Test: Factors And Multiples- 1 - GRE MCQ

• There are total 8 divisor of 222 i.e. 1, 2, 3, 6, 37, 74, 111, 222

• Multiples of 3 exist between -10 and 10 are -9, -6, -3, 0, 3, 6, 9

• Multiples of 2 between -10 and 10 are  -8, -6, -4, -2, 0, 2, 4, 6, 8

• If y is divisible by 24. The rule says if the number is divisible by all factors of other number the last number is a factor of the first
• So, 24=3x2x2x2
• S1. y is divisible by 6, which is 3x2. Do not know about 2*2. Not Sufficient
• S2. y is divisible by 4, which is 2x2. Do not know about 3*2. Not Sufficient
• S1+S2 says that y divisible by 3x2 and 2x2.
• The trick is that it is enough to be 3x2x2=12 to be divisible by 4 and 6 but not enough to be divisible by 24. But if we have 3x2x2x2=24 it is enough for all. Not Sufficient
   

• To determine if integer y is divisible by 16, let's analyze each statement.
• Statement 1: y is divisible by 8.
• This means that y can be expressed as y = 8k for some integer k. However, knowing that y is divisible by 8 does not necessarily imply that y is also divisible by 16. For example, if y = 8, then y is not divisible by 16, but if y = 16, then it is. So, this statement alone is not sufficient.
• Statement 2: 2y is divisible by 16.
• This means 2y = 16m for some integer m, or equivalently, y = 8m. This shows that y is divisible by 8, but does not guarantee that y is divisible by 16. For instance, if m = 1, then y = 8, which is not divisible by 16. However, if m = 2, then y = 16, which is divisible by 16. So, this statement alone is also not sufficient.
• Combining Statements 1 and 2:
• From both statements, we know:
• y is divisible by 8 (from Statement 1).
• y = 8m, where m is an integer (from Statement 2).
• Even with both statements combined, y could be 8 (not divisible by 16) or 16 (divisible by 16), so we still cannot definitively determine if y is divisible by 16.
• Conclusion:
• The answer is (E) Both statements together are not sufficient to determine whether y is divisible by 16.

• To determine if integer y is divisible by 16, let's analyze each statement.
• Statement 1: y² is divisible by 16.
• This means that y² = 16k for some integer k. If y is divisible by 4, then y² would indeed be divisible by 16. However, this does not guarantee that y itself is divisible by 16. For example, if y = 4, then y² = 16, but y is not divisible by 16. So, this statement alone is not sufficient.
• Statement 2: y³ is divisible by 16.
• This means that y³ = 16m for some integer m. If y is divisible by 2, then y³ could be divisible by 16. However, this also does not guarantee that y itself is divisible by 16. For example, if y = 2, then y³ = 8 which is not divisible by 16. Thus, this statement alone is not sufficient.
• Combining Statements 1 and 2:
• With both statements, we know:
• y² is divisible by 16 (from Statement 1).
• y³ is divisible by 16 (from Statement 2).
• Even with both statements combined, y could be 4 (not divisible by 16) or 16 (divisible by 16). Therefore, we cannot conclusively determine if y is divisible by 16.

.

Let’s break down the problem step-by-step:

1. Understanding x:
x is the product of all integers from 1 to 25. This means:

x = 1 × 2 × 3 × 4 × ... × 25

So, x is a very large number that includes every integer from 1 to 25, and therefore, it is divisible by each of these numbers.

2. Finding the Smallest Prime Factor of (x + 1):
We are asked to find the smallest prime factor of (x + 1), which is x plus 1.

3. Why x + 1 is Special:
Since x is the product of all numbers from 1 to 25, x is divisible by each prime number up to 25, such as 2, 3, 5, 7, 11, 13, 17, 19, and 23.
But when we add 1 to x to get x + 1, it no longer divides evenly by any of these primes. Why? Because when you add 1 to x, it leaves a remainder of 1 for any division by these primes.

4. What This Means:
Since x + 1 is not divisible by any prime up to 25, we need to look for the next possible smallest prime factor that hasn’t been included in the product. The first prime number greater than 25 is 29.

5. Conclusion:
Therefore, the smallest prime factor of (x + 1) must be 29.

Answer: The smallest prime factor of (x + 1) is 29 which is greater than 25.

• 
• Now, two numbers h(100)=250∗50!h(100)=250∗50! and h(100)+1=250∗50!+1h(100)+1=250∗50!+1 are consecutive integers.
• Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1.
• For example 20 and 21 are consecutive integers, thus only common factor they share is 1.
• As h(100)=250∗50!h(100)=250∗50! has ALL prime numbers from 1 to 50 as its factors, then, according to the above, h(100)+1=250∗50! +1h(100)+1=250∗50! +1 won't have ANY prime factor from 1 to 50.
• Hence pp (>1>1), the smallest prime factor of h(100)+1h(100)+1 must be more than 50.
• Hence, option D is correct
   

1. Statement (1): (x - 1) is divisible by y

This means there exists an integer k such that:

x - 1 = ky (for some integer k)

Rearranging this, we have:

x = ky + 1

From this equation, we can see that x is equal to ky + 1. Therefore, x cannot be divisible by y unless k is such that 1 is also divisible by y, which can only happen if y = 1. Thus, this statement alone is not sufficient to determine if x is divisible by y.

2. Statement (2): x > y

This statement only tells us that x is greater than y but provides no information about the divisibility of x by y. Therefore, this statement alone is also not sufficient to determine if x is divisible by y.

Combining the Statements:

Now, let's analyze both statements together:

• From statement (1), we know x = ky + 1.
• From statement (2), we know x > y.

We can substitute ky + 1 into the inequality from statement (2):

ky + 1 > y

This simplifies to:

ky > y - 1

or

k > (y - 1)/(y)

However, this still doesn't provide any information about the divisibility of x by y.

To determine whether x is divisible by y, let's analyze the statements provided:

1. Statement (1): x is a multiple of (y + 1)

This means there exists an integer k such that:

x = k(y + 1) (for some integer k)

To see if x is divisible by y, we can rewrite this as:

x = ky + k

For x to be divisible by y, the term k (which is equal to &frac{x}{y + 1}) must also ensure that when you take ky + k, k alone is divisible by y. However, this is not guaranteed. Therefore, this statement alone is not sufficient to determine if x is divisible by y.

2. Statement (2): y > 1

This statement tells us that y is greater than 1, but it does not provide any information about the relationship between x and y. Therefore, this statement alone is also not sufficient to determine if x is divisible by y.

Combining the Statements:

Now, let's analyze both statements together:

• From statement (1), we know x = k(y + 1).
• From statement (2), we know y > 1.

Even when we combine the two statements, we still only have information about x being a multiple of (y + 1) while y is simply greater than 1. This still does not help in determining if x is divisible by y because the fact that x is a multiple of (y + 1) does not imply anything about divisibility by y.

Conclusion:

Neither statement alone is sufficient to determine if x is divisible by y, and combining the two statements does not lead to a definitive answer. Thus, the answer is:

The statements together are not sufficient to determine if x is divisible by y.

• The product of the integers (a), (b), (c), and (d) is (1155)
• a > b > c > d > 1.
• Factorize (1155) into its prime factors and find the smallest difference between two consecutive prime factors.
• 

Statement 1: x and y share only one common factor --> every integer has 1 as a factor, thus since x and y share only one common factor it must be 1. Sufficient.

Statement 2: x and y are unique prime numbers. Two different primes can have only 1 as common factor. Sufficient.

• We are given the equation 6x=8y=14z. To find a possible sum of x, y, and z, let's approach this systematically.
• Step 1: Express x, y, and z in terms of a common variable
• Let k be the constant such that:
• 6x=8y=14z=k
• So,
• x=k6,y=k8,z=k14
• Step 2: Find the least common value of k
• To ensure x, y, and z are integers, k must be a common multiple of 6, 8, and 14. The least common multiple (LCM) of 6, 8, and 14 is 168.
• Thus, let k=168
• Step 3: Calculate the values of x, y, and z
• x=168/6=28
• y=168/8=21
• z=168/14=12
• Step 4: Find the sum of x, y, and z
• The sum is:
• x+y+z=28+21+12=61
• Conclusion:
• A possible sum of the positive integers x, y, and z is 61

• Divisor of a positive integer cannot be more than that integer (for example integer 4 doesn't have a divisor more than 4, the largest divisor it has is 4 itself), so greatest common divisor of two positive integers x and y can not be more than a or b.
• So answer will be a + b.