Number Properties and Arithmetic

232 soru

Soru 161Soru

If n=24×53×71n = 2^4 \times 5^3 \times 7^1, how many positive integer divisors of nn are divisible by 1010 but are not multiples of 3535?

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Cevap: 12

Cevap

12
The value 12 is correct because a divisor 2a×5b×7c2^a \times 5^b \times 7^c of 24×53×712^4 \times 5^3 \times 7^1 is divisible by 10 when a{1,2,3,4}a \in \{1, 2, 3, 4\} (4 options) and b{1,2,3}b \in \{1, 2, 3\} (3 options). To ensure it is not divisible by 35 (5×75 \times 7), cc cannot be 1 when b1b \ge 1. Thus cc must be 0 (1 option). Multiplying the choices yields 4×3×1=124 \times 3 \times 1 = 12.

Adım Adım Çözüm

1
Express the general form of a divisor of nn.
Any divisor dd of n=24×53×71n = 2^4 \times 5^3 \times 7^1 can be written as d=2a×5b×7cd = 2^a \times 5^b \times 7^c, where 0a40 \le a \le 4, 0b30 \le b \le 3, and 0c10 \le c \le 1.
Prime factorization determines the structure of all positive divisors.
2
Apply the condition that dd must be divisible by 1010.
Since 10=21×5110 = 2^1 \times 5^1, we require a1a \ge 1 and b1b \ge 1.
A number is divisible by 10 if and only if it contains at least one factor of 2 and at least one factor of 5.
3
Apply the condition that dd must NOT be a multiple of 3535.
Since 35=51×7135 = 5^1 \times 7^1, a multiple of 35 must have b1b \ge 1 and c1c \ge 1. Since b1b \ge 1 is already satisfied, to avoid being a multiple of 35, we must set c=0c = 0.
If c=1c = 1, then combined with b1b \ge 1, the divisor would automatically be a multiple of 35.
4
Calculate the total number of valid choices for (a,b,c)(a, b, c).
a{1,2,3,4}a \in \{1, 2, 3, 4\} (4 options), b{1,2,3}b \in \{1, 2, 3\} (3 options), c{0}c \in \{0\} (1 option). Total = 4×3×1=124 \times 3 \times 1 = 12.
By the Fundamental Counting Principle, multiplying the independent choices gives the number of valid divisors.

Anahtar Kavram

Counting Divisors with Prime Factor Constraints
Tahmini Süre:1m 30s
Soru 162Soru

A positive integer nn is not divisible by 33. If nn has exactly 1515 positive divisors and 2n2n has exactly 2020 positive divisors, how many positive divisors does 3n3n have?

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Cevap: 30

Cevap

30
The number of positive divisors of an integer with prime factorization p1e1p2e2p_1^{e_1} p_2^{e_2} \dots is (e1+1)(e2+1)(e_1 + 1)(e_2 + 1) \dots. Given d(n)=15d(n) = 15, nn can be p14p^{14} or p4q2p^4 q^2. Given d(2n)=20d(2n) = 20, 2 must be a prime factor of nn with exponent 2, making n=22p4n = 2^2 p^4. Since nn is not divisible by 3, p3p \neq 3. Multiplying nn by 3 introduces 313^1 into the prime factorization, giving 3n=2231p43n = 2^2 \cdot 3^1 \cdot p^4. The total number of positive divisors is (2+1)(1+1)(4+1)=30(2+1)(1+1)(4+1) = 30.

Adım Adım Çözüm

1
Analyze the number of positive divisors of nn.
nn is either p14p^{14} or p4q2p^4 q^2 for distinct primes pp and qq.
The number of positive divisors of n=p1e1p2e2n = p_1^{e_1} p_2^{e_2} \dots is given by (e1+1)(e2+1)(e_1 + 1)(e_2 + 1) \dots. Since 15=15×1=5×315 = 15 \times 1 = 5 \times 3, the exponent structure must be 14 or 4 and 2.
2
Analyze the number of positive divisors of 2n2n.
n=22p4n = 2^2 p^4, where pp is a prime other than 2 and 3.
If n=p4q2n = p^4 q^2, multiplying by 2 increases the exponent of 2 by 1. If q=2q = 2, then n=22p4n = 2^2 p^4 and 2n=23p42n = 2^3 p^4, yielding (3+1)(4+1)=20(3 + 1)(4 + 1) = 20 divisors. Any other prime structure yields a different count.
3
Determine the prime factorization of 3n3n and count its positive divisors.
The number of divisors of 3n3n is 30.
Since nn is not divisible by 3, p3p \neq 3. Thus 3n=2231p43n = 2^2 \cdot 3^1 \cdot p^4, which has (2+1)(1+1)(4+1)=30(2 + 1)(1 + 1)(4 + 1) = 30 positive divisors.

Anahtar Kavram

Divisor Count Formula and Prime Factorization
Soru 163Soru

Let K=2a347bK = 2^a \cdot 3^4 \cdot 7^b, where aa and bb are positive integers. If KK has exactly 4848 positive integer divisors that are multiples of 66, and exactly 3030 positive integer divisors that are multiples of 1414, what is the total number of positive integer divisors of KK?

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Cevap: 70

Cevap

The total number of positive integer divisors of KK is 70.
The prime factorization of KK is 2a347b2^a \cdot 3^4 \cdot 7^b. Any positive divisor of KK takes the form 2x3y7z2^x \cdot 3^y \cdot 7^z, where 0xa0 \le x \le a, 0y40 \le y \le 4, and 0zb0 \le z \le b. A divisor is a multiple of 6=21316 = 2^1 \cdot 3^1 if x1x \ge 1, y1y \ge 1, and z0z \ge 0. The number of such divisors is a×4×(b+1)=48a \times 4 \times (b+1) = 48, which simplifies to a(b+1)=12a(b+1) = 12. A divisor is a multiple of 14=217114 = 2^1 \cdot 7^1 if x1x \ge 1, y0y \ge 0, and z1z \ge 1. The number of such divisors is a×5×b=30a \times 5 \times b = 30, which simplifies to ab=6ab = 6. Substituting ab=6ab = 6 into ab+a=12ab + a = 12 gives 6+a=12    a=66 + a = 12 \implies a = 6, which means b=1b = 1. The total number of positive integer divisors of K=263471K = 2^6 \cdot 3^4 \cdot 7^1 is (6+1)(4+1)(1+1)=7×5×2=70(6+1)(4+1)(1+1) = 7 \times 5 \times 2 = 70.

Adım Adım Çözüm

1
Express the condition for divisors of KK being multiples of 6 in terms of exponents.
a4(b+1)=48    a(b+1)=12a \cdot 4 \cdot (b+1) = 48 \implies a(b+1) = 12
A divisor of K=2a347bK = 2^a \cdot 3^4 \cdot 7^b has the form 2x3y7z2^x \cdot 3^y \cdot 7^z with 0xa0 \le x \le a, 0y40 \le y \le 4, 0zb0 \le z \le b. For it to be a multiple of 6=21316 = 2^1 \cdot 3^1, we must have x1x \ge 1 (aa choices), y1y \ge 1 (44 choices), and z0z \ge 0 (b+1b+1 choices).
2
Express the condition for divisors of KK being multiples of 14 in terms of exponents.
a5b=30    ab=6a \cdot 5 \cdot b = 30 \implies ab = 6
For a divisor to be a multiple of 14=217114 = 2^1 \cdot 7^1, we must have x1x \ge 1 (aa choices), y0y \ge 0 (55 choices), and z1z \ge 1 (bb choices).
3
Solve the system of equations for the positive integer exponents aa and bb.
a=6a = 6 and b=1b = 1
Expanding a(b+1)=12a(b+1) = 12 gives ab+a=12ab + a = 12. Substituting ab=6ab = 6 yields 6+a=12    a=66 + a = 12 \implies a = 6. Then 6b=6    b=16b = 6 \implies b = 1.
4
Calculate the total number of positive integer divisors of K=263471K = 2^6 \cdot 3^4 \cdot 7^1.
(6+1)(4+1)(1+1)=7×5×2=70(6+1)(4+1)(1+1) = 7 \times 5 \times 2 = 70
The total number of divisors of a prime factorization p1e1p2e2pkekp_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).

Anahtar Kavram

Counting Divisors using Prime Factor Exponents
Tahmini Süre:2m 0s
Soru 164Soru

A store owner purchases a batch of books for $150\$150 each and sells them at a price that is 40%40\% higher than the purchase price. During a clearance sale, the selling price is reduced by 10%10\%. What is the final sale price, in dollars, of one book?

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Cevap: 189

Cevap

The final sale price of one book is $189\$189.
First, calculate the price after the 40%40\% markup: $150×1.40=$210\$150 \times 1.40 = \$210. Then apply the 10%10\% discount to this new amount: $210×0.90=$189\$210 \times 0.90 = \$189. The final price is $189\$189.

Adım Adım Çözüm

1
Calculate the price after a 40% markup on the base price of $150.
$210
A 40% markup means multiplying the original purchase price by 1.40.
2
Apply the 10% clearance discount to the marked-up price of $210.
$189
A 10% discount means multiplying the new price by 0.90.

Anahtar Kavram

Successive Percentage Change
Tahmini Süre:45s
Soru 165Soru

At a technology firm, 38\frac{3}{8} of the total annual operating budget was initially allocated to Research & Development, 0.350.35 of the budget was allocated to Marketing, and the remainder was allocated to Operations. Mid-year, the Marketing allocation was increased by 20%20\% of its initial value, and the Operations allocation was decreased by 40%40\% of its initial value, while the Research & Development allocation remained unchanged. By what net percentage did the firm's total annual operating budget change?

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Cevap: A decrease of 4%4\%

Cevap

A net decrease of 4%4\%
Converting 38\frac{3}{8} to 37.5%37.5\% leaves 27.5%27.5\% for Operations (100%37.5%35%100\% - 37.5\% - 35\%). A 20%20\% increase on the 35%35\% Marketing budget adds 7%7\% to the overall total (0.20×35%=+7%0.20 \times 35\% = +7\%). A 40%40\% decrease on the 27.5%27.5\% Operations budget reduces the overall total by 11%11\% (0.40×27.5%=11%0.40 \times 27.5\% = -11\%). Summing these changes gives +7%11%=4%+7\% - 11\% = -4\%, representing a net decrease of 4%4\%.

Adım Adım Çözüm

1
Convert initial fractional and decimal allocations to percentages of the total budget
Research & Development allocation = 38=37.5%\frac{3}{8} = 37.5\%; Marketing allocation = 0.35=35%0.35 = 35\%
Converting all components to a common percentage format enables straightforward comparison and arithmetic.
2
Calculate the initial percentage allocated to Operations
Operations allocation = 100%(37.5%+35%)=100%72.5%=27.5%100\% - (37.5\% + 35\%) = 100\% - 72.5\% = 27.5\%
The remainder of the budget after R&D and Marketing constitutes the Operations department share.
3
Determine the net percentage change contributed by each department's adjustment
Marketing change = +20% of 35%=+7%+20\% \text{ of } 35\% = +7\% of total budget; Operations change = 40% of 27.5%=11%-40\% \text{ of } 27.5\% = -11\% of total budget; R&D change = 0%0\%
Each departmental percentage change must be weighted by that department's portion of the overall budget.
4
Sum the net contributions to find the overall budget change
Net overall change = +7%11%=4%+7\% - 11\% = -4\% (a decrease of 4%4\%)
Combining the positive and negative adjustments yields the overall net percentage change relative to the initial budget.

Anahtar Kavram

Weighted Percent Change across Mixed Fractional and Decimal Sub-components
Soru 166Soru

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 167Soru

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 168Soru

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 169Soru

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 170Soru

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 171Soru

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 172Soru

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 173Soru

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 174Soru

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 175Soru

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 176Soru

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 177Soru

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 178Soru

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 179Soru

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 180Soru

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.
ÖncekiSayfa 9 / 12Sonraki
Number Properties and Arithmetic Alıştırma Soruları — GMAT — Sayfa 9 | Examkin