (x+y)2(x + y)^2 is an odd integer

Cevap

The statements that must be true are 'yxy^x is an even integer', 'xyyx^y - y is an odd integer', and '(x+y)2(x + y)^2 is an odd integer'.
From the given conditions, we deduce that xx is a positive odd integer and yy is a positive even integer with x<yx < y. Therefore:
- Raising the positive even integer yy to the positive power xx yields an even integer (yxy^x is even).
- Raising the odd integer xx to the positive power yy yields an odd integer, and subtracting the even integer yy leaves an odd integer (xyyx^y - y is odd).
- The sum of odd xx and even yy is odd, and squaring an odd integer yields an odd integer ((x+y)2(x + y)^2 is odd).

Adım Adım Çözüm

1
Analyze Condition I: (x)y+1<0(-x)^{y + 1} < 0
x>0x > 0 (a positive integer) and yy is an even integer.
For a number raised to an integer power to be strictly negative, the base must be negative and the exponent must be odd. Hence, x<0    x>0-x < 0 \implies x > 0, and y+1y + 1 is odd     y\implies y is even.
2
Analyze Condition II: x2y+xx^2 y + x is odd
xx is an odd integer.
Factor the expression as x(xy+1)x(xy + 1). A product of two integers is odd if and only if both factors are odd. Thus, xx must be odd. (Additionally, xy+1xy + 1 must be odd     xy\implies xy is even, which holds since yy is even).
3
Analyze Condition III: x<yx < y
yy is a positive even integer.
Since xx is a positive odd integer (x1x \ge 1) and x<yx < y, yy must also be a positive integer (y2y \ge 2).
4
Evaluate the statements based on the derived properties (xx is positive odd, yy is positive even, x<yx < y)
The statements 'yxy^x is an even integer', 'xyyx^y - y is an odd integer', and '(x+y)2(x + y)^2 is an odd integer' are guaranteed to be true.
1) evenpositive odd=even\text{even}^{\text{positive odd}} = \text{even}. 2) oddpositive eveneven=oddeven=odd\text{odd}^{\text{positive even}} - \text{even} = \text{odd} - \text{even} = \text{odd}. 3) (odd+even)2=odd2=odd(\text{odd} + \text{even})^2 = \text{odd}^2 = \text{odd}.

Anahtar Kavram

Deducing parity and signs of variables using exponent rules and arithmetic properties of even and odd numbers.
Tahmini Süre:2m 0s
Soru 43Soru

If xx is a real number such that 2x5=x+4|2x - 5| = |x + 4|, which of the following could be the value of x2x^2? Select all such values.

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Cevap: 19\frac{1}{9}; 8181

Cevap

The possible values of x2x^2 are 19\frac{1}{9} and 8181.
To solve 2x5=x+4|2x - 5| = |x + 4|, set 2x52x - 5 equal to both x+4x + 4 and (x+4)-(x + 4). Solving 2x5=x+42x - 5 = x + 4 gives x=9x = 9, which leads to x2=81x^2 = 81. Solving 2x5=x42x - 5 = -x - 4 gives 3x=13x = 1, so x=13x = \frac{1}{3}, which leads to x2=19x^2 = \frac{1}{9}. Therefore, the options representing 19\frac{1}{9} and 8181 are correct.

Adım Adım Çözüm

1
Set up equations to remove the absolute value bars
Two linear equations: 2x5=x+42x - 5 = x + 4 or 2x5=(x+4)2x - 5 = -(x + 4).
For real numbers AA and BB, A=B|A| = |B| implies A=BA = B or A=BA = -B.
2
Solve the first equation 2x5=x+42x - 5 = x + 4
x=9x = 9.
Subtract xx and add 55 to both sides.
3
Solve the second equation 2x5=x42x - 5 = -x - 4
3x=1    x=133x = 1 \implies x = \frac{1}{3}.
Add xx and add 55 to both sides.
4
Compute x2x^2 for each possible value of xx
If x=9x = 9, then x2=81x^2 = 81. If x=13x = \frac{1}{3}, then x2=19x^2 = \frac{1}{9}.
The question asks for the values of x2x^2, not xx.

Anahtar Kavram

Solving equations with absolute values on both sides requires considering positive and negative case equivalences.
Soru 44Soru

If xx and yy are real numbers such that x43|x - 4| \le 3 and y+25|y + 2| \le 5, what is the maximum possible value of xy|x - y|?

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Cevap: 14

Cevap

The maximum possible value of xy|x - y| is 14.
Solving x43|x - 4| \le 3 gives the closed interval [1,7][1, 7] for xx. Solving y+25|y + 2| \le 5 gives the closed interval [7,3][-7, 3] for yy. The maximum possible value of xy|x - y| is the maximum distance between a point in [1,7][1, 7] and a point in [7,3][-7, 3], which is 7(7)=147 - (-7) = 14.

Adım Adım Çözüm

1
Determine the range of possible values for xx.
1x71 \le x \le 7
The inequality x43|x - 4| \le 3 represents all numbers within distance 3 of 4 on the number line.
2
Determine the range of possible values for yy.
7y3-7 \le y \le 3
The inequality y+25|y + 2| \le 5 represents all numbers within distance 5 of -2 on the number line.
3
Find the maximum distance between any point xx in [1,7][1, 7] and any point yy in [7,3][-7, 3].
14
The maximum absolute difference xy|x - y| occurs between the upper endpoint of the xx-interval (77) and the lower endpoint of the yy-interval (7-7).

Anahtar Kavram

Absolute value inequalities as distance intervals on the real number line and maximizing differences between bounded variables
Soru 45Soru

If nn is a positive integer such that 2n2+2n1+2n+2n+1=4802^{n-2} + 2^{n-1} + 2^n + 2^{n+1} = 480, what is the value of nn?

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Cevap: 7

Cevap

The value of nn is 7.
Factoring out 2n22^{n-2} converts the sum into 2n2(1+2+4+8)=152n2=4802^{n-2}(1 + 2 + 4 + 8) = 15 \cdot 2^{n-2} = 480. Dividing 480 by 15 gives 2n2=322^{n-2} = 32. Since 32=2532 = 2^5, setting n2=5n - 2 = 5 yields n=7n = 7.

Adım Adım Çözüm

1
Factor out the lowest power of 2, which is 2n22^{n-2}, from all terms on the left side of the equation.
2n2(1+2+4+8)=4802^{n-2}(1 + 2 + 4 + 8) = 480
Factoring out a common exponential term simplifies the addition of powers into a product of a single exponential term and a constant.
2
Evaluate the constant factor inside the parentheses and solve for the exponential expression 2n22^{n-2}.
152n2=480    2n2=3215 \cdot 2^{n-2} = 480 \implies 2^{n-2} = 32
Summing 1+2+4+81 + 2 + 4 + 8 yields 15. Dividing both sides by 15 isolates the base-2 term.
3
Express 32 as a power with base 2 and equate the exponents.
2n2=25    n2=5    n=72^{n-2} = 2^5 \implies n - 2 = 5 \implies n = 7
Since 32=2532 = 2^5 and the bases are identical, the exponents must be equal.

Anahtar Kavram

Factoring sum of exponential terms with common bases
Soru 46Soru

On the real number line, xx is a negative real number such that 52x=11|5 - 2x| = 11, and yy is a positive real number such that 3y+1=13|3y + 1| = 13. What is the distance on the real number line between xx and yy?

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Cevap: 7

Cevap

The distance between xx and yy on the real number line is 7.
Solving 52x=11|5 - 2x| = 11 with x<0x < 0 yields x=3x = -3. Solving 3y+1=13|3y + 1| = 13 with y>0y > 0 yields y=4y = 4. The distance between 3-3 and 44 on the number line is 4(3)=7|4 - (-3)| = 7.

Adım Adım Çözüm

1
Solve for the negative real number xx using the equation 52x=11|5 - 2x| = 11.
x=3x = -3
The equation splits into 52x=115 - 2x = 11 (yielding x=3x = -3) and 52x=115 - 2x = -11 (yielding x=8x = 8). Since xx must be negative, x=3x = -3 is selected.
2
Solve for the positive real number yy using the equation 3y+1=13|3y + 1| = 13.
y=4y = 4
The equation splits into 3y+1=133y + 1 = 13 (yielding y=4y = 4) and 3y+1=133y + 1 = -13 (yielding y=143y = -\frac{14}{3}). Since yy must be positive, y=4y = 4 is selected.
3
Compute the distance between xx and yy on the number line.
7
The distance between two points on the number line is given by yx=4(3)=7|y - x| = |4 - (-3)| = 7.

Anahtar Kavram

Distance on the real number line between two points aa and bb is given by ab|a - b|, solved by evaluating absolute value equations under given sign constraints.

Alternatif Yöntem

Plot the candidate solutions for xx (x=3x = -3 and x=8x = 8) and yy (y=4y = 4 and y=14/3y = -14/3) on a number line, then directly count units between the valid points x=3x = -3 and y=4y = 4.
Tahmini Süre:1m 15s
Soru 47Soru

In a school club of 4040 students, the ratio of the number of boys to the number of girls is 3:53 : 5. Which of the following statements must be true? Select all such statements.

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Cevap: The number of boys in the club is 1515.; The ratio of the number of boys to the total number of students is 3:83 : 8.; The number of girls in the club exceeds the number of boys by 1010.

Cevap

The correct statements are those indicating that there are 15 boys in the club, that the ratio of boys to total students is 3:8, and that there are 10 more girls than boys in the club.
The total number of ratio parts is 3+5=83 + 5 = 8, representing the entire group of 4040 students. Dividing 4040 by 88 gives 55 students per part. Therefore, the number of boys is 3×5=153 \times 5 = 15, the number of girls is 5×5=255 \times 5 = 25, and the ratio of boys to total students is 15:40=3:815:40 = 3:8. Subtracting the number of boys from girls (2515=1025 - 15 = 10) shows there are 1010 more girls than boys.

Adım Adım Çözüm

1
Calculate total ratio parts and determine the value of one part.
Total ratio parts = 3+5=83 + 5 = 8. Each part corresponds to 408=5\frac{40}{8} = 5 students.
Converting ratio terms into parts allows calculation of actual quantities from the total group size.
2
Compute the total number of boys and girls.
Number of boys = 3×5=153 \times 5 = 15. Number of girls = 5×5=255 \times 5 = 25.
Multiply each component's ratio share by the value of one part.
3
Evaluate the given statements against calculated quantities.
There are 1515 boys; the ratio of boys to total students is 15:40=3:815:40 = 3:8; and the difference between girls and boys is 2515=1025 - 15 = 10.
Determines which options represent true statements about the group.

Anahtar Kavram

Part-to-Part vs. Part-to-Whole Ratios
Soru 48Soru

Let xx and yy be real numbers such that x24=2y|x^2 - 4| = 2y and y31|y - 3| \le 1. Which of the following could be the value of xx? Select all such values.

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Cevap: 3-3; 00; 33

Cevap

The values that xx could take are 3-3, 00, and 33.
The inequality y31|y - 3| \le 1 restricts yy to the closed interval [2,4][2, 4]. Consequently, 2y2y lies in [4,8][4, 8], which means 4x2484 \le |x^2 - 4| \le 8. Breaking this compound inequality into cases yields x=0x = 0 (from x2=0x^2 = 0) or x[12,8][8,12]x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]. Since 82.83\sqrt{8} \approx 2.83 and 123.46\sqrt{12} \approx 3.46, the integer values 3-3 and 33 fall in these intervals, alongside 00. Thus, 3-3, 00, and 33 are all valid choices.

Adım Adım Çözüm

1
Determine the acceptable range for yy from the given absolute value inequality.
Solving y31|y - 3| \le 1 gives 1y31-1 \le y - 3 \le 1, which simplifies to 2y42 \le y \le 4.
The inequality ycr|y - c| \le r represents all values of yy within distance rr from cc.
2
Relate the range of yy to the absolute value expression x24|x^2 - 4|.
Since 2y42 \le y \le 4, multiplying by 22 yields 42y84 \le 2y \le 8. Therefore, 4x2484 \le |x^2 - 4| \le 8.
Substitute 2y=x242y = |x^2 - 4| into the inequality derived for yy.
3
Analyze the two cases for the absolute value equation x24|x^2 - 4|.
Case 1: 4x248    8x2124 \le x^2 - 4 \le 8 \implies 8 \le x^2 \le 12, which gives x[12,8][8,12]x \in [-\sqrt{12}, -\sqrt{8}] \cup [\sqrt{8}, \sqrt{12}]. Case 2: 4(x24)8    44x28    4x24    4x204 \le -(x^2 - 4) \le 8 \implies 4 \le 4 - x^2 \le 8 \implies -4 \le -x^2 \le 4 \implies -4 \le x^2 \le 0. Since x20x^2 \ge 0 for all real numbers, x2=0    x=0x^2 = 0 \implies x = 0.
Absolute value A|A| splits into AA when A0A \ge 0 and A-A when A<0A < 0.
4
Evaluate the test values against the valid domains for xx.
For x=3x = -3, x2=9x^2 = 9, which satisfies 89128 \le 9 \le 12. For x=0x = 0, x2=0x^2 = 0, which satisfies x2=0x^2 = 0. For x=3x = 3, x2=9x^2 = 9, which satisfies 89128 \le 9 \le 12. The values x=2x = 2 and x=4x = 4 give x2=4x^2 = 4 and x2=16x^2 = 16, neither of which fall into the valid intervals.
Checking test options confirms which values fall within the solution intervals.

Anahtar Kavram

Absolute Value Equations and System Constraints on Real Numbers
Soru 49Soru

Let xx and yy be positive integers such that gcd(x,y)=14\gcd(x, y) = 14 and lcm(x,y)=420\text{lcm}(x, y) = 420. Which of the following values could be the sum x+yx + y? Select all such values.

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Cevap: 154; 182; 238

Cevap

The possible values for the sum x+yx + y are 154, 182, and 238.
By writing x=14ax = 14a and y=14by = 14b with gcd(a,b)=1\gcd(a, b) = 1, the relation lcm(x,y)=14ab=420\text{lcm}(x, y) = 14ab = 420 requires ab=30ab = 30. The positive coprime factor pairs of 30 are (1,30)(1, 30), (2,15)(2, 15), (3,10)(3, 10), and (5,6)(5, 6). Multiplying these pairs by 14 gives the possible sums 434, 238, 182, and 154. Therefore, the options equal to 154, 182, and 238 are all correct.

Adım Adım Çözüm

1
Express xx and yy in terms of their greatest common divisor.
Let x=14ax = 14a and y=14by = 14b, where aa and bb are positive integers such that gcd(a,b)=1\gcd(a, b) = 1.
Factoring out the greatest common divisor leaves coprime quotient factors aa and bb.
2
Relate the least common multiple to aa and bb.
\text{lcm}(x, y) = 14ab = 420 \implies ab = \frac{420}{14} = 30$.
The least common multiple of two numbers sharing a GCD of gg is given by gabg \cdot a \cdot b.
3
Find all coprime pairs (a,b)(a, b) with aba \le b whose product is 30.
The prime factorization of 30 is 2×3×52 \times 3 \times 5. The valid coprime pairs (a,b)(a, b) are (1,30)(1, 30), (2,15)(2, 15), (3,10)(3, 10), and (5,6)(5, 6).
Since gcd(a,b)=1\gcd(a, b) = 1, all factors of 30 split into pairs of coprime integers.
4
Calculate the corresponding values of xx, yy, and their sum x+yx + y for each pair.
Pair (1, 30): x=14,y=420    x+y=434x = 14, y = 420 \implies x + y = 434.
Pair (2, 15): x=28,y=210    x+y=238x = 28, y = 210 \implies x + y = 238.
Pair (3, 10): x=42,y=140    x+y=182x = 42, y = 140 \implies x + y = 182.
Pair (5, 6): x=70,y=84    x+y=154x = 70, y = 84 \implies x + y = 154.
Multiplying each pair by the GCD of 14 yields the original integers xx and yy.

Anahtar Kavram

Relationship between GCD, LCM, and prime factorization of quotient factors
Soru 50Soru

A logistics company received a shipment of identical freight containers. On Monday, the crew unloaded 27\frac{2}{7} of the total shipment. On Tuesday, they unloaded 35\frac{3}{5} of the remaining containers. On Wednesday, they unloaded 12\frac{1}{2} of the containers that remained after Tuesday's work. If 30 containers remained unloaded at the end of Wednesday, what was the total number of containers in the original shipment?

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Cevap: 210

Cevap

The total number of containers in the original shipment was 210.
The correct answer is 210. Working forward, after Monday 57\frac{5}{7} of the initial shipment NN remains. On Tuesday, 25\frac{2}{5} of that remainder stays unloaded, which equals 25×57N=27N\frac{2}{5} \times \frac{5}{7}N = \frac{2}{7}N. On Wednesday, 12\frac{1}{2} of that remainder stays unloaded, yielding 12×27N=17N\frac{1}{2} \times \frac{2}{7}N = \frac{1}{7}N. Setting 17N=30\frac{1}{7}N = 30 gives N=210N = 210.

Adım Adım Çözüm

1
Determine the fraction of containers remaining after Monday.
Since 27\frac{2}{7} of the total shipment NN was unloaded on Monday, the fraction remaining is 127=571 - \frac{2}{7} = \frac{5}{7} of NN.
Subtracting the fraction unloaded on Monday from 1 gives the remaining fraction.
2
Calculate the fraction of containers remaining after Tuesday.
On Tuesday, 35\frac{3}{5} of the remaining 57N\frac{5}{7}N was unloaded, which is 35×57N=37N\frac{3}{5} \times \frac{5}{7}N = \frac{3}{7}N. The remaining fraction after Tuesday is 57N37N=27N\frac{5}{7}N - \frac{3}{7}N = \frac{2}{7}N.
Alternatively, if 35\frac{3}{5} of the remainder was unloaded, then 135=251 - \frac{3}{5} = \frac{2}{5} of the remainder was left: 25×57N=27N\frac{2}{5} \times \frac{5}{7}N = \frac{2}{7}N.
3
Calculate the fraction of containers remaining after Wednesday.
On Wednesday, 12\frac{1}{2} of the remaining 27N\frac{2}{7}N was unloaded, leaving 112=121 - \frac{1}{2} = \frac{1}{2} of that remainder. Thus, the final fraction remaining is 12×27N=17N\frac{1}{2} \times \frac{2}{7}N = \frac{1}{7}N.
Multiplying the remaining fraction after Tuesday by the fraction left unhandled on Wednesday yields the overall fraction of the original shipment remaining.
4
Solve for the total initial number of containers NN.
Set 17N=30\frac{1}{7}N = 30, which gives N=30×7=210N = 30 \times 7 = 210.
Equating the calculated final remaining fraction to the given numerical count allows solving for the total original quantity.

Anahtar Kavram

Sequential Fraction of Remaining Quantities
Soru 51Soru

If kk and mm are integers such that k<0<mk < 0 < m, (1)k=1(-1)^k = -1, and (1)m=1(-1)^m = 1, which of the following expressions MUST be a negative odd integer?

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Cevap: kmk - m

Cevap

The expression kmk - m MUST be a negative odd integer.
Given k<0<mk < 0 < m, kk is negative and mm is positive. The relation (1)k=1(-1)^k = -1 shows kk is odd, while (1)m=1(-1)^m = 1 shows mm is even. Subtracting a positive even integer mm from a negative odd integer kk yields kmk - m, which must be less than 0 (negative) and odd (odd minus even).

Adım Adım Çözüm

1
Determine the parity and sign of kk
kk is a negative odd integer.
k<0k < 0 specifies that kk is negative, and (1)k=1(-1)^k = -1 implies that the exponent kk must be odd.
2
Determine the parity and sign of mm
mm is a positive even integer.
m>0m > 0 specifies that mm is positive, and (1)m=1(-1)^m = 1 implies that the exponent mm must be even.
3
Evaluate the sign and parity of kmk - m
kmk - m is strictly negative and odd.
Since k<0k < 0 and m>0m > 0, km=k+(m)<0k - m = k + (-m) < 0. By parity rules, oddeven=odd\text{odd} - \text{even} = \text{odd}.

Anahtar Kavram

Even-Odd Properties and Sign Rules
Soru 52Soru

On the real number line, the distance between two real numbers xx and yy is dd. If the midpoint of xx and yy is 77 and x3=2y3|x - 3| = 2|y - 3|, what is the maximum possible value of dd?

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Cevap: 24

Cevap

The maximum possible value of dd is 24.
The midpoint condition dictates that x+y=14x + y = 14, so y=14xy = 14 - x and the distance between them is d=xy=2x7d = |x - y| = 2|x - 7|. Substituting y=14xy = 14 - x into x3=2y3|x - 3| = 2|y - 3| gives x3=211x|x - 3| = 2|11 - x|. Solving the positive case x3=2(x11)x - 3 = 2(x - 11) gives x=19x = 19 and y=5y = -5, resulting in distance d=19(5)=24d = |19 - (-5)| = 24. Solving the negative case x3=2(x11)x - 3 = -2(x - 11) gives x=25/3x = 25/3 and y=17/3y = 17/3, resulting in distance d=8/3d = 8/3. Thus, 24 is the maximum possible value of dd.

Adım Adım Çözüm

1
Express yy in terms of xx using the midpoint formula
Since the midpoint of xx and yy is 77, x+y2=7\frac{x + y}{2} = 7, which gives y=14xy = 14 - x. The distance d=xy=x(14x)=2x14=2x7d = |x - y| = |x - (14 - x)| = |2x - 14| = 2|x - 7|.
Relating yy to xx reduces the problem to a single variable.
2
Substitute y=14xy = 14 - x into the absolute value equation
x3=2(14x)3    x3=211x=2x11|x - 3| = 2|(14 - x) - 3| \implies |x - 3| = 2|11 - x| = 2|x - 11|.
Setting up the single-variable absolute value equation allows finding all possible values for xx.
3
Solve the absolute value equation for all possible cases
Case 1: x3=2(x11)    x3=2x22    x=19x - 3 = 2(x - 11) \implies x - 3 = 2x - 22 \implies x = 19. Then y=1419=5y = 14 - 19 = -5.
Case 2: x3=2(x11)    x3=2x+22    3x=25    x=253x - 3 = -2(x - 11) \implies x - 3 = -2x + 22 \implies 3x = 25 \implies x = \frac{25}{3}. Then y=14253=173y = 14 - \frac{25}{3} = \frac{17}{3}.
Absolute value equations A=B|A| = B split into A=BA = B and A=BA = -B.
4
Calculate the distance dd for each case and select the maximum
For Case 1 (x=19,y=5x = 19, y = -5): d=19(5)=24d = |19 - (-5)| = 24.
For Case 2 (x=25/3,y=17/3x = 25/3, y = 17/3): d=25/317/3=8/3d = |25/3 - 17/3| = 8/3.
The maximum possible value of dd is 2424.
Comparing the distance values determined in each case identifies the maximum distance.

Anahtar Kavram

Absolute Value as Distance and Multi-Case Equations on the Real Number Line
Tahmini Süre:2m 30s
Soru 53Soru

For how many integer values of nn in the interval 15n15-15 \le n \le 15 is the value of the expression (1)n2+n(1)3n(-1)^{n^2 + n} - (-1)^{3n} equal to 22?

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Cevap: 16

Cevap

The correct answer is 16.
Because n2+n=n(n+1)n^2 + n = n(n + 1) is the product of two consecutive integers, it is guaranteed to be even for every integer nn. Consequently, (1)n2+n=1(-1)^{n^2 + n} = 1. Substituting this into the given equation yields 1(1)3n=21 - (-1)^{3n} = 2, which reduces to (1)3n=1(-1)^{3n} = -1. A power of 1-1 equals 1-1 if and only if the exponent is odd, so 3n3n must be odd, which requires nn itself to be odd. In the interval [15,15][-15, 15], there are 16 odd integers: 8 negative odd integers and 8 positive odd integers.

Adım Adım Çözüm

1
Determine the parity of n2+nn^2 + n
The expression n2+n=n(n+1)n^2 + n = n(n + 1) represents the product of two consecutive integers. Because one of any two consecutive integers is even, their product is always even. Therefore, (1)n2+n=1(-1)^{n^2 + n} = 1 for all integers nn.
Simplifying the exponent with a known parity rule reduces the expression to a constant.
2
Isolate (1)3n(-1)^{3n} in the equation
Substituting 11 into the original equation gives 1(1)3n=21 - (-1)^{3n} = 2, which simplifies to (1)3n=1(-1)^{3n} = -1.
Isolating the exponential term reveals the sign condition required for the equality to hold.
3
Find the parity condition for nn
For (1)3n(-1)^{3n} to equal 1-1, the exponent 3n3n must be an odd integer. Since 33 is odd, the product 3n3n is odd if and only if nn is odd.
Applying the product parity rule (odd×odd=odd\text{odd} \times \text{odd} = \text{odd}) relates the condition on 3n3n back to nn.
4
Count the odd integers in the interval [15,15][-15, 15]
The odd integers in the interval are 15,13,11,9,7,5,3,1,1,3,5,7,9,11,13,15-15, -13, -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9, 11, 13, 15. There are 16 such integers.
Counting all qualifying values within the specified range yields the final numeric answer.

Anahtar Kavram

Parity rules for consecutive integers and exponents of negative numbers
Soru 54Soru

At the beginning of Year 1, a commercial logistics company allocated its total storage space between two facilities, Facility X and Facility Y, such that Facility X contained 60%60\% of the total space and Facility Y contained the remaining 40%40\%. During Year 1, the storage space of Facility X was increased by x%x\%, while the storage space of Facility Y was increased by 25%25\%. During Year 2, the storage space of Facility X was decreased by 20%20\% from its Year 1 level, while the storage space of Facility Y was increased by x%x\% from its Year 1 level. If the total combined storage space of both facilities at the end of Year 2 was 17.6%17.6\% greater than the total combined storage space at the beginning of Year 1, what is the value of xx?

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Cevap: 20

Cevap

The value of xx is 20.
Let the initial total space be AA. Facility X initially has 0.60A0.60A space and Facility Y has 0.40A0.40A space. After Year 1, Facility X space becomes 0.60A(1+x/100)0.60A(1 + x/100) and Facility Y space becomes 0.40A(1.25)=0.50A0.40A(1.25) = 0.50A. At the end of Year 2, Facility X space decreases by 20%20\% to 0.48A(1+x/100)0.48A(1 + x/100), while Facility Y space increases by x%x\% to 0.50A(1+x/100)0.50A(1 + x/100). Factoring out (1+x/100)(1 + x/100), the total final combined space is (0.48A+0.50A)(1+x/100)=0.98A(1+x/100)(0.48A + 0.50A)(1 + x/100) = 0.98A(1 + x/100). Setting this equal to 1.176A1.176A (a 17.6%17.6\% increase over AA) yields 1+x/100=1.176/0.98=1.201 + x/100 = 1.176 / 0.98 = 1.20, which gives x=20x = 20.

Adım Adım Çözüm

1
Define initial storage spaces for Facility X and Facility Y in terms of total initial area AA.
Facility X initial space = 0.60A0.60A, Facility Y initial space = 0.40A0.40A.
Establishing explicit algebraic representations based on the given 60%/40%60\% / 40\% ratio.
2
Calculate the storage space of each facility after Year 1 changes.
Facility X after Year 1 = 0.60A(1+x100)0.60A \left(1 + \frac{x}{100}\right); Facility Y after Year 1 = 0.40A(1+0.25)=0.50A0.40A (1 + 0.25) = 0.50A.
Facility X increased by x%x\% and Facility Y increased by 25%25\% of its original base.
3
Calculate the storage space of each facility at the end of Year 2.
Facility X final = 0.60A(1+x100)(10.20)=0.48A(1+x100)0.60A \left(1 + \frac{x}{100}\right) (1 - 0.20) = 0.48A \left(1 + \frac{x}{100}\right); Facility Y final = 0.50A(1+x100)0.50A \left(1 + \frac{x}{100}\right).
Facility X decreased by 20%20\% from its Year 1 amount, and Facility Y increased by x%x\% from its Year 1 amount.
4
Sum the final spaces of both facilities and set equal to the total space given (17.6%17.6\% overall increase over AA).
Total final space = 0.48A(1+x100)+0.50A(1+x100)=0.98A(1+x100)=1.176A0.48A \left(1 + \frac{x}{100}\right) + 0.50A \left(1 + \frac{x}{100}\right) = 0.98A \left(1 + \frac{x}{100}\right) = 1.176A.
The total space at the end of Year 2 is 100%+17.6%=117.6%100\% + 17.6\% = 117.6\% of initial total space AA.
5
Solve the linear equation for xx.
1+x100=1.1760.98=1.20    x100=0.20    x=201 + \frac{x}{100} = \frac{1.176}{0.98} = 1.20 \implies \frac{x}{100} = 0.20 \implies x = 20.
Dividing both sides by 0.98A0.98A isolates the percentage factor.

Anahtar Kavram

Successive Percent Change and Multi-Base Percentage Problems
Tahmini Süre:2m 30s
Soru 55Soru

A positive integer nn has the prime factorization n=2a3b5cn = 2^a \cdot 3^b \cdot 5^c, where aa, bb, and cc are positive integers. If nn is divisible by 3636 and is a divisor of 54005{}400, which of the following values could be the total number of positive divisors of nn? Select all such values.

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Cevap: 18; 27; 32

Cevap

18, 27, and 32 are all possible total numbers of positive divisors for nn.
Prime factorization gives 36=223236 = 2^2 \cdot 3^2 and 5400=2333525{}400 = 2^3 \cdot 3^3 \cdot 5^2. For n=2a3b5cn = 2^a \cdot 3^b \cdot 5^c to be a multiple of 36 and a divisor of 5,400 with positive integer exponents, a{2,3}a \in \{2,3\}, b{2,3}b \in \{2,3\}, and c{1,2}c \in \{1,2\}. The total number of divisors is (a+1)(b+1)(c+1)(a+1)(b+1)(c+1). The possible values for this product are 18, 24, 27, 32, 36, and 48. Among the choices, 18, 27, and 32 are valid values.

Adım Adım Çözüm

1
Find the prime factorizations of the boundary numbers 36 and 5,400.
36=223236 = 2^2 \cdot 3^2 and 5400=2333525{}400 = 2^3 \cdot 3^3 \cdot 5^2.
Establishing the prime factor bounds determines the range of possible values for exponents aa, bb, and cc.
2
Determine the constraints on exponents aa, bb, and cc.
Since 36n36 \mid n, a2a \ge 2, b2b \ge 2, and c1c \ge 1 (given cc is a positive integer). Since n5400n \mid 5{}400, a3a \le 3, b3b \le 3, and c2c \le 2. Thus, a{2,3}a \in \{2, 3\}, b{2,3}b \in \{2, 3\}, and c{1,2}c \in \{1, 2\}.
Divisibility rules require prime factor exponents of a multiple to be greater than or equal to those of the divisor, and exponents of a divisor to be less than or equal to those of the multiple.
3
Calculate all possible total divisor counts using the formula d(n)=(a+1)(b+1)(c+1)d(n) = (a+1)(b+1)(c+1).
Possible factor values are (a+1){3,4}(a+1) \in \{3, 4\}, (b+1){3,4}(b+1) \in \{3, 4\}, and (c+1){2,3}(c+1) \in \{2, 3\}. Evaluating all combinations yields: 332=183 \cdot 3 \cdot 2 = 18, 333=273 \cdot 3 \cdot 3 = 27, 342=243 \cdot 4 \cdot 2 = 24, 343=363 \cdot 4 \cdot 3 = 36, 442=324 \cdot 4 \cdot 2 = 32, and 443=484 \cdot 4 \cdot 3 = 48.
The total number of positive integer divisors is found by adding 1 to each exponent in the prime factorization and multiplying the results.
4
Compare the calculated divisor counts with the options provided.
The values 18, 27, and 32 appear in the calculated set of possible total divisors.
Direct matching identifies all valid options.

Anahtar Kavram

Prime factor exponent bounds and total number of positive divisors formula
Soru 56Soru

If xx is a real number such that x29=5x3|x^2 - 9| = 5|x - 3|, which of the following could be the value of xx? Select all such values.

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Cevap: 8-8; 22; 33

Cevap

The values of xx that satisfy the equation are 8-8, 22, and 33.
Factoring the left side of x29=5x3|x^2 - 9| = 5|x - 3| yields x3x+3=5x3|x - 3||x + 3| = 5|x - 3|. Setting the common factor x3=0|x - 3| = 0 gives x=3x = 3. Dividing both sides by the non-zero quantity x3|x - 3| leaves x+3=5|x + 3| = 5, which splits into x+3=5    x=2x + 3 = 5 \implies x = 2 and x+3=5    x=8x + 3 = -5 \implies x = -8. Therefore, 8-8, 22, and 33 are all valid solutions.

Adım Adım Çözüm

1
Apply the product rule for absolute values to factor the left-hand side.
x29=(x3)(x+3)=x3x+3|x^2 - 9| = |(x - 3)(x + 3)| = |x - 3| \cdot |x + 3|, rewriting the equation as x3x+3=5x3|x - 3| \cdot |x + 3| = 5|x - 3|.
The absolute value of a product is equal to the product of the individual absolute values.
2
Evaluate the case where the shared factor is zero: x3=0|x - 3| = 0.
x3=0    x=3x - 3 = 0 \implies x = 3. Both sides equal 00, making x=3x = 3 a valid solution.
Dividing by x3|x - 3| without checking if it can be zero would cause the loss of the root x=3x = 3.
3
Evaluate the case where x30|x - 3| \neq 0 by dividing both sides of the equation by x3|x - 3|.
x+3=5|x + 3| = 5.
Since x3>0|x - 3| > 0, we can safely divide both sides by this non-zero quantity.
4
Solve the remaining absolute value equation x+3=5|x + 3| = 5.
x+3=5    x=2x + 3 = 5 \implies x = 2, and x+3=5    x=8x + 3 = -5 \implies x = -8.
An expression inside an absolute value equal to a positive number kk can equal either kk or k-k.

Anahtar Kavram

Factoring absolute value expressions using ab=ab|ab| = |a||b| and systematically considering all cases to avoid dropping zero-roots or negative solutions.
Soru 57Soru

Two positive integers mm and nn satisfy gcd(m,n)=20\gcd(m, n) = 20 and lcm(m,n)=4200\text{lcm}(m, n) = 4{}200. It is given that mm is divisible by 77 and has exactly 2424 positive integer divisors. If nn is a multiple of 33, what is the value of nn?

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Cevap: 6060

Cevap

The value of nn is 6060.
Prime factorization reveals that gcd(m,n)=22305170\gcd(m,n) = 2^2 \cdot 3^0 \cdot 5^1 \cdot 7^0 and lcm(m,n)=23315271\text{lcm}(m,n) = 2^3 \cdot 3^1 \cdot 5^2 \cdot 7^1. The condition that nn is a multiple of 33 fixes the exponent of 33 in nn to 11 (so mm has exponent 00). The condition that mm is a multiple of 77 fixes the exponent of 77 in mm to 11 (so nn has exponent 00). For m=2a305c71m = 2^a \cdot 3^0 \cdot 5^c \cdot 7^1, its divisor count equation 2(a+1)(c+1)=242(a+1)(c+1) = 24 simplifies to (a+1)(c+1)=12(a+1)(c+1) = 12. Since a{2,3}a \in \{2, 3\} and c{1,2}c \in \{1, 2\}, the only valid solution is a=3a=3 and c=2c=2. This leaves nn with exponents 22 for prime 22, 11 for prime 33, 11 for prime 55, and 00 for prime 77, giving n=22315170=60n = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^0 = 60.

Adım Adım Çözüm

1
Express the GCD and LCM in prime factorized form.
gcd(m,n)=20=22305170\gcd(m, n) = 20 = 2^2 \cdot 3^0 \cdot 5^1 \cdot 7^0 and lcm(m,n)=4200=23315271\text{lcm}(m, n) = 4{}200 = 2^3 \cdot 3^1 \cdot 5^2 \cdot 7^1.
The prime factor exponents of mm and nn must have minimums equal to the GCD exponents and maximums equal to the LCM exponents.
2
Determine the exponents of prime factors 33 and 77 for mm and nn.
Since nn is divisible by 33, exponent of 33 in nn is 11, so exponent of 33 in mm is 00. Since mm is divisible by 77, exponent of 77 in mm is 11, so exponent of 77 in nn is 00.
Each prime exponent in lcm(m,n)\text{lcm}(m, n) must belong to at least one of the numbers.
3
Use the divisor count of mm to find its remaining exponents for primes 22 and 55.
Let m=2a305c71m = 2^a \cdot 3^0 \cdot 5^c \cdot 7^1, where a{2,3}a \in \{2, 3\} and c{1,2}c \in \{1, 2\}. The number of positive divisors is (a+1)(0+1)(c+1)(1+1)=2(a+1)(c+1)=24(a+1)(0+1)(c+1)(1+1) = 2(a+1)(c+1) = 24, giving (a+1)(c+1)=12(a+1)(c+1) = 12. Testing a=2    c=3a=2 \implies c=3 (invalid range). Testing a=3    c=2a=3 \implies c=2 (valid). Thus m=23305271=1400m = 2^3 \cdot 3^0 \cdot 5^2 \cdot 7^1 = 1{}400.
The total number of positive integer divisors of p1e1p2e2p_1^{e_1} p_2^{e_2} \cdots is given by (e1+1)(e2+1)(e_1+1)(e_2+1)\cdots.
4
Determine the prime exponents for nn and calculate its value.
Since a=3a=3, nn gets 222^2. Since c=2c=2, nn gets 515^1. Together with 313^1 and 707^0, n=22315170=60n = 2^2 \cdot 3^1 \cdot 5^1 \cdot 7^0 = 60.
For each prime pp, the exponent in nn must match the bound opposite to mm to satisfy both gcd\gcd and lcm\text{lcm}.

Anahtar Kavram

Prime exponent analysis of GCD and LCM alongside the divisor count formula
Soru 58Soru

Consider three non-zero integers xx, yy, and zz that satisfy all of the following conditions:

I. (1)x2y+z=1(-1)^{x^2 y + z} = -1
II. xyz2<0x y z^2 < 0
III. x+yx + y is an even integer

Which of the following expressions MUST be an odd integer?

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Cevap: (x+z)(y+z)(x + z)(y + z)

Cevap

The expression (x+z)(y+z)(x + z)(y + z) MUST be an odd integer.
Condition I dictates that x2y+zx^2 y + z is odd. Condition III states that x+yx + y is even, meaning xx and yy share the same parity. If xx and yy are both even, x2yx^2 y is even, forcing zz to be odd. If xx and yy are both odd, x2yx^2 y is odd, forcing zz to be even. Consequently, zz always has the opposite parity of both xx and yy. Therefore, (x+z)(x + z) is always odd and (y+z)(y + z) is always odd. The product of two odd integers, (x+z)(y+z)(x + z)(y + z), is guaranteed to be odd.

Adım Adım Çözüm

1
Analyze Condition I for exponent parity
x2y+zx^2 y + z must be an odd integer
For (1)k=1(-1)^k = -1, the exponent kk must be odd.
2
Analyze Condition III for shared parity of xx and yy
xx and yy are either both even or both odd
The sum of two integers is even if and only if they share the same parity.
3
Deduce parity relationship for zz across cases
In Case 1 (x,yx, y even), x2yx^2 y is even, so zz must be odd. In Case 2 (x,yx, y odd), x2yx^2 y is odd, so zz must be even.
To satisfy x2y+z=oddx^2 y + z = \text{odd}, x2yx^2 y and zz must have opposite parities.
4
Evaluate the parity of (x+z)(y+z)(x + z)(y + z)
(x+z)(x + z) is odd and (y+z)(y + z) is odd, so their product is odd
In both cases, zz has opposite parity to both xx and yy. Adding two integers of opposite parity always yields an odd integer, and the product of two odd integers is always odd.

Anahtar Kavram

Parity rules under exponentiation and algebraic combination
Tahmini Süre:2m 0s
Soru 59Soru

If xx is a negative real number, which of the following expressions are equivalent to x3\sqrt{-x^3}? Select all such expressions.

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Cevap: xx-x\sqrt{-x}; xx|x|\sqrt{-x}; (x)3/2(-x)^{3/2}

Cevap

The expressions equivalent to x3\sqrt{-x^3} are xx-x\sqrt{-x}, xx|x|\sqrt{-x}, and (x)3/2(-x)^{3/2}.
Because xx is negative, x-x is a positive quantity. We can express x3-x^3 as (x)3=(x)2(x)(-x)^3 = (-x)^2 \cdot (-x). Taking the principal square root yields (x)2(x)=(x)2x=xx\sqrt{(-x)^2 \cdot (-x)} = \sqrt{(-x)^2} \cdot \sqrt{-x} = -x\sqrt{-x}. Because x=x|x| = -x for negative numbers, the expression xx|x|\sqrt{-x} is identical to xx-x\sqrt{-x}. Furthermore, converting to rational exponents gives (x)3/2=(x)3=x3(-x)^{3/2} = \sqrt{(-x)^3} = \sqrt{-x^3}. Thus, all three of these expressions are mathematically equivalent to the original radical expression.

Adım Adım Çözüm

1
Analyze the sign of the base and inside of the radical
Since x<0x < 0, the quantity x-x is strictly positive (x>0 -x > 0 ). Consequently, x3=(x)3>0-x^3 = (-x)^3 > 0, ensuring x3\sqrt{-x^3} is a real, non-negative number.
Principal square roots require a non-negative radicand and yield a non-negative result in real arithmetic.
2
Simplify the radical expression using perfect squares
x3=(x)2(x)=(x)2x=(x)x=xx\sqrt{-x^3} = \sqrt{(-x)^2 \cdot (-x)} = \sqrt{(-x)^2} \cdot \sqrt{-x} = (-x)\sqrt{-x} = -x\sqrt{-x}.
Because x>0-x > 0, (x)2=x\sqrt{(-x)^2} = -x.
3
Evaluate equivalent representations using absolute value and rational exponents
Since x<0x < 0, x=x|x| = -x, so xx=xx|x|\sqrt{-x} = -x\sqrt{-x}. Also, (x)3/2=(x)3=x3(-x)^{3/2} = \sqrt{(-x)^3} = \sqrt{-x^3}.
Both rewrite rules preserve both magnitude and non-negative sign for all x<0x < 0.

Anahtar Kavram

Simplifying radicals and fractional exponents with negative variable bases
Soru 60Soru

A pharmaceutical facility uses two liquid compounding lines, Line AA and Line BB, to produce a standardized saline solution. Line AA operates at a constant flow rate of 120120 liters per hour, producing a solution with active compound and purified water in a ratio of 1:31 : 3 by volume. Line BB operates at a constant flow rate of 180180 liters per hour, producing a solution with active compound and purified water in a ratio of 2:32 : 3 by volume. If Line AA runs for 44 hours and Line BB runs for 55 hours, and all output is collected into a single storage tank, what is the ratio of active compound to purified water in the storage tank?

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Cevap: 8:158 : 15

Cevap

The ratio of active compound to purified water in the storage tank is 8:158 : 15.
The correct answer is derived by first converting the flow rates and operating hours into total volumes (480480 L for Line A and 900900 L for Line B). Using the part-to-whole fraction conversion, Line A contributes 120120 L of active compound and 360360 L of water, while Line B contributes 360360 L of active compound and 540540 L of water. Adding these quantities yields 480480 L of active compound and 900900 L of water, which simplifies to the ratio 8:158 : 15.

Adım Adım Çözüm

1
Calculate total output volume and component quantities from Line A
Total Line A Volume = 120 L/hr×4 hrs=480 liters120 \text{ L/hr} \times 4 \text{ hrs} = 480 \text{ liters}. Active Compound = 480×11+3=120 liters480 \times \frac{1}{1+3} = 120 \text{ liters}. Purified Water = 480120=360 liters480 - 120 = 360 \text{ liters}.
Line A produces a total volume of 480 liters, and a 1:31 : 3 ratio means active compound constitutes 14\frac{1}{4} of the total volume.
2
Calculate total output volume and component quantities from Line B
Total Line B Volume = 180 L/hr×5 hrs=900 liters180 \text{ L/hr} \times 5 \text{ hrs} = 900 \text{ liters}. Active Compound = 900×22+3=360 liters900 \times \frac{2}{2+3} = 360 \text{ liters}. Purified Water = 900360=540 liters900 - 360 = 540 \text{ liters}.
Line B produces a total volume of 900 liters, and a 2:32 : 3 ratio means active compound constitutes 25\frac{2}{5} of the total volume.
3
Sum the components from both lines in the storage tank
Total Active Compound = 120+360=480 liters120 + 360 = 480 \text{ liters}. Total Purified Water = 360+540=900 liters360 + 540 = 900 \text{ liters}.
Combining the output of both lines adds the respective volumes of active compound and purified water.
4
Compute and simplify the final ratio of active compound to purified water
Ratio = 480900=815\frac{480}{900} = \frac{8}{15}, expressed as 8:158 : 15.
Dividing both numerator and denominator by their greatest common divisor, 60, yields the simplest integer ratio.

Anahtar Kavram

Weighted mixture ratios combining multi-stage rates and part-to-whole fraction conversions
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