Number Properties - Free MCQ Practice Test with solutions, GMAT

Since 13,689,000 is a multiple of 13 (the worth of a red chip), the product must have at least one red chip. Divide 13,689,000 by 13 to find the product of the other chips: 13,689,000 / 13 = 1,052,000.
Now, notice that 1,052,000 is a multiple of 5 (the worth of a green chip). Divide 1,052,000 by 5 to find the product of the other chips: 1,052,000 / 5 = 210,400.
Since all remaining chips are blue and purple, and blue chips are worth 2 points, the product of all remaining chip points must be a power of 2. The highest power of 2 that divides 210,400 is 2^4 = 16. Divide 210,400 by 16 to find the product of the purple chips: 210,400 / 16 = 13,150.
Now we just need to find what x could be for the purple chips. Since x must be greater than 5 and less than 13, our possible values for x are 6, 7, 8, 9, 10, 11, and 12. The only value of x that divides 13,150 without any remainder is 12. Divide 13,150 by 12 to see how many purple chips there are: 13,150 / 12 = 1095 / 6 = 365 / 2 = 182.5.
Since there must be a whole number of purple chips, there were 2 purple chips selected. The answer is B.

To solve this problem, we need to find the prime factorization of 10080 and then determine which perfect square factors are present in the prime factorization of 10!.

First, we find the prime factorization of 10080 by dividing it by the smallest prime numbers starting from 2:

10080 ÷ 2 = 5040
5040 ÷ 2 = 2520
2520 ÷ 2 = 1260
1260 ÷ 2 = 630
630 ÷ 2 = 315
315 ÷ 3 = 105
105 ÷ 3 = 35
35 ÷ 5 = 7

So, the prime factorization of 10080 is 2^5 * 3² * 5 * 7.

Now, let's find the prime factorization of 10!:

10! = 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1
= 2⁸ * 3⁴ * 5² * 7

Since 10! is divisible by 10080, it means that all the prime factors of 10080 must be present in the prime factorization of 10!.

Therefore, the perfect square factors of 10! are 2⁴, 3², and 5².

The greatest possible value of h is the largest perfect square factor, which is 5².

Therefore, the answer is B: 36.

To find the last two digits of the product 63 * 35 * 37 * 82 * 71 * 41, we can perform the multiplication and observe the pattern of the last two digits.

Let's calculate it step by step:

63 * 35 = 2205
2205 * 37 = 81585
81585 * 82 = 6697770
6697770 * 71 = 474247370
474247370 * 41 = 19449363970

Now, let's focus on the last two digits of the result: 19449363970.

The last two digits of a number can be obtained by taking the remainder when divided by 100. So, we'll divide 19449363970 by 100:

19449363970 ÷ 100 = 194493639 remainder 70

Therefore, the last two digits of the product 63 * 35 * 37 * 82 * 71 * 41 are 70.

Hence, the correct answer is option D: 70.

To maximize the number of employees who take the subway, we want to maximize both (1/3) of the part-time employees and (1/4) of the full-time employees.

Let's start by finding the maximum number of part-time employees who take the subway. Since (1/3) of the part-time employees take the subway, the maximum number of part-time employees who take the subway is (1/3) * (48 - 48/4) = (1/3) * (48 - 12) = (1/3) * 36 = 12.

Next, let's find the maximum number of full-time employees who take the subway. Since (1/4) of the full-time employees take the subway, the maximum number of full-time employees who take the subway is (1/4) * (48/4) = (1/4) * 12 = 3.

Therefore, the greatest possible number of employees who take the subway to work is 12 + 3 = 15.

The correct answer is D.

To determine the number of prime numbers between 260 and 280, we can simply check each number in that range to see if it is prime.

260 is not a prime number because it is divisible by 2, so we can eliminate it.

261 is also not a prime number because it is divisible by 3.

262 is divisible by 2.

263 is a prime number.

264 is divisible by 2.

265 is divisible by 5.

266 is divisible by 2.

267 is divisible by 3.

268 is divisible by 2.

269 is a prime number.

270 is divisible by 2 and 3.

271 is a prime number.

272 is divisible by 2.

273 is divisible by 3.

274 is divisible by 2.

275 is divisible by 5.

276 is divisible by 2 and 3.

277 is a prime number.

278 is divisible by 2.

279 is divisible by 3.

280 is divisible by 2 and 5.

Therefore, there are four prime numbers between 260 and 280.

The answer is E: Four.

Total Number of even numbers in 10-100 inclusive = 46 = 50(total in 100) - 4 (2,4,6,8).

Total numbers divisible by 3 in range 10 - 100 =  = 30 of which half will be even numbers. = 15.

Total even numbers not divisible by 3 are = 46 - 15 = 31.

The given condition states that 14N/60 is an integer. We can simplify this expression further:

14N/60 = 7N/30

For 7N/30 to be an integer, N must have prime factors that cancel out the prime factors in the denominator, which are 2 and 5.

If N has additional prime factors other than 2 and 5, they would not affect the divisibility by 30 since 7N/30 can still be an integer. Therefore, the number of different positive prime factors of N cannot be determined.

Hence, the correct answer is option E: cannot be determined.

The sum of consecutive integers can be calculated using the formula: sum = (first term + last term) * number of terms / 2

Let's calculate the number of terms first. The range from -42 to n inclusive consists of (n - (-42) + 1) = (n + 42 + 1) = (n + 43) terms.

Now we can substitute the given sum and calculate:

372 = (-42 + n) * (n + 43) / 2

Simplifying the equation:

372 = (n - 42 + n) * (n + 43) / 2

372 = (2n - 42) * (n + 43) / 2

Multiplying both sides of the equation by 2 to eliminate the fraction:

744 = (2n - 42) * (n + 43)

Expanding the equation:

744 = 2n² + 86n - 42n - 1806

Rearranging the equation and simplifying:

2n² + 44n - 2550 = 0

Dividing the equation by 2:

n² + 22n - 1275 = 0

To factorize the quadratic equation, we need to find two numbers whose product is -1275 and whose sum is 22.

After trying different combinations, we find that the numbers are 45 and -3.

Therefore, we can rewrite the equation as:

(n + 45)(n - 3) = 0

Setting each factor to zero:

n + 45 = 0 or n - 3 = 0

Solving each equation:

n = -45 or n = 3

Since the range is from -42 to n (inclusive), n cannot be -45. Hence, the value of n is:

n = 3

Therefore, the correct answer is option D: 50.

We can represent the two-digit numbers as "ab" and "cd", where "a", "b", "c", and "d" are digits.

The values of the numbers can be written as 10a + b and 10c + d.

By solving the equation (10a + b)(10c + d) = (10b + a)(10d + c), we find that ac = bd.

Since ac represents a composite single digit, the possible values for ac are 4, 6, 8, and 9.

Out of these four options, we can eliminate 4 and 9 because the digits must be distinct.

Therefore, ac can be either 6 or 8.

For ac = 6, there are 2 possibilities for a and c: (2, 3) or (3, 2).

For ac = 8, there are 2 possibilities for a and c: (2, 4) or (4, 2).

In total, we have 2 * 2 = 4 possibilities for the pairs (a, c) and (b, d) that satisfy the conditions.

Hence, the correct answer is 16.