Tüm alıştırma soruları

2195 soru

Soru 641Soru

What is the remainder when 3423^{42} is divided by 55?

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Cevap: 4

Cevap

The remainder when 3423^{42} is divided by 5 is 4.
The remainders of powers of 3 divided by 5 follow a repeating pattern of length 4: (3, 4, 2, 1). To find the remainder of 342÷53^{42} \div 5, divide the exponent 42 by 4. Since 42=4×10+242 = 4 \times 10 + 2, the remainder of the exponent is 2. The 2nd term in the repeating sequence is 4, so 342(mod5)=43^{42} \pmod 5 = 4.

Adım Adım Çözüm

1
Determine the remainder pattern for consecutive powers of 3 divided by 5.
The remainders for 31,32,33,34,353^1, 3^2, 3^3, 3^4, 3^5 are 3,4,2,1,33, 4, 2, 1, 3, establishing a cycle of length 4.
Powers of integers divided by a positive integer yield repeating sequence patterns (cyclicity).
2
Divide the target exponent by the cycle length.
Dividing 42 by 4 gives a quotient of 10 and a remainder of 2.
The remainder indicates the specific term within the 4-step cycle.
3
Evaluate the value corresponding to the 2nd position in the cycle.
The 2nd term in the cycle (3, 4, 2, 1) is 4.
A remainder of 2 in the exponent position corresponds to the same remainder as 323^2 divided by 5.

Anahtar Kavram

Remainders and Units Digit Cyclicity
Soru 642Soru

Three automated signal beacons emit flashes at regular time intervals of xx, yy, and zz seconds, respectively, where xx, yy, and zz are positive integers. If gcd(x,y)=18\text{gcd}(x, y) = 18, gcd(y,z)=24\text{gcd}(y, z) = 24, and gcd(x,z)=30\text{gcd}(x, z) = 30, what is the minimum possible value of lcm(x,y,z)\text{lcm}(x, y, z)?

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Cevap: 360360

Cevap

360360
By breaking each given greatest common divisor into its prime factorization, we establish minimum requirements for the exponents of prime factors 22, 33, and 55 across xx, yy, and zz. Specifically, gcd(y,z)=24=23×3\text{gcd}(y, z) = 24 = 2^3 \times 3 dictates that both yy and zz must contain 232^3, making the maximum exponent of 22 equal to 33. Similarly, gcd(x,y)=18=2×32\text{gcd}(x, y) = 18 = 2 \times 3^2 forces the maximum exponent of 33 to be 22, and gcd(x,z)=30=2×3×5\text{gcd}(x, z) = 30 = 2 \times 3 \times 5 forces the maximum exponent of 55 to be 11. Multiplying these prime powers together gives 23×32×5=3602^3 \times 3^2 \times 5 = 360.

Adım Adım Çözüm

1
Express the given pairwise greatest common divisors in prime factorized form.
gcd(x,y)=18=21×32\text{gcd}(x, y) = 18 = 2^1 \times 3^2, gcd(y,z)=24=23×31\text{gcd}(y, z) = 24 = 2^3 \times 3^1, gcd(x,z)=30=21×31×51\text{gcd}(x, z) = 30 = 2^1 \times 3^1 \times 5^1.
GCD conditions determine the minimum prime factor exponents across pairs of variables.
2
Analyze the exponents for prime factor 22.
min(v2(x),v2(y))=1\min(v_2(x), v_2(y)) = 1, min(v2(y),v2(z))=3\min(v_2(y), v_2(z)) = 3, and min(v2(x),v2(z))=1\min(v_2(x), v_2(z)) = 1. This implies v2(y)3v_2(y) \ge 3 and v2(z)3v_2(z) \ge 3, while v2(x)=1v_2(x) = 1. To minimize the LCM, set v2(x)=1v_2(x) = 1, v2(y)=3v_2(y) = 3, v2(z)=3v_2(z) = 3. Thus, max(v2(x),v2(y),v2(z))=3\max(v_2(x), v_2(y), v_2(z)) = 3.
The exponent of 22 in lcm(x,y,z)\text{lcm}(x, y, z) is the maximum of the individual exponents of 22.
3
Analyze the exponents for prime factor 33.
min(v3(x),v3(y))=2\min(v_3(x), v_3(y)) = 2, min(v3(y),v3(z))=1\min(v_3(y), v_3(z)) = 1, and min(v3(x),v3(z))=1\min(v_3(x), v_3(z)) = 1. This implies v3(x)2v_3(x) \ge 2 and v3(y)2v_3(y) \ge 2, while v3(z)=1v_3(z) = 1. To minimize the LCM, set v3(x)=2v_3(x) = 2, v3(y)=2v_3(y) = 2, v3(z)=1v_3(z) = 1. Thus, max(v3(x),v3(y),v3(z))=2\max(v_3(x), v_3(y), v_3(z)) = 2.
The exponent of 33 in lcm(x,y,z)\text{lcm}(x, y, z) is the maximum of the individual exponents of 33.
4
Analyze the exponents for prime factor 55.
min(v5(x),v5(y))=0\min(v_5(x), v_5(y)) = 0, min(v5(y),v5(z))=0\min(v_5(y), v_5(z)) = 0, and min(v5(x),v5(z))=1\min(v_5(x), v_5(z)) = 1. This requires v5(x)1v_5(x) \ge 1 and v5(z)1v_5(z) \ge 1, while v5(y)=0v_5(y) = 0. To minimize the LCM, set v5(x)=1v_5(x) = 1, v5(y)=0v_5(y) = 0, v5(z)=1v_5(z) = 1. Thus, max(v5(x),v5(y),v5(z))=1\max(v_5(x), v_5(y), v_5(z)) = 1.
The exponent of 55 in lcm(x,y,z)\text{lcm}(x, y, z) is the maximum of the individual exponents of 55.
5
Calculate the minimum value of lcm(x,y,z)\text{lcm}(x, y, z).
lcm(x,y,z)=23×32×51=8×9×5=360\text{lcm}(x, y, z) = 2^3 \times 3^2 \times 5^1 = 8 \times 9 \times 5 = 360.
No additional prime factors are required to satisfy the conditions, yielding the minimum possible LCM.

Anahtar Kavram

Prime Exponent Constraints on Pairwise GCD and LCM
Tahmini Süre:2m 0s
Soru 643Soru

A real estate developer purchased two tracts of land. Tract 1 consists of 1515 acres purchased at an average cost of $24,000\$24,000 per acre. Tract 2 consists of 3535 acres purchased at an average cost of $40,000\$40,000 per acre. What is the average cost per acre, in thousands of dollars, for the combined land purchase?

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Cevap: 35.2

Cevap

The average cost per acre for the combined land purchase is 35.235.2 thousand dollars (or $35,200\$35,200).
The overall weighted average is calculated by dividing the sum of all individual costs by the total number of acres across both tracts. Since 1515 acres cost $24,000\$24,000 per acre and 3535 acres cost $40,000\$40,000 per acre, the total cost is 15(24,000)+35(40,000)=360,000+1,400,000=1,760,00015(24,000) + 35(40,000) = 360,000 + 1,400,000 = 1,760,000 dollars. Dividing by the total of 5050 acres yields $35,200\$35,200 per acre, or 35.235.2 thousand dollars.

Adım Adım Çözüm

1
Determine the total financial expenditure for each tract
Tract 1 cost = 15×24,000=$360,00015 \times 24,000 = \$360,000; Tract 2 cost = 35×40,000=$1,400,00035 \times 40,000 = \$1,400,000
Total value of a set equals the number of items multiplied by the mean of the items.
2
Calculate total combined cost and total combined acreage
Total cost = $360,000+$1,400,000=$1,760,000\$360,000 + \$1,400,000 = \$1,760,000; Total acres = 15+35=5015 + 35 = 50
Combined sets require summing all total values and all individual counts.
3
Compute the weighted average cost per acre
Weighted Mean = 1,760,00050=35,200\frac{1,760,000}{50} = 35,200 dollars, which is 35.235.2 thousand dollars.
Divide total combined cost by total combined number of units.

Anahtar Kavram

Weighted Average of Combined Sets
Soru 644Soru

What is the greatest prime factor of 25232^5 - 2^3?

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Cevap: 3

Cevap

3
Evaluating 25232^5 - 2^3 yields 328=2432 - 8 = 24. The prime factorization of 24 is 23×32^3 \times 3. The distinct prime factors are 2 and 3, making 3 the greatest prime factor.

Adım Adım Çözüm

1
Evaluate the arithmetic expression
2523=328=242^5 - 2^3 = 32 - 8 = 24
Calculate the numerical value of the expression before finding prime factors.
2
Find the prime factorization of 24
24=23×324 = 2^3 \times 3
Break down 24 into a product of prime numbers.
3
Identify the greatest prime factor
The prime factors are 2 and 3. The greatest prime factor is 3.
Compare the distinct prime factors (2 and 3) to find the largest one.

Anahtar Kavram

Prime Factorization
Tahmini Süre:45s
Soru 645Soru

If the sum of five consecutive integers is 115115, what is the value of the largest integer in this set?

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Cevap: 25

Cevap

The largest integer in the set is 25.
For any evenly spaced set with an odd number of elements, the arithmetic mean equals the median. Dividing the sum of 115115 by 55 gives a median of 2323. The five consecutive integers are 21,22,23,24,2521, 22, 23, 24, 25, so the largest integer is 2525.

Adım Adım Çözüm

1
Find the average (and median) of the five consecutive integers.
Dividing the sum 115115 by 55 yields a median of 2323.
For an odd number of consecutive integers, the average of the set is equal to its median (the middle term).
2
Determine the five consecutive integers centered around the median.
The integers are 21,22,23,24,2521, 22, 23, 24, 25.
Consecutive integers differ by 11, so the two terms above the median 2323 are 2424 and 2525.
3
Identify the largest integer.
The largest integer is 2525.
The question asks specifically for the maximum value in the set.

Anahtar Kavram

Average and Median Equivalence in Consecutive Integer Sets
Tahmini Süre:45s
Soru 646Soru

In number theory, the units digit of a positive integer raised to successive positive integer powers follows a repeating cyclic pattern. What is the units digit of 4254^{25}?

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Cevap: 4

Cevap

The units digit of 4254^{25} is 4.
The units digit of powers of 4 alternates between 4 (for odd powers) and 6 (for even powers). Because 25 is an odd number, 4254^{25} has a units digit of 4.

Adım Adım Çözüm

1
Identify the units digit pattern for powers of 4
The units digits cycle between 4 (for odd exponents) and 6 (for even exponents), giving a cycle length of 2.
Units digits of positive integer powers follow a periodic pattern determined by the base digit.
2
Determine the parity of the exponent 25
25 is an odd integer (remainder 1 when divided by 2).
The exponent's remainder modulo 2 determines which position in the 2-element cycle [4, 6] the number falls into.
3
Select the corresponding units digit from the cycle
Since 25 is odd, the units digit is 4.
Odd powers of 4 always have a units digit of 4.

Anahtar Kavram

Units Digit Cyclicity
Soru 647Soru

If nn is a positive integer with exactly two distinct prime factors such that nn is a multiple of 1212 and n2n^2 has exactly 2727 positive integer divisors, what is the value of nn?

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Cevap: 48

Cevap

48
Since nn is a multiple of 12=22312 = 2^2 \cdot 3 and has only two prime factors, n=2a3bn = 2^a \cdot 3^b with a2a \ge 2 and b1b \ge 1. The square n2=22a32bn^2 = 2^{2a} \cdot 3^{2b} has (2a+1)(2b+1)=27(2a+1)(2b+1) = 27 divisors. The only factor pair of 27 satisfying 2a+152a+1 \ge 5 is 9×39 \times 3, yielding 2a+1=9    a=42a+1=9 \implies a=4 and 2b+1=3    b=12b+1=3 \implies b=1. Thus, n=2431=48n = 2^4 \cdot 3^1 = 48.

Adım Adım Çözüm

1
Express nn in terms of its prime factorization.
Since nn is a multiple of 12=22312 = 2^2 \cdot 3 and has exactly two distinct prime factors, its prime factorization must be n=2a3bn = 2^a \cdot 3^b, where a2a \ge 2 and b1b \ge 1.
Divisibility by 12 requires at least two factors of 2 and at least one factor of 3.
2
Formulate the expression for the number of positive divisors of n2n^2.
n2=(2a3b)2=22a32bn^2 = (2^a \cdot 3^b)^2 = 2^{2a} \cdot 3^{2b}. The number of positive divisors is (2a+1)(2b+1)=27(2a + 1)(2b + 1) = 27.
The total number of positive integer divisors of a prime power product pxqyp^x q^y is (x+1)(y+1)(x+1)(y+1).
3
Solve for integer values of aa and bb using factor pairs of 27.
The factor pairs of 27 into two integers greater than 1 are 9×39 \times 3. Since a2a \ge 2, 2a+152a + 1 \ge 5. Thus, 2a+1=92a + 1 = 9 and 2b+1=32b + 1 = 3, giving a=4a = 4 and b=1b = 1.
Setting 2a+1=32a + 1 = 3 yields a=1a = 1, which violates the requirement that a2a \ge 2 for divisibility by 12.
4
Calculate nn.
n=2431=163=48n = 2^4 \cdot 3^1 = 16 \cdot 3 = 48.
Substituting a=4a=4 and b=1b=1 into the prime factorization of nn gives 48.

Anahtar Kavram

Divisibility rules and prime factorization count of positive divisors
Soru 648Soru

If 3(x2)+4=2x+113(x - 2) + 4 = 2x + 11, what is the value of xx?

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Cevap: 13

Cevap

The value of xx is 13.
Expanding the left side yields 3x6+4=3x23x - 6 + 4 = 3x - 2. Equating this to the right side gives 3x2=2x+113x - 2 = 2x + 11. Subtracting 2x2x from both sides gives x2=11x - 2 = 11, and adding 22 to both sides yields x=13x = 13.

Adım Adım Çözüm

1
Expand the left side of the equation
3x6+4=2x+113x - 6 + 4 = 2x + 11
Apply the distributive property 3(x2)=3x63(x - 2) = 3x - 6.
2
Combine like constant terms on the left side
3x2=2x+113x - 2 = 2x + 11
Combine 6-6 and +4+4 to get 2-2.
3
Subtract 2x2x from both sides
x2=11x - 2 = 11
Move variable terms to one side of the equation.
4
Add 2 to both sides
x=13x = 13
Isolate the variable xx.

Anahtar Kavram

Solving Linear Equations in One Variable
Soru 649Soru

At a biotechnology facility, a solution tank initially contains a mixture of ethanol, water, and stabilizer liquid. By volume, 0.360.36 of the initial mixture is ethanol and 25\frac{2}{5} is water, with the remainder consisting of stabilizer liquid. During a purification process, 25%25\% of the ethanol and 18\frac{1}{8} of the water evaporate, while the volume of the stabilizer liquid remains unchanged. If the volume of the remaining stabilizer liquid is 132132 liters, by what percentage did the total volume of the liquid mixture in the tank decrease?

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Cevap: 14%

Cevap

The total volume of the liquid mixture in the tank decreased by 14%.
The initial mixture consists of 36% ethanol, 40% water, and 24% stabilizer by volume. Ethanol loses 25% of its volume, which corresponds to 0.25×36%=9%0.25 \times 36\% = 9\% of the total initial mixture volume. Water loses 18\frac{1}{8} of its volume, corresponding to 18×40%=5%\frac{1}{8} \times 40\% = 5\% of the total initial mixture volume. Combining these losses gives 9%+5%=14%9\% + 5\% = 14\% net decrease relative to the initial total volume.

Adım Adım Çözüm

1
Express initial component proportions as unified decimals or fractions.
Ethanol fraction = 0.36=9250.36 = \frac{9}{25}. Water fraction = 25=0.40=1025\frac{2}{5} = 0.40 = \frac{10}{25}. Stabilizer fraction = 1(0.36+0.40)=0.24=6251 - (0.36 + 0.40) = 0.24 = \frac{6}{25}.
Establishing all initial parts relative to the total initial volume allows step-by-step proportion tracking.
2
Calculate the fraction of total volume lost due to evaporation for each component.
Ethanol loss = 25%×0.36=0.25×0.36=0.0925\% \times 0.36 = 0.25 \times 0.36 = 0.09 of initial total volume. Water loss = 18×0.40=0.125×0.40=0.05\frac{1}{8} \times 0.40 = 0.125 \times 0.40 = 0.05 of initial total volume.
Applying the component-specific percentage decrease to each component's fraction yields its contribution to the overall volume loss.
3
Sum the component volume losses to find the total percentage decrease.
Total volume loss fraction = 0.09+0.05=0.140.09 + 0.05 = 0.14, which equals 14%14\% of the total initial volume.
The stabilizer experience no loss, so the net change in total volume is the direct sum of the losses in ethanol and water relative to the original base.

Anahtar Kavram

Multi-step arithmetic operations combining decimals, fractions, and weighted percent changes using a consistent base value.
Tahmini Süre:1m 50s
Soru 650Soru

A positive integer nn is a multiple of 66, but is neither a multiple of 44 nor a multiple of 99. If nn has exactly 1212 positive integer divisors and n2n^2 has exactly 4545 positive integer divisors, how many positive integer divisors of n2n^2 are divisible by nn?

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Cevap: 12

Cevap

The number of positive integer divisors of n2n^2 that are divisible by nn is 1212.
By analyzing the prime factorization n=2131p2n = 2^1 \cdot 3^1 \cdot p^2, we find n2=2232p4n^2 = 2^2 \cdot 3^2 \cdot p^4. For a divisor 2x3ypz2^x \cdot 3^y \cdot p^z of n2n^2 to be divisible by nn, the exponents must satisfy x{1,2}x \in \{1, 2\}, y{1,2}y \in \{1, 2\}, and z{2,3,4}z \in \{2, 3, 4\}, yielding 2×2×3=122 \times 2 \times 3 = 12 valid divisors.

Adım Adım Çözüm

1
Analyze the prime factorization of nn for prime factors 22 and 33.
The exponents of 22 and 33 in nn are both 11.
Since nn is a multiple of 66, it must contain at least one factor of 22 and one factor of 33. Because nn is not a multiple of 44, the exponent of 22 cannot exceed 11. Because nn is not a multiple of 99, the exponent of 33 cannot exceed 11.
2
Determine the remaining prime factors of nn using the total divisor count of nn.
n=2131p2n = 2^1 \cdot 3^1 \cdot p^2 for some prime p>3p > 3.
The number of positive divisors is given by d(n)=(1+1)(1+1)(ci+1)=12d(n) = (1+1)(1+1)\prod(c_i+1) = 12, which simplifies to 4(ci+1)=124\prod(c_i+1) = 12, so (ci+1)=3\prod(c_i+1) = 3. Since 33 is prime, there is exactly one additional prime factor pp with exponent c=2c = 2.
3
Verify with the divisor count of n2n^2.
d(n2)=(2(1)+1)(2(1)+1)(2(2)+1)=335=45d(n^2) = (2(1)+1)(2(1)+1)(2(2)+1) = 3 \cdot 3 \cdot 5 = 45.
Squaring nn doubles all prime exponents, so n2=2232p4n^2 = 2^2 \cdot 3^2 \cdot p^4, which has (2+1)(2+1)(4+1)=45(2+1)(2+1)(4+1) = 45 positive divisors, consistent with the given information.
4
Calculate the number of divisors of n2n^2 that are multiples of nn.
1212
Any divisor of n2n^2 has the form 2x3ypz2^x \cdot 3^y \cdot p^z with 0x20 \le x \le 2, 0y20 \le y \le 2, and 0z40 \le z \le 4. For this divisor to be a multiple of n=2131p2n = 2^1 \cdot 3^1 \cdot p^2, the exponents must satisfy 1x21 \le x \le 2 (22 choices), 1y21 \le y \le 2 (22 choices), and 2z42 \le z \le 4 (33 choices). Multiplying the choices gives 223=122 \cdot 2 \cdot 3 = 12.

Anahtar Kavram

Divisor counting formula and prime factor exponent constraints
Tahmini Süre:2m 0s
Soru 651Soru

If mm and nn are positive two-digit integers such that m<nm < n, GCD(m,n)=12\text{GCD}(m, n) = 12, and LCM(m,n)=420\text{LCM}(m, n) = 420, what is the value of nmn - m?

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Cevap: 2424

Cevap

The value of nmn - m is 2424.
Writing m=12xm = 12x and n=12yn = 12y for coprime integers x<yx < y gives LCM(m,n)=12xy=420\text{LCM}(m, n) = 12xy = 420, which simplifies to xy=35xy = 35. The coprime factor pairs of 3535 are (1,35)(1, 35) and (5,7)(5, 7). The pair (1,35)(1, 35) yields m=12m = 12 and n=420n = 420, but 420420 is not a two-digit integer. The pair (5,7)(5, 7) gives two-digit integers m=60m = 60 and n=84n = 84. The difference is 8460=2484 - 60 = 24.

Adım Adım Çözüm

1
Express mm and nn in terms of their GCD.
Let m=12xm = 12x and n=12yn = 12y, where xx and yy are coprime positive integers with x<yx < y.
Factoring out the greatest common divisor GCD(m,n)=12\text{GCD}(m, n) = 12 leaves coprime factors xx and yy.
2
Relate GCD and LCM to determine xyx \cdot y.
LCM(m,n)=12xy=420    xy=35\text{LCM}(m, n) = 12 \cdot x \cdot y = 420 \implies x \cdot y = 35.
For integers written as 12x12x and 12y12y with GCD(x,y)=1\text{GCD}(x,y)=1, the least common multiple is 12xy12xy.
3
Determine the valid pair (x,y)(x, y) that satisfies the two-digit constraint.
The coprime factor pairs of 3535 with x<yx < y are (1,35)(1, 35) and (5,7)(5, 7). For (1,35)(1, 35), m=12m = 12 and n=420n = 420 (a three-digit number). For (5,7)(5, 7), m=60m = 60 and n=84n = 84 (both are two-digit numbers). Thus, (x,y)=(5,7)(x, y) = (5, 7).
The problem specifies that both mm and nn are two-digit positive integers.
4
Compute nmn - m.
nm=8460=24n - m = 84 - 60 = 24.
Subtracting m=60m = 60 from n=84n = 84 gives 2424.

Anahtar Kavram

Properties of Greatest Common Divisor (GCD) and Least Common Multiple (LCM)
Soru 652Soru

A total of 360 identical notebooks are to be distributed equally among nn students, where n>1n > 1, such that each student receives an integer number of notebooks strictly greater than 1. If nn must be a multiple of 4, how many different values of nn are possible?

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Cevap: 11

Cevap

The total number of possible values for nn is 11.
To find the number of valid values of nn, prime factorize 360=23×32×51360 = 2^3 \times 3^2 \times 5^1. For nn to be a factor of 360 and a multiple of 4, the exponent of 2 must be 2 or 3 (2 choices), the exponent of 3 can be 0, 1, or 2 (3 choices), and the exponent of 5 can be 0 or 1 (2 choices), yielding 2×3×2=122 \times 3 \times 2 = 12 total factors of 360 that are multiples of 4. Excluding n=360n = 360, which results in 1 notebook per student, leaves 121=1112 - 1 = 11 valid values.

Adım Adım Çözüm

1
Prime factorize the total number of notebooks, 360.
360=23×32×51360 = 2^3 \times 3^2 \times 5^1
Prime factorization allows us to systematically count factors meeting specific divisibility conditions.
2
Calculate the number of factors of 360 that are multiples of 4.
2 choices for the exponent of 2 (222^2 or 232^3), 3 choices for the exponent of 3 (30,31,323^0, 3^1, 3^2), and 2 choices for the exponent of 5 (50,515^0, 5^1). Total factors = 2×3×2=122 \times 3 \times 2 = 12.
A factor is a multiple of 4 if and only if its prime factorization contains at least two factors of 2.
3
Apply the constraint that each student receives strictly more than 1 notebook.
If n=360n = 360, each student receives 360360=1\frac{360}{360} = 1 notebook, which violates the condition. Subtracting this case yields 121=1112 - 1 = 11 valid values.
The problem requires each student to receive an integer number of notebooks strictly greater than 1.

Anahtar Kavram

Counting constrained factors using prime factorization
Soru 653Soru

What is the value of 26+26+26+26\sqrt{2^6 + 2^6 + 2^6 + 2^6}?

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Cevap: 16

Cevap

16
The four identical terms 26+26+26+262^6 + 2^6 + 2^6 + 2^6 sum to 4×264 \times 2^6. Since 4=224 = 2^2, the expression inside the radical becomes 22×26=282^2 \times 2^6 = 2^8. Taking the square root yields 28=24=16\sqrt{2^8} = 2^4 = 16.

Adım Adım Çözüm

1
Express the sum of the four identical terms as a product.
26+26+26+26=4×262^6 + 2^6 + 2^6 + 2^6 = 4 \times 2^6
Adding a number to itself 4 times is equivalent to multiplying that number by 4.
2
Convert 4 to a power of 2 and combine exponents.
4×26=22×26=22+6=284 \times 2^6 = 2^2 \times 2^6 = 2^{2+6} = 2^8
When multiplying powers with the same base, add their exponents.
3
Simplify the square root of 282^8.
28=(28)1/2=24=16\sqrt{2^8} = (2^8)^{1/2} = 2^4 = 16
Taking the square root of a power means dividing the exponent by 2.

Anahtar Kavram

Factoring identical exponential terms and applying exponent laws to simplify radicals.
Soru 654Soru

If 3x92=3103^x \cdot 9^2 = 3^{10}, what is the value of xx?

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Cevap: 6

Cevap

The value of xx is 6.
Rewriting 929^2 as (32)2=34(3^2)^2 = 3^4 converts the expression to a common base of 33. Applying the product rule gives 3x34=3x+43^x \cdot 3^4 = 3^{x+4}. Setting the exponent x+4x+4 equal to 1010 yields x=6x = 6.

Adım Adım Çözüm

1
Express all powers using the common base 3
92=(32)2=349^2 = (3^2)^2 = 3^4
Converting terms to a common base allows application of exponent rules.
2
Apply the product rule for exponents
3x34=3x+43^x \cdot 3^4 = 3^{x+4}
When multiplying terms with identical bases, add the exponents.
3
Equate the exponents and solve for x
x+4=10    x=6x + 4 = 10 \implies x = 6
If bm=bnb^m = b^n for b>0b > 0 and b1b \neq 1, then m=nm = n.

Anahtar Kavram

Exponent rules with base conversion and product of powers
Soru 655Soru

If the sum of a set of 77 consecutive odd integers is 105105, what is the median integer of the set?

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Cevap: 15

Cevap

The median integer of the set is 15.
For any set of evenly spaced numbers, such as consecutive odd integers, the arithmetic mean is equal to the median. Since the sum of the 77 integers is 105105, the mean is 105÷7=15105 \div 7 = 15. Therefore, the median of the set is 1515.

Adım Adım Çözüm

1
Apply the property of evenly spaced sets.
For consecutive odd integers, the mean of the set is equal to its median.
The numbers in an arithmetic sequence are symmetrically distributed around the middle value.
2
Calculate the arithmetic mean.
Mean = SumNumber of terms=1057=15\frac{\text{Sum}}{\text{Number of terms}} = \frac{105}{7} = 15.
Dividing the sum of the terms by the count gives the average value.
3
Determine the median.
Median = 1515.
Because mean equals median for evenly spaced sets, the median must be 15.

Anahtar Kavram

Average and Median Equivalence in Evenly Spaced Sets
Tahmini Süre:45s
Soru 656Soru

Two positive integers mm and nn satisfy GCD(m,n)=60\text{GCD}(m, n) = 60 and LCM(m,n)=75600\text{LCM}(m, n) = 75{}600. Given that mm is divisible by 6363 but not by 189189, and nn is divisible by 400400, what is the value of mm?

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Cevap: 1260

Cevap

1260
By finding the prime factorizations of the GCD (2231512^2 \cdot 3^1 \cdot 5^1) and LCM (243352712^4 \cdot 3^3 \cdot 5^2 \cdot 7^1), we determine the prime powers for mm and nn using min\min and max\max rules. The constraint that nn is divisible by 400=2452400 = 2^4 \cdot 5^2 fixes e2(n)=4e_2(n)=4 and e5(n)=2e_5(n)=2, which forces e2(m)=2e_2(m)=2 and e5(m)=1e_5(m)=1. The constraint that mm is divisible by 63=327163 = 3^2 \cdot 7^1 but not 189=3371189 = 3^3 \cdot 7^1 fixes e3(m)=2e_3(m)=2 and e7(m)=1e_7(m)=1. Evaluating m=22325171m = 2^2 \cdot 3^2 \cdot 5^1 \cdot 7^1 gives 12601{}260.

Adım Adım Çözüm

1
Prime factorize the GCD and LCM
GCD = 2^2 * 3^1 * 5^1 * 7^0 and LCM = 2^4 * 3^3 * 5^2 * 7^1
GCD represents the minimum exponent of each prime shared by m and n, whereas LCM represents the maximum exponent.
2
Determine the prime exponents for m using divisibility conditions
e_2(m) = 2, e_3(m) = 2, e_5(m) = 1, e_7(m) = 1
Divisibility of n by 400 forces e_2(n)=4 and e_5(n)=2, leaving e_2(m)=2 and e_5(m)=1. Divisibility of m by 63 but not 189 fixes e_3(m)=2 and e_7(m)=1.
3
Compute the product of prime powers for m
m = 4 * 9 * 5 * 7 = 1260
Multiplying the determined prime factors yields the exact value of integer m.

Anahtar Kavram

Greatest Common Divisor (GCD) and Least Common Multiple (LCM) Prime Exponent Rules
Tahmini Süre:2m 0s
Soru 657Soru

Two positive integers xx and yy satisfy x<yx < y, GCD(x,y)=15\text{GCD}(x, y) = 15, and LCM(x,y)=1,800\text{LCM}(x, y) = 1,800. If xx is not a multiple of 99 and yy is not a multiple of 2525, what is the value of yxy - x?

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Cevap: 285

Cevap

The value of yxy - x is 285.
Expressing xx and yy as 15a15a and 15b15b yields ab=120ab = 120. Testing coprime pairs (a,b)(a,b) shows that only (5,24)(5, 24) gives values x=75x = 75 and y=360y = 360 that satisfy both constraints (75 is not a multiple of 9, and 360 is not a multiple of 25). The difference is 36075=285360 - 75 = 285.

Adım Adım Çözüm

1
Set up coprime factor representation
x=15ax = 15a and y=15by = 15b with GCD(a,b)=1\text{GCD}(a, b) = 1 and a<ba < b
Any two numbers can be expressed as their GCD multiplied by coprime integer quotients.
2
Solve for the product of coprime quotient factors aba \cdot b
ab=120a \cdot b = 120
The LCM of two numbers divided by their GCD equals the product of their coprime quotients.
3
Identify candidate coprime pairs (a,b)(a, b)
(1,120),(3,40),(5,24),(8,15)(1, 120), (3, 40), (5, 24), (8, 15)
Since 120=233151120 = 2^3 \cdot 3^1 \cdot 5^1, all prime factor powers must be assigned entirely to either aa or bb to maintain GCD(a,b)=1\text{GCD}(a, b) = 1.
4
Filter pairs using divisibility constraints on xx and yy
Only (75,360)(75, 360) satisfies x≢0(mod9)x \not\equiv 0 \pmod 9 and y≢0(mod25)y \not\equiv 0 \pmod{25}
75=32575 = 3 \cdot 25 (not divisible by 9) and 360=895360 = 8 \cdot 9 \cdot 5 (not divisible by 25).
5
Calculate the target difference yxy - x
285
36075=285360 - 75 = 285.

Anahtar Kavram

Relationship between GCD, LCM, and prime factorization distribution in coprime quotients.
Soru 658Soru

A cloud data center processes an incoming raw data stream through sequential filtering stages. In Stage 1, 38\frac{3}{8} of the incoming raw data volume is discarded as noise, and 0.200.20 of the remaining data is flagged for long-term archiving. In Stage 2, 45\frac{4}{5} of the data not flagged for long-term archiving is processed into active storage, while the rest is discarded. If the volume of data processed into active storage in Stage 2 is 5050 terabytes greater than the volume of data discarded in Stage 1, what was the initial volume of the raw data stream, in terabytes?

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Cevap: 2,000

Cevap

2,000 terabytes
Let the initial data volume be XX. Discarded in Stage 1 is 38X\frac{3}{8}X, leaving 58X\frac{5}{8}X. Archiving takes 20%20\% (0.20=150.20 = \frac{1}{5}) of this remainder, which is 15×58X=18X\frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X. The data entering Stage 2 is 58X18X=12X\frac{5}{8}X - \frac{1}{8}X = \frac{1}{2}X. Active storage in Stage 2 is 45×12X=25X\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X. Subtracting Stage 1 discarded volume from Stage 2 active storage volume gives 25X38X=1640X1540X=140X\frac{2}{5}X - \frac{3}{8}X = \frac{16}{40}X - \frac{15}{40}X = \frac{1}{40}X. Setting 140X=50\frac{1}{40}X = 50 yields X=2,000X = 2,000 terabytes.

Adım Adım Çözüm

1
Define the variable and compute Stage 1 discarded volume and remaining volume.
Discarded in Stage 1 = 38X\frac{3}{8}X; Remaining after Stage 1 discard = X38X=58XX - \frac{3}{8}X = \frac{5}{8}X.
Establishing quantities in terms of the total initial volume XX allows setting up a single-variable linear equation.
2
Calculate the volume flagged for archiving and the volume available for Stage 2.
Archived volume = 0.20×58X=15×58X=18X0.20 \times \frac{5}{8}X = \frac{1}{5} \times \frac{5}{8}X = \frac{1}{8}X. Available for Stage 2 = \frac{5}{8}X - \frac{1}{8}X = \frac{4}{8}X = \frac{1}{2}X$.
The 0.200.20 decimal must be converted to a fraction (15\frac{1}{5}) and applied to the remaining 58X\frac{5}{8}X base.
3
Calculate the volume processed into active storage in Stage 2.
Active storage volume = 45×12X=25X\frac{4}{5} \times \frac{1}{2}X = \frac{2}{5}X.
Stage 2 processes 45\frac{4}{5} of the data that entered Stage 2.
4
Set up the algebraic equation comparing active storage volume in Stage 2 to discarded volume in Stage 1.
25X38X=50\frac{2}{5}X - \frac{3}{8}X = 50.
The problem states that active storage volume in Stage 2 is 5050 terabytes greater than the Stage 1 discarded volume.
5
Find a common denominator and solve for XX.
(16401540)X=50    140X=50    X=2,000\left(\frac{16}{40} - \frac{15}{40}\right)X = 50 \implies \frac{1}{40}X = 50 \implies X = 2,000.
Converting fractions to a denominator of 4040 yields 140X=50\frac{1}{40}X = 50, so multiplying by 4040 gives X=2,000X = 2,000 terabytes.

Anahtar Kavram

Multi-step arithmetic operations involving sequential fractions, decimals, and percent bases
Tahmini Süre:2m 0s
Soru 659Soru

Let K=783442K = 7^{83} - 4^{42}. What is the remainder when K-K is divided by 55?

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Cevap: 3

Cevap

3
To find the remainder of K-K divided by 5, first calculate K(mod5)K \pmod 5. By analyzing powers of 7 modulo 5, 712,724,733,7417^1 \equiv 2, 7^2 \equiv 4, 7^3 \equiv 3, 7^4 \equiv 1, so 783=74(20)+33(mod5)7^{83} = 7^{4(20)+3} \equiv 3 \pmod 5. For powers of 4 modulo 5, 414,4214^1 \equiv 4, 4^2 \equiv 1, so 4421(mod5)4^{42} \equiv 1 \pmod 5. Therefore, K31=2(mod5)K \equiv 3 - 1 = 2 \pmod 5. Consequently, K2(mod5)-K \equiv -2 \pmod 5. Adding the divisor 5 gives a non-negative remainder of 3.

Adım Adım Çözüm

1
Find the remainder of 7837^{83} when divided by 55.
7127^1 \equiv 2, 7247^2 \equiv 4, 7337^3 \equiv 3, 741(mod5)7^4 \equiv 1 \pmod 5. Since 83=4×20+383 = 4 \times 20 + 3, 783733(mod5)7^{83} \equiv 7^3 \equiv 3 \pmod 5.
Powers of 7 follow a 4-term repeating remainder cycle modulo 5: (2, 4, 3, 1).
2
Find the remainder of 4424^{42} when divided by 55.
4144^1 \equiv 4, 421(mod5)4^2 \equiv 1 \pmod 5. Since 42 is even, 4421(mod5)4^{42} \equiv 1 \pmod 5.
Powers of 4 follow a 2-term repeating remainder cycle modulo 5: (4, 1).
3
Compute K(mod5)K \pmod 5.
K31=2(mod5)K \equiv 3 - 1 = 2 \pmod 5.
Modular subtraction allows evaluating terms independently.
4
Determine the positive remainder of K(mod5)-K \pmod 5.
K22+5=3(mod5)-K \equiv -2 \equiv -2 + 5 = 3 \pmod 5.
Standard division algorithm requires the remainder rr when dividing by 5 to satisfy 0r<50 \le r < 5.

Anahtar Kavram

Modular cyclicity and negative remainder conversion
Tahmini Süre:2m 0s
Soru 660Soru

If the sum of three consecutive even integers is 4242, what is the smallest of these integers?

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Cevap: 1212

Cevap

1212
The arithmetic mean (average) of three consecutive even integers is equal to the middle integer. Dividing the total sum 4242 by 33 yields 1414, which is the middle integer. Subtracting 22 gives the smallest integer, 1212.

Adım Adım Çözüm

1
Define the three consecutive even integers algebraically
Let the three consecutive even integers be nn, n+2n + 2, and n+4n + 4, where nn represents the smallest integer.
Consecutive even integers differ by 22.
2
Set up the equation for their sum
n+(n+2)+(n+4)=42    3n+6=42n + (n + 2) + (n + 4) = 42 \implies 3n + 6 = 42
The problem states that the sum of the three integers equals 4242.
3
Solve for nn
3n=36    n=123n = 36 \implies n = 12
Subtract 66 from both sides and divide by 33 to find the smallest integer.

Anahtar Kavram

Properties and Sums of Consecutive Even Integers
Tahmini Süre:45s
ÖncekiSayfa 33 / 110Sonraki
Tüm alıştırma soruları — GMAT | Examkin