Test: Divisibility/Multiples/Factors - GMAT MCQ

If xy = 0, it means that either x or y (or both) must be equal to 0, since the product of any number and 0 is always 0.

Given that y is not divisible by 6, it means y cannot be equal to 0. Therefore, x must be equal to 0.

The greatest integer less than x is -1.

Therefore, the correct answer is C.

Prime factorizing all the multiples of 5 up to 25, we have:

25 = 5²

20 = 5¹ x 2²

15 = 5¹ x 3

10 = 5¹ x 2

5 = 5¹

Thus, we see that there are 6 factors of 5 in 25! and thus the maximum value of x is 6.

To find the least common multiple (LCM) of 18 and 60, we can factorize the numbers and take the highest power of each prime factor.

Factorizing 18: 18 = 2 * 3^2

Factorizing 60: 60 = 2^2 * 3 * 5

Taking the highest power of each prime factor: LCM(18, 60) = 2^2 * 3^2 * 5 = 4 * 9 * 5 = 180

Now let's check which of the given factors, 24, 36, and 45, divide 180.

24 is not a factor of 180 (180 ÷ 24 = 7.5).

36 is a factor of 180 (180 ÷ 36 = 5).

45 is a factor of 180 (180 ÷ 45 = 4).

Therefore, the correct answer is E. Both II (36) and III (45) are factors of the least common multiple of 18 and 60.

To find the number of positive integers that are factors of 360, we can determine the prime factorization of 360:

360 = 2³ * 3² * 5¹

To find the number of factors, we add 1 to the exponent of each prime factor and multiply these numbers together:

(3 + 1) * (2 + 1) * (1 + 1) = 4 * 3 * 2 = 24

Therefore, there are 24 positive integers that are factors of 360.

The correct answer is A.

If a positive integer n is divisible by both 5 and 7, it must also be divisible by their least common multiple (LCM), which is 35. This means that n is also divisible by 35.

On the other hand, n being divisible by 5 and 7 does not necessarily imply that it is divisible by 12 or 70.

Therefore, the correct answer is C, "II only," as n must be divisible by 35.

To determine if n is divisible by a certain number, we need to check if n leaves a remainder of 0 when divided by that number.

I. To check if n is divisible by 13, we need to find the remainder of n when divided by 13.

Remainder of 19! when divided by 13 can be found using the property of factorial: n! ≡ 0 (mod p) for p > n.

Since 13 > 19, we can conclude that 19! ≡ 0 (mod 13).

Adding 26 to both sides: 19! + 26 ≡ 0 + 26 ≡ 26 (mod 13).

Since 26 leaves a remainder of 0 when divided by 13, n is divisible by 13.

II. To check if n is divisible by 19, we need to find the remainder of n when divided by 19.

Remainder of 19! when divided by 19 is 0 since 19 is included in the factorial.

Adding 26 to both sides: 19! + 26 ≡ 0 + 26 ≡ 26 (mod 19).

Since 26 leaves a remainder of 7 when divided by 19, n is not divisible by 19.

III. To check if n is divisible by 26, we need to find the remainder of n when divided by 26.

Remainder of 19! when divided by 26 can be found by calculating the factorial and finding the remainder.

By calculating 19!, we find that it leaves a remainder of 24 when divided by 26.

Adding 26 to both sides: 19! + 26 ≡ 24 + 26 ≡ 50 (mod 26).

Since 50 leaves a remainder of 24 when divided by 26, n is not divisible by 26.

Therefore, n is divisible by 13 and not divisible by 19 and 26.

The correct answer is D.

To find the times when both of your alarms could go off, we need to find the common multiples of 12 minutes and 18 minutes.

The multiples of 12 minutes are: 12, 24, 36, 48, 60, ...

The multiples of 18 minutes are: 18, 36, 54, 72, 90, ...

From the list, we can see that the common multiples are 36, 72, and so on.

Among the given options: A: 6:30 am -> Not a common multiple B: 6:48 am -> Not a common multiple C: 7:00 am -> Not a common multiple D: 7:48 am -> Common multiple (72 minutes = 1 hour and 12 minutes after 6:00 am) E: 8:12 am -> Not a common multiple

Therefore, the only time when both of your alarms could go off is 7:48 am.

The correct answer is D.

To find the lowest positive integer that is divisible by 8, 9, 10, 11, and 12, we need to find their least common multiple (LCM).

Factorizing the given numbers: 8 = 2^3 9 = 3^2 10 = 2 × 5 11 = 11^1 12 = 2^2 × 3

To find the LCM, we take the highest power of each prime factor: LCM(8, 9, 10, 11, 12) = 2^3 × 3^2 × 5 × 11 = 8 × 9 × 5 × 11 = 3,960

Therefore, the lowest positive integer that is divisible by 8, 9, 10, 11, and 12 is 3,960.

The correct answer is C.

Let's analyze the given information:

If riders are let on only in groups of 5, there will be 2 riders that do not get on. This means the total number of riders is not divisible by 5.

If riders are let on only in groups of 6, all riders will be able to get on. This means the total number of riders is divisible by 6.

To find the largest possible value of the number of people in line, we need to find the least common multiple (LCM) of 5 and 6.

Factorizing 5: 5 = 5^1 Factorizing 6: 6 = 2 * 3^1

Taking the highest power of each prime factor: LCM(5, 6) = 2 * 3 * 5^1 = 30

Since the total number of riders must be divisible by 6, and less than 112 people are in the line, the largest possible value of the number of people in line is a multiple of 30 that is less than 112.

The largest possible value of the number of people in line is 102 (30 × 3 + 12).

Therefore, the correct answer is D.