Property of Numbers

Mistake 5

1 − 0.000001 ? Here I have to recognise the difference of squares 1^2 - 0.000001^2 (1 - 0.001) (1+0.001) (0.999) (1.001)

Mistake 24

A set of numbers has the property that for any number t in the set, t + 2 is in the set. If -1 is in the set, which of the following must also be in the set? a) -3 b) 1 c) 5 If -1 is in the set, 1,3,5,7 and further must be in the set. Because we don't have any information about numbers preceding -1, we can't make the assumption that 3 should be in the set Answer is B and C

Mistake 39

For any positive integer n , the length of n is defined as the number of prime factors whose product is n . For example, the length of 75 is 3, since 75 = 3 × 5 × 5. How many two-digit positive integers have length 6 ? Since the number must have two digit, the product of the prime numbers must be less than 100 The following numbers meet the criteria: 2^6= 64 Then we proceed with the next smallest number 2^5 x 3 = 96 If we proceed with any other number we notice that they exceed 100 Hence, only 2 numbers meet this criteria

Mistake 23

For every even positive integer m, f(m) represents the product of all even integers from 2 to m, inclusive. For example, f(12)=2x4x6x8x10x12. What is the greatest prime factor of f(24)? F(24) = 2x4x6x8x10x12x14x16x18x20x22x24 We need to find the greatest prime factor, so we have to check from right to reach the solution faster 24 = 8 x 3 22 = 11 x 2 11 must be the greatest prime factor you will have in this product. Every term smaller than 22 will certainly have 2 as a factor since they are all even so they cannot have another factor greater than 11.

Mistake 6

How many integers x satisfy both 2<x≤4 and 0≤x≤3? The inequality we have is 2<x≤3 x must be greater than 2 and smaller or equal than 3 Hence the only possible number is 3

Mistake 46

If S is the sum of the reciprocals of the consecutive integers from 91 to 100, inclusive, which of the following is less than S? I. 1/8 II. 1/9 III. 1/10 1/91 is the largest term and 1/100 is the smallest term If all 10 terms were equal to 1/91, then the sum would be 10/91, but since actual sum is less than that, then we have that S<10/91 If all 10 terms were equal to 1/100, then the sum would be 10/100=1/10, but since actual sum is more than that, then we have that S>1/10. Therefore, 1/10 < S < 10/91. From this statement we know that 1/10 is already smaller than S Also, notice that 10/91 < 1/9 < 1/8, thus we have that 1/10 < S < 10/91 < 1/9 < 1/8. Therefore only 1/10 is less than S. An alternative way to check the relative sizes of the fractions is the following What is greater 10/91 or 1/90 10/91 x 1/9 = 90<91 Hence 1/9 is greater than 10/91

Mistake 36

If a, b, and c are consecutive integers and 0 < a < b < c, is the product abc a multiple of 8 ? (1) The product ac is even. (2) The product bc is a multiple of 4. Since a,b,and c are multipled there are two possible cases Case 1: Odd - Even - Odd Case 2: Even - Odd - Even Statement 1 This will be case 1 If a is any given number, we can represent ac as 2a x 2(a+1) 4a(a+1) If a is odd, then (a+1) will be even and abc will be a multiple of 8 If a is even, then 4a will be a multiple of 8 and abc will be a multiple of 8 Statement 2 Case 1: a= 3 b=4 c= 5 abc = 60 is not a multiple of 8 Case 2: a = 7 b = 8 c = 9 abc = 7 x 8 x 9 is a multiple of 8 Since we have contradictory findings, this is insufficient

Mistake 17

If a= 1 + 1/4 + 1/16 + 1/64 and b= 1 + 1/4a, what is a - b? a - 1 - 1/4a 3/4a - 1 3/4(1 + 1/4 + 1/16 + 1/64) - 1 3/4 (85/64) - 1 255/256 - 256/256 - 1/256

Mistake 48

If an integer n is to be selected at random from 1 to 100, inclusive, what is probability n(n+1) will be divisible by 4? n(n+1), one part will be odd and the other part will be even. Therefore, we need to identify how many numbers are divisible by 4 from 1 to 100 100/4 = 25 There are 25 numbers that are a multiple of 4 If we think closely for every pair that divisible by 4, there will be a second pair The first pair: 3x4 4x5 The second pair 7x8 8x9 And so on Hence, there will be 50 numbers that are divisble by 4 50/100 = 1/2

Mistake 45

If k = (n + 2)(n - 2), where n is an integer greater than 2, what is the value of k ? (1) k is the product of two primes. (2) k < 100 k= n^2 - 4 Statement 1 n^2-4 could be 1. 3^2 - 4 = 1 2. 4^2 - 4= 12 3. 5^2 - 4 = 21 4. 6^2 - 4 = 32 5. 7^2 - 4 = 45 6. 8^2 - 4 = 60 7. 9^2 - 4 = 77 8. 10^2 - 4 = 96 And so on... Since we know that k is the product of two primes k could be either 21 (7x3) or 77 (11x7) Not sufficient Statement 2 Not sufficient Both statements combined Since there are two possible answers where k is less than 100, both statements combined are not sufficient

Mistake 47

If k is a positive integer and n = k(k + 7), is n divisible by 6 ? (1) k is odd. (2) When k is divided by 3, the remainder is 2. For n to be divisible by 6, it must be divisible by 3 and divisible by 2 Statement 1 k= 1, n=8 Not divisible k=3, n=30 Divisible Not sufficient Statement 2 k= 3q + 2 n = (3q + 2)(3q + 9) n = 3(q+3)(3q+2) n is a multiple of 3 and hence is divisible by 3 Now we need to confirm that n is divisible by 2 If q is even: 3(5)(8), n is divisible by 2 if q is odd: 3(4)(5), n is divisible by 2 We can confirm that n is divisible by 6 and that this statement is sufficient

Mistake 32

If m and n are nonzero integers, is m^n an integer? (1) n^m is positve (2) n^m is an integer. If n is a positive integer then m^n will be an integer for any value of m (taking into account that both are nonzero integers). If n is negative then m^n will be an integer if and only m=1 or m=-1, for example: (-1)^(-2)=1/(-1)^2=1 So basically we are asked: is n positive or m=|1|? (1) n^m is positive --> either m=even (and in this case n can take any value) or n=positive (and in this case m can take any value). Not sufficient. (2) n^m is an integer --> either m=positive (and in this case n can take any value) or m=negative and in this case n=1 or -1. Not sufficient. (1)+(2) If n^m=(-1)^2=positive integer, then the answer will be NO as m^n=2^(-1)=1/2 but if n^m=1^2=positive integer, then the answer will be YES as m^n=2^1=2. Not Sufficient.

Mistake 28

If m is an integer greater than 1, is m an even integer? (1) 32 is a factor of m. (2) m is a factor of 32. Statement 1 M is divisible by 2 and hence is even. Sufficient Statement 2 32= 2^5 x 1 Since m must be greater than 1, m will always be even. Sufficient.

Mistake 42

If n is a positive integer and r is the remainder when 4 + 7n is divided by 3, what is the value of r ? (1) n + 1 is divisible by 3. (2) n > 20 Since the divisor is 3, we know that the remainder can be either 0, 1, or 2 Statement 1 Since n+1 is divisible by 3, n will be any multiple of 3 minus 1 n(2,5,8,11..) Let's see what is the remainder of all these numbers n=2, 4+7(2)=18, r=0 n=5, 4+7(5)=39, r=0 n=8, 4+7(8)=60, r=0 We can see that the remainder will always be 0 Hence, the statement is sufficient Statement 2 Clearly insufficient

Mistake 14

If n is a positive integer, what is the remainder when x is divided by 4? 1) x = 3^81n 2) n = 2k where k is a positive integer Here we need to use the rule of the unit digits to calculate the remainder under different scenarios Statement 1 Since we don't know the value of n, the power of n can be even and odd Statement 2 Related directly to Statement 1, not sufficient Both combined Since the exponent n, will always be even, the units digit of x will always be either 9 or 1. Remember the unit digit pattern of 3 (3,9,7,1) Since any power of 3 with unit digit 9 or 1 when is divided by 4 has a remainder of 1, both statements are sufficient 9/4 = 2 + 1/4 81/4 = 20 + 1/4

Mistake 26

If n=9!-6^4 which of the following is the greatest integer k such that 3^kis a factor of n ? 9x8x7x6x5x4x3x2x1 - (3^4x2^4) 3^4x2^6x7x10 - 3^4x2^4 3^4x2^4(2^2x7x10-1) 3^4x2^4(4x7x10-1) 3^4x2^4(280-1) 3^4x2^4(279) Prime factorise 279 279= 3^2 x 31 3^4x2^4x3^2x31 3^6x2^4x31 6 is the greatest integer k

Mistake 3

If the smaller of 2 consecutive odd integers is a multiple of 5, which of the following could NOT be the sum of these 2 integers? Our smaller integer is an odd multiple of 5, so is equal to 5x where x is odd. Our numbers are consecutive odd numbers, so the other is 5x+2. Their sum is thus 10x + 2, where x is odd. So if the answer is going to be positive (we can look at answer A later if we need to), the tens digit will be odd (since when we find 10x, we're just sticking a 0 on the end of some odd number), and the units digit will be 2. So we can't possibly get 22 as our answer.

Mistake 29

If x is a positive integer, is x^1/2 an integer? (1) (4x)^1/2 is an integer. (2) (3x)^1/2 is not an integer. Statement 1 4^1/2 x x^1/2 must be an integer 2 x (x^1/2) must be an integer (x^1/2) must be an integer Sufficient Statement 2 Case 1: if x = 4, (12)^1/2 is not an integer and 4^1/2 is an integer Case 2: If x=7, 21^1/2 is not an integer, nad 7^1/2 is not an integer Not sufficient

Mistake 19

If x is a positive integer, r is the remainder when x is divided by 4, and R is the remainder when x is divided by 9, what is the greatest possible value of r^2 + R ? Remember that the remainder must always be lower than the divisor If 7/2, the remainder will always be less than 2. In this case it's 1 If 12/5, the remainder will always be less than 5. In this case it's 2 Since r can be at most 3 and R can be at most 8, r^2 + R = 3^2 + 8 = 17

Mistake 41

If x is a positive integer, what is the least common multiple of x, 6, and 9 ? (1) The least common multiple of x and 6 is 30. (2) The least common multiple of x and 9 is 45. LCM (x,6,9) LCM(x,18) Statement 1 x can be 5, 15, and 30 If x = 5, LCM(5,6,9) = 90 If x = 15, LCM(15,6,9) = 90 If x = 30, LCM (30,6,9) = 90 Sufficient Statement 2 x can be 5,15,45 We know that if x is 5 or 15 the LCM is 90 if x = 45, LCM(6,9,945) = 90 Sufficient

Mistake 15

If x is an integer greater than 1 but less than 101 such that x/2= k, where k is an integer, is 1/3x an integer? 1) x = s^2 where s is an integer 2) x = t^4 where t is an integer The question is asking whether x is divisible by 3, considering that is even/multiple of 2. Statement 1 We can see that x must be a perfect square. 4 is not divisible by 3 but 36 is. Not sufficient Statement 2 We know that the fourth square of t must be an even integer. Hence the t must be 16, and x must be 2. 2^4 = 16. X can't be 3 because 3^4 is 81 and is uneven. And x can't be 4 because 4^4 = 256 which is more than 100. Hence the only possible value for x is 2, and x is not divisible by 2

Mistake 44

If x= u^2-v^2, y=2uv, and z=u^2 + v^2, and if x=11, what is the value of z? (1) y = 60 (2) u = 6 Before immersing into the statements, we know that: x=(u + v)(u - v) = 11 (u+v)(u-v)= 11 z + y = 2uv + u^2 + v^2 z + y = (u+v)^2 z = (u+v)^2 - y Statement 1 If y=60 uv= 30 u=30/v We can replace u by 30/v in the following equation (u+v)(u-v)= 11 and we would be able to obtain u and v Finally we can use z=u^2 + v^2 to get the value of z Sufficient Statement 2 If u=6 We can substitute u by 6 in the following equation (u+v)(u-v)= 11 and we would be able to get the value of v With the value of v and u can use z=u^2 + v^2 to get the value of z Sufficient Alternative reasoning approach Statement 1 If y=60 uv= 30 We know that u^2 and v^2 are separated by 11 units in the number line. At the same time we know that u x v must equal 30. Hence, there must be only one position where u^2 and v^2 are separated by 11 and their product is 30. Sufficient Statement 2 If u=6 We can easily follow the approach outlined above

Mistake 11

If x^2+bx+5=(x+c)^2 for all numbers x, where b and c are positive constants, what is the value of b ? x^2+bx+5=x^2 + 2cx + c^2 Comparing both sides c^2= 5 2c=b c= sqr(5) b = 2c b = 2sqr(5)

Mistake 35

If x^2=2^x, what is the value of x ? (1) 2x=(x/2)^3 (2) x=2^(x−2) Statement 1 2x = x^3/2^3 2x = x^3/8 16x = x^3 0 = x^3-16x 0= x(x^2-16) x=0 x=-4 x=4 Although we have 3 different equations, if we plug them in the master equation only x=4 yields an equivalent result. Hence, the statement is sufficient Statement 2 x=2^x/4 4x = 2^x Since the master equation tells us that x^2=2^x, we can replace one side of the equation 4x = x^2 x^2-4x = 0 x(x-4)=0 x= 0 x=4 Because we know that 0 does't fit in the master equation but 4 does, this statement is sufficient

Mistake 33

If xy = - 6, what is the value of xy(x + y)? (1) x−y=5 (2) xy^2=18 xy(x + y) x^2y + xy^2 Statement 1 Clearly insufficient Statement 2 xy^2= xyy xy= -6 -6y=18 y= -3 if y=-3 xy=-6 x(-3)=-6 x= 2 Since we have the value of x and y individually, this statement is sufficient

Mistake 1

Is N odd? (1) N is a factor of 100 (2) N is a factor of 135 1) We can quickly disregard this one because there are both odd and even multiples of 100 Odd = 5, 25 Even = 2, 10, 20, 30... 2) Think of the prime numbers of 135 135 = 3^3 x 5 Because it only has odd prime factors, and we know that in order to get an even number you need at least even number, no factors of 135 won't be even,

Mistake 2

Is x an integer? (1) x + 5 is an integer. (2) x / 4 is an integer. 1) For the sum of two numbers to be an integer, each number must be an integer i.e 2.5 + 5 will never be an integer Because of this we know that x is an integer 2) If x = 2, (x / (1/4)) will be an integer => 2 x (4/1) = 8 But if x = 1/4, x/4 will be an integer, and x will not be an integer (1 / 4) / 4 = 4 / 4 = 1 Thus, under statement 2, x can be both an integer, a integer

Mistake 30

Look at the picture What I need to remember is that the division signs which is in the same height as the equal, is the one in charge of separating the numbers. On the left a/b/c is equivalent to: a/b/c/1 a/b x 1/c a/bc On the right a/b/c is equivalent to: a/1/b/c a/1 x c/b ac/b

Mistake 43

Of the numbers r + s, r - s, r*s, and r/s, which is greatest? (1) r = s =1 (2) r - s is the least of the numbers Statement 1 a) r + s = 1 + 1 = 2 b) r - s = 1 - 1 = 0 c) r x s = 1 x 1 = 1 d) r / s = 1/1 = 1 a is clearly the largest number. Sufficient Statement 2 From the statement above we see that b is the smallest number while b is the largest. However, we need to test with other numbers to confirm that this relationship holds true To ensure that r - s is the lowest possible, the absolute value of s should be much bigger than the absolute value of r r = 2 and s = 6 a) r + s = 8 b) r - s = - 4 c) r x s = 12 d) r / s = 2/6 In this case, the largest number is c. Hence, insufficient

Mistake 13

On a certain plane, 2/5 of the passengers speak Farsi, and 3/4 speak Hebrew, If all of the passengers on the plane speak at least one of these languages, what is the smallest number of passengers that could be on the plane? Since to 2/3 speak Farsi and 3/4 speak Hebrew, the smallest number of passengers on the plane is the LCM of 5 and 4. Therefore 20 is the minimum number of passengers that could be on the plane

Mistake 8

The expression n! is defined as the product of the integers from 1 through n. If p is the product of the integers from 100 through 299 and q is the product of the integers from 200 through 299, which of the following is equal to p/q? P/Q= 100 x 101...x299 / 200 x 201...x299 First 200 x 201...299 cancels out and we are left with 100 x 101...x199 Looking at the answers, we can re-express this as 199!/99! By having this expression we know that 1 x 2..x99 will cancel out and we will only be left with 100 x 101...x199

Mistake 9

The "prime sum" of an integer n greater than 1 is the sum of all the prime factors of n, including repetitions. For example , the prime sum of 12 is 7, since 12 = 2 x 2 x 3 and 2 +2 + 3 = 7. For which of the following integers is the prime sum greater than 35 ? Think of what makes the prime sum large, and conclude that you need large prime factors. Every number has small prime factors but 620. Prime factorise 620 into 1) 62 x 10 2) 31 x 2 x 10 3) 31 x 2 x 5 x 2 The sum is 40

Mistake 27

The integer 120 has many factorizations. For example, 120=(2)(60)120=(2)(60), 120=(3)(4)(10)120=(3)(4)(10), and 120=(-1)(-3)(4)(10)120=(-1)(-3)(4)(10). In how many of the factorizations of 120 are the factors consecutive integers in ascending order? 120= 2^3 x 3 x 5 Looking at the factorization, we have 2,3,5 and power of 2. Hence we can deduce that 2,3,4,5 are factors of 120 1. 1*2*3*4*5 2. 2*3*4*5 3. -5*-4*-3*-2 4. 4*5*6

Mistake 12

The set of solutions for the equation (x^2-25)^2=x^2-10x+25 contains how many real numbers? (x-5)^2(x+5)^2=(x-5)^2 From this we can see that one solution must be 5 Next We can divide the whole equation by (x-5)^2 and we get (x+5)^2=1 x^2 + 10x + 25= 1 x^2 + 10x - 24 (x+6)(x+4) There are 3 real solution: -6,-4 and 5

Mistake 25

What is the greatest positive integer n such that 5^n divides 10!-(2)(5!)210!-(2)(5!)^2 ? 10! - 2(5!)(5!) 10*9*8*7*6*5! - 2(5!)(5!) 5! [10*9*8*7*6 - 2(5*4*3*2)] 5! [10*9*8*7*6 - 10*8*3] 5! [80 (9*7*6) - 80(3)] 5! [80(375)] Now, 5! has one 5; 80 has one 5; 375 has three 5s. So, the max number of 5s are 5 i.e. n = 5

Mistake 20

Each of the nine digits 0, 1, 1, 4, 5, 6, 8, 8, and 9 is used once to form 3 three-digit integers. What is the greatest possible sum of the 3 integers? Since we want the greatest possible sum of the 3 integers, we should use 8, 8 and 9 as the hundreds digits of the 3 integers, 4, 5 and 6 as the tens digits, and 0, 1 and 1 as the units digits. Therefore, the greatest possible sum of the 3 integers is: (900 + 800 + 800) + (40 + 50 + 60) + ( 0 + 1 + 1) = 2652

Mistake 40

For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, then p is We can start by playing with the numbers h(100)= 2 x 4 x 6 x 8...x100 We notice that because they are even they have 2 as a common factor h(100) = 2(1 x 2 x 3 x 4...x50) Hence we can translate the above as h(100) = 2(50!) Since we want to find the lowest prime of 2(50!)+1 and we know that two consecutive numbers don't share prime numbers other than 1, then 2(50!) and 2(50!)+1 won't have any shared prime factors. As a result, the lowest prime factor of 2(50!)+1 will be a number greater than 50

Mistake 4

For positive integers a and b, the remainder when a is divided by b is equal to the remainder when b is divided by a. Which of the following could be a value of ab? The only way the remainder when a is divided by b is equal to the remainder when b is divided by a is a = b since then the remainder in both cases is 0. That is, if a = b, a/b = 1 R 0 and b/a = 1 R 0.Since a = b, we see that ab must be a perfect square. So only 36 could meets this premise

Mistake 21

Given that 1^2+2^2+3^2+...+10^2=385, what is the value of 3^2+6^2+9^2+...+30^2 ? 3^2+6^2+9^2+...+30^2 can be expressed as (3 x 1)^2 + (3 x 2)^2 + (3 x 3)^2 ... (10 x 3)^2 This can be re-expressed as: 3^2 x 1^2 + 3^2 x 2^2 + 3^2 x 3^2 ... 10^2 x 3^2 We can get 3^2 as a common factor 3^2(1^2 + 2^2 + 3^2... + 10^2) The value of (1^2 + 2^2 + 3^2... + 10^2) is given in the statement 3^2 (385) = 3465

Mistake 7

How many integers between 1 and 16, inclusive, have exactly 3 different positive integer factors? (Note: 6 is NOT such an integer because 6 has 4 different positive integer factors: 1, 2, 3, and 6.) The thing you need to recognize is that numbers that have exactly 3 prime numbers are perfect squares made of prime factors. From 1 to 16, only 3 and 4 are perfect squares made from prime factors.

Mistake 31

How many integers n are there such that r < n < s ? (1) s - r = 5 (2) r and s are not integers. Statement 1 s= 5 r=0 0 < n < 5 N can be 1,2,3 and 4 4 integers s=5.5 r=0.5 0.5 < n < 5.5 N can be 1,2,3,4 and 5 5 integers Since results are contradictory, not sufficient Statement 2 Not sufficient Both statements combined If they are not integers, then n can be 5 integers. Sufficient

Mistake 16

If x and y are positive integers, is the largest prime factor of x greater than the largest prime factor of y? 1) x = 7^n where n is a positive integer 2) LCM of x and y is 105 Statement 1 Not sufficient, no information about y Statement 2 Not sufficient, we don't know which factors belong to x and y Both combined Initially, it seems that because z has a 7 on its prime factors and y has just a 5, the greatest prime factor of x will always be greater than the greatest prime factor of y. However, it could be the case that x=7 and y=105, and the LCM would still be 105. In this case, the greatest prime factor of x would be no greater than the greatest prime factor of y

Mistake 38

If x is positive, is x > 3 ? (1) (x - 1)^2 > 4 (2) (x - 2)^2 > 9 Statement 1 x^2 -2x + 1 > 4 x^2 - 2x - 3 > 0 (x-3)(x+1)>0 z>3 z<-1 Sine x must be positive, x>3 Sufficient Statement 2 x^2 - 4x + 4 > 9 x^2 -4x - 5 > 0 (x-5)(x+1)>0 x>5 x<-1 Not sufficient

Mistake 22

Two numbers differ by 2 and sum to S. Which of the following is the greater of the numbers in terms of S ? x: small number x + 2: largest number x + x + 2 = s 2x + 2 = s 2x= s - 2 x = (s-2)/2 Since the above is the smaller number, the larger number will be x + 2 = (s-2)/2 + 2 s/2 - 1 + 2 s/2 + 1 A different way to solve this problem is a - b = 2 a + b = S Summing both equations a - b + a + b = 2 + s 2a = 2 + s a = (2 + s)/2 a = 1 + s/2 An alternative way of solving these problems that you should always keep in mind is substituting with real values. In this case, we can let a= 3 and b= 1 and s= 3 + 1 = 4 and check for the correct answer

Mistake 37

What is the remainder when the positive integer x is divided by 3 ? (1) When x is divided by 6, the remainder is 2. (2) When x is divided by 15, the remainder is 2. Statement 1 x = 6q + 2 If we plug in 6q + 2 in the original equation (6q + 2) / 3 2q + 2/3 Hence the remainder is 2 Sufficient Statement 2 x= 15q + 2 If we plug in 15q + 2 in the original equation (15q + 2)/3 5q + 2/3 The remainder is 2 Sufficient

Mistake 49

What is the tens digit of the positive integer n ? (1) The hundreds digit of 10n is 2. (2) The tens digit of n - 9 is 1. N= A B C D We want to find out the value of C Statement 1 10N = A B C D 0 Since the hundred digit when N is multiplied by 10 is C, then the statement is giving us the answer The tens digit of the positive integer n is 2 Statement 2 N - 9 = A B 1 (D-9) We have two possible cases under this set-up If D=9, then the tens digit won't be altered If D<9, then the tends digit will go down by 1 E.g 227-9= 218 -> See how the tens digit moves from 2 to 1 229 - 9 = 220 -> See how the tens digit doesn't change Because we don't know the unit digit, this statement is insufficent

Mistake 34

What is the value of x^2-y^2? 1) (x-y)^2=9 2) x + y = 6 x^2-y^2 = (x+y)(x-y) Statement 1 x - y = 3 x-y = -3 Since we are taking the square root, x-y can be both positive and negative Insufficient Statement 2 Insufficient Both statements combined (x+y)(x-y) Case 1: 6x3=18 Case 2: 6(-3)= -18 Since we have two contradictory results, this is insufficient

Mistake 10

What values of x have a corresponding value of y that satisfies both xy > 0 and xy = x + y ? A. x≤1≤1 B. −1<x≤0 C. 0<x≤1 D. x>1 E. All real numbers The main mistake that I did here I that I didn't break down xy = x + y Since we are interested in having x as the input, and y as the output, let's isolate y in terms of x xy - y = x y(x-1) = x y = x /(x-1) From this equation we can see that x can't be 1 because if so, the equation would be undefined. At this point we can eliminate answer choices A, C and E, since they all allow for x to equal 1. Next, we can try out with numbers to decide between B and D If x = 2 y = 2/(2-1) y = 2 This satisfies the equation, and hence the right answer is D

Mistake 18

When 24 is divided by the positive integer n, the remainder is 4. Which of the following statements about n must be true? I. n is even II. n is a multiple of 5 III. n is a factor of 20 24 / n = q + 4/n 24 = qn + 4 20 = qn Since the divisor must always be greater than the reminder, or the reminder must always be less than the divisor, n>4 Hence the possible options are: Q = 1 N = 20 Q = 4 N = 5 Q = 2 N = 10 We can see that n is not always even (it can be 5), n is a multiple of 5 and n is a factor of 20