Test: Divisibility/Multiples/Factors - GMAT MCQ

To determine which of the given options must be true, let's consider the factors of 50:

The factors of 50 are 1, 2, 5, 10, 25, and 50.

We need to find a and b, which are both factors of 50.

Option (A) ab ≥ 2: This option is not necessarily true because a and b could be 1, and 1 multiplied by 1 is 1, which is less than 2.

Option (B) a + b ≤ 50: This option is not necessarily true either. For example, if a = 25 and b = 25, then a + b = 50, which is equal to 50, not less than.

Option (C) ab is a factor of 100: This option is not necessarily true. For example, if a = 10 and b = 5, then ab = 10 * 5 = 50, which is not a factor of 100.

Option (D) ab is a factor of 2500: This option is true. Since a and b are factors of 50, their product, ab, will also be a factor of 50. As 2500 is equal to 50 * 50, ab must be a factor of 2500.

Option (E) ab is even: This option is not necessarily true. For example, if a = 5 and b = 10, then ab = 5 * 10 = 50, which is not an even number.

Therefore, the correct answer is (D) ab is a factor of 2500.

A = {3k, 3k+3, 3k+6}; k>0; k is a positive integer
B = {5m, 5m+5, 5m + 10, 5m+15, 5m+20}; m>0; m is a positive integer

3k + 3k + 3 + 3k + 6 = 5m + 5m+5 + 5m + 10 + 5m+15 + 5m+20
9k + 9 = 25m + 50
9k = 25m + 41

k_min = 24
3k_min = 72

A. c + d:
The GCF of c and d may or may not divide c + d evenly. For example, if c = 2 and d = 4, their GCF is 2, but 2 does not divide 2 + 4 evenly. Therefore, we cannot conclude that the GCF of c and d is also the GCF of c and c + d.

B. 2 + d:
Similarly, the GCF of c and d may or may not divide 2 + d evenly. For example, if c = 3 and d = 5, their GCF is 1, but 1 does not divide 2 + 5 evenly. Therefore, we cannot conclude that the GCF of c and d is also the GCF of c and 2 + d.

C. cd:
The product of c and d, cd, will always be divisible by the GCF of c and d. This is because the GCF is a factor of both c and d, and their product will contain the factors of both numbers. Therefore, the GCF of c and d is also the GCF of c and cd.

D. 2d:
The GCF of c and d may or may not divide 2d evenly. For example, if c = 3 and d = 4, their GCF is 1, but 1 does not divide 2 * 4 evenly. Therefore, we cannot conclude that the GCF of c and d is also the GCF of c and 2d.

E. d²:
The GCF of c and d may or may not divide d^2 evenly. For example, if c = 2 and d = 3, their GCF is 1, but 1 does not divide 3^2 evenly. Therefore, we cannot conclude that the GCF of c and d is also the GCF of c and d^2.

By analyzing the options, we can see that the only possible answer is A) c + d. Since we cannot determine that the GCF of c and d is also the GCF of any of the other options, the correct answer is A) c + d.

Let's assume the smallest number drawn by one of the friends is x. Since the range is 10, the largest number drawn would be x + 10.

The sum of evenly spaced integers can be found using the formula:

Sum = (Number of terms / 2) * (First term + Last term)

In this case, the number of terms is 6 (since there are 6 friends) and the first term is x, while the last term is x + 10.

So, the sum of the numbers would be:

Sum = (6 / 2) * (x + (x + 10))
= 3 * (2x + 10)
= 6x + 30

Now, let's check each option to see if it can be obtained as the sum of the numbers:

A) 1,254:
6x + 30 = 1,254
6x = 1,224
x = 204
The sum can be obtained.

B) 1,428:
6x + 30 = 1,428
6x = 1,398
x = 233
The sum can be obtained.

C) 3,972:
6x + 30 = 3,972
6x = 3,942
x = 657
The sum can be obtained.

D) 4,316:
6x + 30 = 4,316
6x = 4,286
x ≈ 714.333
The sum cannot be obtained with evenly spaced integers since x is not an integer.

E) 8,010:
6x + 30 = 8,010
6x = 7,980
x = 1,330
The sum can be obtained.

Therefore, the sum that could NOT be the sum of their numbers is D) 4,316.

Since an even integer divisible by 7 must also be divisible by 14, we need to find the number of integers between 100 and 200 that are divisible by 14. By considering multiples of 14 instead of 7, we avoid counting the odd multiples of 7, which would need to be subtracted later.

Using the formula (largest - smallest) / 14 + 1, we can calculate the count of such integers:

(196 - 112) / 14 + 1 = 7

Therefore, there are 7 integers in the range of 100 to 200 that are divisible by 14.

Hence, the answer remains E.

To determine the largest 3-digit number with an odd number of factors, we need to consider the prime factorization of the numbers given as options.

Let's analyze each option:

A: 625 = 5^4. It has 5 factors: 1, 5, 25, 125, and 625. The number of factors is even, so it's not the answer.

B: 729 = 3^6. It has 7 factors: 1, 3, 9, 27, 81, 243, and 729. The number of factors is odd, but it is not a 3-digit number.

C: 841 = 29^2. It has 3 factors: 1, 29, and 841. The number of factors is odd, but it is not a 3-digit number.

D: 943 is a prime number, so it only has 2 factors: 1 and 943. The number of factors is even, so it's not the answer.

E: 961 = 31^2. It has 3 factors: 1, 31, and 961. The number of factors is odd, and it is a 3-digit number.

Therefore, the largest 3-digit number with an odd number of factors is E: 961.

First, let's consider the condition for x. We know that x is equal to the sum of the squares of two distinct prime numbers. The prime numbers less than 60 are: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, and 59.

We need to find two distinct prime numbers, square them, and add them together to get x.

Now, let's consider the condition for y. It is mentioned that y is a multiple of 17, which means y can be any integer multiple of 17.

Analyzing the options:

A: -19
B: -7
C: 0
D: 4
E: 9

To evaluate x - y, we need to check if any of the given options can be obtained by subtracting a possible value of y from x.

Let's analyze each option:

A: x - y = (sum of squares of two primes) - (-19)
B: x - y = (sum of squares of two primes) - (-7)
C: x - y = (sum of squares of two primes) - 0
D: x - y = (sum of squares of two primes) - 4
E: x - y = (sum of squares of two primes) - 9

Since y can be any multiple of 17, it will not affect the possibility of x - y being a specific value. Therefore, we only need to focus on the possible values of x.

By analyzing all the prime numbers less than 60, we find that the only possible values for x (sum of squares of two primes) that can be obtained are:

x = 2² + 3² = 4 + 9 = 13
x = 2² + 5² = 4 + 25 = 29
x = 2² + 7² = 4 + 49 = 53
x = 3² + 7² = 9 + 49 = 58

Out of these possible values of x, let's evaluate x - y for each option:

A: 13 - (-19) = 32 (not equal to C)
B: 29 - (-7) = 36 (not equal to C)
C: 13 - 0 = 13 (equal to C)
D: 53 - 4 = 49 (not equal to C)
E: 29 - 9 = 20 (not equal to C)

Therefore, the only option for which x - y could be equal is option C.

Hence, the answer is C.

64/7 = Rem 1
2⁶ = 64
Now, 2²⁰⁵⁰ = 2^341∗6+2
Thus, 2² = 4  must be subtracted from the number to make it completely divisible by 7, Answer must be (C) 4

Given that both a and b are perfect squares and their product, a * b, is divisible by both 10 and 15, we can conclude that a * b must also be divisible by the highest common factor (HCF) of 10 and 15. Additionally, since a and b are perfect squares, their product, a * b, must also be a perfect square.

We can express a * b as (5 * 2 * 3)² * (Integer)^2, where the common factors of 10 and 15, i.e., 5 and 3, are squared. Therefore, a * b can be written as 900 * I², where I represents any integer.

Considering the given options, the only number that may not divide the product a * b is 120.

Hence, the answer is option C.

To determine the option that could NOT be the greatest common divisor (GCD) of 30x and 15y, we need to analyze each option:

A: The greatest common divisor (GCD) of 30x and 30x is 30x itself. Therefore, A could be the GCD of 30x and 15y.

B: The GCD of 15y and 15y is 15y itself. Hence, B could be the GCD of 30x and 15y.

C: The GCD of 15(x + y) can be found by factoring out the common factor of 15. We have GCD(15(x + y), 15) = 15. Therefore, C could also be the GCD of 30x and 15y.

D: The GCD of 15(x - y) can be found by factoring out the common factor of 15. We have GCD(15(x - y), 15) = 15. Hence, D could be the GCD of 30x and 15y.

E: The GCD of 15,000 and any number is 15, as 15 is a factor of 15,000. Therefore, E could be the GCD of 30x and 15y.

Based on the analysis, the option that could NOT be the GCD of 30x and 15y is option C.

Therefore, the answer is C.