Test: Exponents/Powers - GMAT MCQ

To determine the greatest possible value of m in the expression 17!/7m, we need to find the highest power of 7 that divides evenly into 17!.

The prime factorization of 17! can be determined by breaking down each number from 1 to 17 into its prime factors and multiplying them together. However, instead of going through the entire prime factorization process, we can count the number of factors of 7 that appear in the prime factorization of each number from 1 to 17.

We have the following numbers in the range 1 to 17 that are divisible by 7: 7, 14.

Each occurrence of 7 in the prime factorization contributes one factor of 7 to the overall count. Therefore, we have 2 factors of 7 in the prime factorization of 17!.

To ensure that the expression 17!/7m is an integer, we want to divide 17! by the highest power of 7, which is 7². Therefore, m should be set to the exponent of the highest power of 7, which is 2.

Hence, the greatest possible value of m is 2.

The correct answer is A.

To find the largest power of 5! that can divide 41!, we need to determine the number of times 5! (which is equal to 5 * 4 * 3 * 2 * 1 = 120) can be divided evenly into 41!.

Let's calculate the power of 5 in the prime factorization of 41!. We know that each multiple of 5 contributes at least one power of 5, and multiples of 25 contribute an additional power of 5.

To calculate the power of 5, we divide 41 by 5 and take the floor value:

41 ÷ 5 = 8

This means there are at least 8 powers of 5 in the prime factorization of 41!.

To account for the multiples of 25, we divide 41 by 25 and take the floor value:

41 ÷ 25 = 1

This means there is an additional power of 5 from the multiples of 25.

Therefore, the largest power of 5! that can divide 41! is 8 + 1 = 9.

The correct answer is A.

4^17 can be written as 2^34, which gives us:
2^34 - 2^28
I can factor out a 2^28 to get:
2^28(2^6-1)
which is basically a bunch of 2's multiplied by (2^6 - 1)
The first part (the bunch of 2's) has 2 as the only prime factor.
The second part (2^6 - 1) = 64 -1 = 63. 63's prime factorization is 3*3*7. So 7 is the largest prime factor.

To find the decimal equivalent of (2/5)⁵, we need to raise 2/5 to the power of 5 and evaluate the result.

(2/5)⁵ = (2⁵)/(5⁵) = 32/3125

Dividing 32 by 3125, we get:

32 ÷ 3125 = 0.01024

Therefore, the decimal equivalent of (2/5)⁵ is 0.01024.

The correct answer is E.

To solve this problem, we need to determine the number of time intervals it takes for the population to double.

The population doubles every 5 years, so we can set up the following equation:

100 * 2ⁿ = 25000

Dividing both sides by 100, we get:

2ⁿ = 250

To solve for n, we can take the logarithm (base 2) of both sides:

n = log2(250)

Using a calculator, we find that log2(250) is approximately 7.97.

Since n represents the number of 5-year intervals, we round up to the nearest whole number to get 8.

Therefore, it will take approximately 8 * 5 = 40 years for the population to grow from 100 to 25000.

The correct answer is E, 40.

To find the sum of the digits of a number, we need to add up each individual digit.

Given that x = 1010 - 47, we can simplify it as follows:

x = 1010 - 47
x = 963

To find the sum of the digits of 963, we add 9 + 6 + 3, which equals 18.

Therefore, the correct answer is E, 80.