Number Properties and Arithmetic

232 soru

Soru 81Soru

If xx is a negative integer and yy is a positive integer, what is the value of xyyxx+y+100\frac{|x - y| - |y - x|}{x + y + 100}?

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Cevap: 0

Cevap

0
Given x<0x < 0 and y>0y > 0, the quantity xyx - y is strictly negative, which means xy=(xy)=yx|x - y| = -(x - y) = y - x. The quantity yxy - x is strictly positive, so yx=yx|y - x| = y - x. Thus, the numerator simplifies to (yx)(yx)=0(y - x) - (y - x) = 0. Dividing zero by any non-zero denominator yields 00.

Adım Adım Çözüm

1
Determine the signs of the terms inside the absolute value functions
Since x<0x < 0 and y>0y > 0, xy<0x - y < 0 and yx>0y - x > 0.
Subtracting a positive number from a negative number yields a negative value, while subtracting a negative number from a positive number yields a positive value.
2
Simplify the absolute value expressions
xy=yx|x - y| = y - x and yx=yx|y - x| = y - x.
The absolute value of a negative number is its negation, and the absolute value of a positive number is the number itself.
3
Compute the difference in the numerator
xyyx=(yx)(yx)=0|x - y| - |y - x| = (y - x) - (y - x) = 0.
Subtracting an algebraic expression from an identical expression results in zero.
4
Evaluate the entire fraction
0x+y+100=0\frac{0}{x + y + 100} = 0.
Zero divided by any non-zero real number is zero.

Anahtar Kavram

Positive and Negative Number Properties with Absolute Value
Soru 82Soru

At the beginning of a fiscal year, a municipal transit authority allocated its capital expenditure budget among three projects: Bus Rapid Transit, Rail Modernization, and Station Upgrades. Exactly 0.300.30 of the total budget was allocated to Bus Rapid Transit. Of the remaining budget, exactly 37\frac{3}{7} was allocated to Rail Modernization, and the rest was allocated to Station Upgrades. By the end of the year, expenditures on Bus Rapid Transit exceeded its initial allocation by 25%25\%, expenditures on Rail Modernization were 15%15\% below its initial allocation, and expenditures on Station Upgrades exceeded its initial allocation by 10%10\%. By what percent did the transit authority's total expenditures across all three projects exceed its initial total budget?

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Cevap: 7

Cevap

The total expenditures across all three projects exceeded the initial total budget by 7%7\%.
To solve this problem, represent the total initial budget as BB. The Bus Rapid Transit allocation is 0.30B0.30B, leaving 0.70B0.70B. Rail Modernization receives 37\frac{3}{7} of 0.70B0.70B, which equals 0.30B0.30B. The remaining portion for Station Upgrades is 0.70B0.30B=0.40B0.70B - 0.30B = 0.40B. End-of-year expenditures are calculated by multiplying each allocation by its respective growth multiplier: Bus Rapid Transit is 0.30B×1.25=0.375B0.30B \times 1.25 = 0.375B, Rail Modernization is 0.30B×0.85=0.255B0.30B \times 0.85 = 0.255B, and Station Upgrades is 0.40B×1.10=0.44B0.40B \times 1.10 = 0.44B. Summing these expenditures gives 0.375B+0.255B+0.44B=1.07B0.375B + 0.255B + 0.44B = 1.07B. Comparing 1.07B1.07B to the initial 1.00B1.00B reveals an overall increase of 0.07B0.07B, or 7%7\%.

Adım Adım Çözüm

1
Express the initial allocations for each project as fractions of the total budget B
Bus Rapid Transit = 0.30B0.30B, Rail Modernization = 0.30B0.30B, Station Upgrades = 0.40B0.40B
Bus Rapid Transit is explicitly 0.30B0.30B. The remaining 0.70B0.70B is split such that Rail Modernization receives 37×0.70B=0.30B\frac{3}{7} \times 0.70B = 0.30B, leaving 0.70B0.30B=0.40B0.70B - 0.30B = 0.40B for Station Upgrades.
2
Apply the individual percentage changes to determine end-of-year expenditures
Bus Rapid Transit = 0.375B0.375B, Rail Modernization = 0.255B0.255B, Station Upgrades = 0.44B0.44B
A 25%25\% increase corresponds to a multiplier of 1.251.25, a 15%15\% decrease corresponds to a multiplier of 0.850.85, and a 10%10\% increase corresponds to a multiplier of 1.101.10.
3
Sum the project expenditures and calculate the net percent change relative to B
Total expenditure = 1.07B1.07B, corresponding to a 7%7\% net increase
Adding 0.375B+0.255B+0.44B0.375B + 0.255B + 0.44B yields 1.07B1.07B. Subtracting the original budget 1.00B1.00B gives 0.07B0.07B, which is 7%7\% of BB.

Anahtar Kavram

Weighted Percentage Changes and Sequential Fraction-Decimal Operations
Soru 83Soru

If xx is a negative real number and yy is a positive real number, what is the value of xx+yy\frac{x}{|x|} + \frac{|y|}{y}?

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Cevap: 0

Cevap

The value of the expression is 0.
For any negative number xx, the ratio xx\frac{x}{|x|} evaluates to 1-1 because x=x|x| = -x. For any positive number yy, the ratio yy\frac{|y|}{y} evaluates to 11 because y=y|y| = y. Summing 1-1 and 11 results in 00.

Adım Adım Çözüm

1
Evaluate the first term for a negative variable
-1
By definition of absolute value, if x<0x < 0, then x=x|x| = -x, making xx=xx=1\frac{x}{|x|} = \frac{x}{-x} = -1.
2
Evaluate the second term for a positive variable
1
If y>0y > 0, then y=y|y| = y, making yy=yy=1\frac{|y|}{y} = \frac{y}{y} = 1.
3
Sum the simplified values
0
Combining 1-1 and 11 yields 1+1=0-1 + 1 = 0.

Anahtar Kavram

Properties of Absolute Value and Signs of Numbers
Soru 84Soru

If kk is a positive integer such that 810+223+47=2k+128\sqrt{8^{10} + 2^{23} + 4^7} = 2^k + 128, what is the value of kk?

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Cevap: 15

Cevap

The value of kk is 15.
Converting all terms under the square root to base 2 produces 230+223+214\sqrt{2^{30} + 2^{23} + 2^{14}}. Recognizing that (215+27)2=(215)2+2(215)(27)+(27)2=230+223+214(2^{15} + 2^7)^2 = (2^{15})^2 + 2(2^{15})(2^7) + (2^7)^2 = 2^{30} + 2^{23} + 2^{14}, taking the square root yields 215+27=215+1282^{15} + 2^7 = 2^{15} + 128. Matching this with 2k+1282^k + 128 yields k=15k = 15.

Adım Adım Çözüm

1
Convert terms under the square root to base 2.
The radical expression becomes 230+223+214\sqrt{2^{30} + 2^{23} + 2^{14}}.
Expressing terms with the same base allows exponent rules and algebraic identity recognition.
2
Identify the expression under the radical as a perfect square of the form (a+b)2(a + b)^2.
Setting a=215a = 2^{15} and b=27b = 2^7 gives 2ab=221527=2232ab = 2 \cdot 2^{15} \cdot 2^7 = 2^{23}, so 230+223+214=(215+27)22^{30} + 2^{23} + 2^{14} = (2^{15} + 2^7)^2.
The middle term 2232^{23} satisfies 22302142 \cdot \sqrt{2^{30}} \cdot \sqrt{2^{14}}.
3
Evaluate the square root and solve for kk.
(215+27)2=215+128\sqrt{(2^{15} + 2^7)^2} = 2^{15} + 128. Setting 215+128=2k+1282^{15} + 128 = 2^k + 128 yields k=15k = 15.
Comparing terms directly after evaluating 27=1282^7 = 128 isolates 2k=2152^k = 2^{15}.

Anahtar Kavram

Application of exponent rules combined with perfect square algebraic identities under radicals
Soru 85Soru

If nn is a positive integer such that 140n140n is a perfect square and 105n105n is a perfect cube, what is the minimum possible number of positive divisors of nn?

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Cevap: 108

Cevap

108
To minimize the number of divisors of nn, express n=2a3b5c7dn = 2^a \cdot 3^b \cdot 5^c \cdot 7^d. The condition that 140n=2a+23b5c+17d+1140n = 2^{a+2} \cdot 3^b \cdot 5^{c+1} \cdot 7^{d+1} is a perfect square implies aa and bb must be even, while cc and dd must be odd. The condition that 105n=2a3b+15c+17d+1105n = 2^a \cdot 3^{b+1} \cdot 5^{c+1} \cdot 7^{d+1} is a perfect cube requires aa to be a multiple of 3, and b,c,db, c, d to leave a remainder of 2 when divided by 3. The minimum non-negative integers satisfying both requirements are a=0a=0 (since 0 is even and a multiple of 3), b=2b=2 (even and leaves remainder 2 mod 3), c=5c=5 (odd and leaves remainder 2 mod 3), and d=5d=5 (odd and leaves remainder 2 mod 3). The minimum number of positive divisors is (0+1)(2+1)(5+1)(5+1)=108(0+1)(2+1)(5+1)(5+1) = 108.

Adım Adım Çözüm

1
Find the prime factorizations of 140 and 105
140=2257140 = 2^2 \cdot 5 \cdot 7 and 105=357105 = 3 \cdot 5 \cdot 7
Decomposing into prime factors allows analyzing exponent constraints for perfect powers.
2
Express nn in terms of prime factors n=2a3b5c7dn = 2^a \cdot 3^b \cdot 5^c \cdot 7^d and determine constraints for 140n140n to be a perfect square
140n=2a+23b5c+17d+1140n = 2^{a+2} \cdot 3^b \cdot 5^{c+1} \cdot 7^{d+1} requires aa to be even, bb to be even, cc to be odd, and dd to be odd.
All prime factor exponents in a perfect square must be even numbers.
3
Determine constraints for 105n105n to be a perfect cube
105n=2a3b+15c+17d+1105n = 2^a \cdot 3^{b+1} \cdot 5^{c+1} \cdot 7^{d+1} requires a0(mod3)a \equiv 0 \pmod 3, b2(mod3)b \equiv 2 \pmod 3, c2(mod3)c \equiv 2 \pmod 3, and d2(mod3)d \equiv 2 \pmod 3.
All prime factor exponents in a perfect cube must be multiples of 3.
4
Find the smallest non-negative integers satisfying both sets of constraints for each exponent
a=0a = 0, b=2b = 2, c=5c = 5, and d=5d = 5
For aa: smallest non-negative even multiple of 3 is 0. For bb: smallest non-negative even number congruent to 2(mod3)2 \pmod 3 is 2. For cc and dd: smallest odd numbers congruent to 2(mod3)2 \pmod 3 are 5.
5
Calculate the total number of positive divisors of n=325575n = 3^2 \cdot 5^5 \cdot 7^5
(2+1)(5+1)(5+1)=3×6×6=108(2 + 1)(5 + 1)(5 + 1) = 3 \times 6 \times 6 = 108
The divisor counting formula multiplies (ei+1)(e_i + 1) for each prime exponent eie_i.

Anahtar Kavram

Prime factor exponent constraints for perfect powers and the divisor counting formula
Soru 86Soru

For any positive integer nn, let SnS_n denote the units digit of the sum 2n+3n+4n+7n2^n + 3^n + 4^n + 7^n. What is the remainder when the sum T=n=1102SnT = \sum_{n=1}^{102} S_n is divided by 99?

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Cevap: 1

Cevap

The remainder when the sum T is divided by 9 is 1.
Evaluating the units digits of each exponent term reveals that SnS_n follows a repeating 4-term cycle (6,8,2,4)(6, 8, 2, 4) with a sum of 20 per cycle. For 102 terms, there are 25 full cycles plus the first two terms (S1=6S_1=6 and S2=8S_2=8). The total sum is T=25×20+6+8=514T = 25 \times 20 + 6 + 8 = 514. Dividing 514 by 9 yields 514=9×57+1514 = 9 \times 57 + 1, so the remainder is 1.

Adım Adım Çözüm

1
Find the cyclicity of the units digit of each term 2n,3n,4n,2^n, 3^n, 4^n, and 7n7^n.
Units digits repeat in patterns of length 4: for 2n2^n (2, 4, 8, 6), for 3n3^n (3, 9, 7, 1), for 4n4^n (4, 6, 4, 6), and for 7n7^n (7, 9, 3, 1).
Units digits of positive integer powers cycle with periodicities that divide 4.
2
Compute SnS_n for n=1,2,3,4n = 1, 2, 3, 4 and find the sum of one 4-term period.
S1=units(2+3+4+7=16)=6S_1 = \text{units}(2+3+4+7=16) = 6, S2=units(4+9+6+9=28)=8S_2 = \text{units}(4+9+6+9=28) = 8, S3=units(8+7+4+3=22)=2S_3 = \text{units}(8+7+4+3=22) = 2, S4=units(6+1+6+1=14)=4S_4 = \text{units}(6+1+6+1=14) = 4. Sum of one period = 6+8+2+4=206+8+2+4 = 20.
The sum of the units digits of individual terms determines the units digit of the total expression.
3
Calculate the total sum T=n=1102SnT = \sum_{n=1}^{102} S_n.
Since 102=25×4+2102 = 25 \times 4 + 2, the sequence consists of 25 complete cycles of 4 terms plus the first 2 terms (S1=6S_1=6 and S2=8S_2=8). Thus, T=25×20+6+8=500+14=514T = 25 \times 20 + 6 + 8 = 500 + 14 = 514.
Dividing the total number of terms by the period length gives the number of full cycles and remaining initial terms.
4
Compute the remainder when T=514T = 514 is divided by 9.
The sum of the digits of 514 is 5+1+4=105 + 1 + 4 = 10, and 101(mod9)10 \equiv 1 \pmod 9. Alternatively, 514=9×57+1514 = 9 \times 57 + 1. Thus, the remainder is 1.
A positive integer and the sum of its digits leave the same remainder when divided by 9.

Anahtar Kavram

Units digit cyclicity of exponential terms and modular arithmetic on sequence sums
Tahmini Süre:2m 0s
Soru 87Soru

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

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Cevap: 37

Cevap

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

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

Anahtar Kavram

Counting inclusive terms in a sequence of consecutive integers
Soru 88Soru

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

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Cevap: 15

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

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

Alternatif Yöntem

Let the set be {a,a+1,,a+n1}\{a, a+1, \dots, a+n-1\}. The sum is n(2a+n1)2=675    n(2a+n1)=1350\frac{n(2a + n - 1)}{2} = 675 \implies n(2a + n - 1) = 1350. The product of endpoints is a(a+n1)=1976a(a + n - 1) = 1976. Solving the system of equations for integer values of aa and nn yields a=38a = 38 and n=15n = 15.
Tahmini Süre:2m 30s
Soru 89Soru

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

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Cevap: 8

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Calculating total positive integer divisors from prime factorizations and analyzing exponent constraints derived from divisibility conditions.
Tahmini Süre:2m 0s
Soru 90Soru

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

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Cevap: 180

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Divisor count formula and prime factorization constraints
Soru 91Soru

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

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Cevap: 44

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic Exponent Rules
Tahmini Süre:2m 0s
Soru 92Soru

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

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Cevap: 2

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Units Digit Cyclicity and Negative Remainder Rules
Tahmini Süre:2m 30s
Soru 93Soru

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

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Cevap: 292^9

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Exponents, Roots, and Powers of Integers
Tahmini Süre:2m 0s
Soru 94Soru

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

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Cevap: 0

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Modular Arithmetic and Exponent Parity
Tahmini Süre:1m 30s
Soru 95Soru

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

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Cevap: 8

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Units Digit Cyclicity and Modular Arithmetic
Tahmini Süre:2m 0s
Soru 96Soru

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

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Cevap: 36

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

The total number of positive divisors of a positive integer n=p1e1p2e2pkekn = p_1^{e_1} p_2^{e_2} \dots p_k^{e_k} is given by (e1+1)(e2+1)(ek+1)(e_1 + 1)(e_2 + 1) \dots (e_k + 1).
Soru 97Soru

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

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Cevap: 18

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Factoring difference of powers using fundamental exponent rules and equating even/odd prime components.
Soru 98Soru

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

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Cevap: 45

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Finding the exponent of a prime factor in a factorial using Legendre's formula
Tahmini Süre:2m 0s
Soru 99Soru

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

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Cevap: 243

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Sum of consecutive integers set formula and positivity constraints
Soru 100Soru

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

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Cevap: 11

Cevap

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

Adım Adım Çözüm

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

Anahtar Kavram

Modular Arithmetic and Polynomial Remainders
ÖncekiSayfa 5 / 12Sonraki
Number Properties and Arithmetic Alıştırma Soruları — GMAT — Sayfa 5 | Examkin