Test: Exponents/Powers - GMAT MCQ

The final population of water lilies covering the lake can be calculated using the formula Final Population = Starting Population * Progression^{(Time/Intervals)}, where Starting Population (S) is the initial number of water lilies, Progression (P) represents the rate of growth (doubling, tripling, etc), Time (t) is the duration in days, and Intervals (l) indicates the time intervals for growth.

We are given that X represents the final number of water lilies after 60 days, which can be calculated as X = 1 * 2^(60/2) since the population doubles every 2 days. Simplifying further, X = 2^30, which is the total number of lilies needed to cover the lake.

Considering the scenario when the lake starts with 2 lilies, we can set up the equation 2³⁰ = 2 * 2^(t/2) to determine the time needed to cover the lake. By comparing the exponents, we find that 30 = t/2 + 1, leading to 29 = t/2. Solving for t, we find t = 58, which represents the time needed to cover the lake when starting with 2 lilies.

Hence, the lake will be fully covered with water lilies after 58 days if initially there were two lilies.

Therefore, the answer is option D.

The powers of 7 follow a pattern:
7¹ = 7
7² = 49
7³ = 343
7⁴ = 2401
7⁵ = 16807
7⁶ = 117649

By observing this pattern, we can identify a cycle of four values: 0, 4, 4, 0. Since we are only interested in the ten's digit, we can focus on the last two digits.

For example, when we have the number 1415, we can calculate its remainder when divided by 4: 1415 = 4 * 353 + 3. Therefore, the ten's digit of 7¹⁴¹⁵ will be 4.

To determine the greatest value of n such that 18n is a factor of 18!, we need to find the highest power of 18 that appears in the prime factorization of 18!.

The prime factorization of 18 is 2 * 3².

Now, let's consider the prime factors 2 and 3. For 18n to be a factor of 18!, the highest power of 2 and 3 that appears in the prime factorization of 18! should be equal to or greater than their respective powers in the prime factorization of 18n.

Prime factor 2:
The highest power of 2 that appears in the prime factorization of 18! can be calculated using the formula:
power of 2 = floor(18/2) + floor(18/4) + floor(18/8) + ...
By calculating the sum, we find that the power of 2 in the prime factorization of 18! is 9.

Prime factor 3:
The highest power of 3 that appears in the prime factorization of 18! can be calculated using the formula:
power of 3 = floor(18/3) + floor(18/9) + floor(18/27) + ...
By calculating the sum, we find that the power of 3 in the prime factorization of 18! is 8.

To satisfy the condition, the highest power of 2 and 3 in the prime factorization of 18n should be less than or equal to the respective powers in the prime factorization of 18!

Since the prime factorization of 18n is 2 * 3² * n, the highest power of 2 and 3 in the prime factorization of 18n is 1 (corresponding to the single factor of 2) and 2 (corresponding to the factor of 3²), respectively.

Comparing the powers, we find that the highest power of 2 and 3 in the prime factorization of 18n is less than the respective powers in the prime factorization of 18!.

Therefore, the greatest value of n such that 18n is a factor of 18! is n = 4 (option D).

Given that b is equal to a plus 4, we are asked to determine which of the provided values of x will result in the smallest value for the expression (x−a)² + (x−b)².

By simplifying the expression, we can rewrite it as 2(x - (a+b)/2)² - (a+b)^2/2 + a² + b².

To find the smallest value for the expression, x should be equal to (a+b)/2, which can be calculated as (2a + 4)/2 or simply a + 2.

Let’s start with the greatest number 66!
Number of 2 present in 66! = 66/2 + 66/4 + 66/8 + 66/16 + 66/32 + 66/64
So greatest option which is NOT divisible by 2⁶⁴ is 65!

333²²² = (329+4)²²² = (7∗47+4)²²². Now if we expand this, all terms but the last one will have 7*47 as a multiple and thus will be divisible by 7. The last term will be 4²²² = 2⁴⁴⁴. So we should find the remainder when 2⁴⁴⁴ is divided by 7.

2¹ divided by 7 yields remainder of 2;
2² divided by 7 yields remainder of 4;
2³ divided by 7 yields remainder of 1;

2⁴ divided by 7 yields remainder of 2;
2⁵ divided by 7 yields remainder of 4;
2⁶ divided by 7 yields remainder of 1;
...

The remainder repeats in blocks of three: {2-4-1}. So, the remainder of 2⁴⁴⁴ divided by 7 would be the same as 2³ divided by 7 (444 is a multiple of 3). 2323 divided by 7 yields remainder of 1.

(a²)³ = a⁶ = (a³)²
Given: M = 0 < a² < 100
Values of a2 can be --> 1²,2²,2⁴,2⁶,3²,3⁴,5²,7²,(2²∗3²)
There are 9 possible values.

There are four possible combinations (7 & 3, 9 & 1, 3 & 7, and 1 & 9) where the sum (7^a + 7^b) will be divisible by 5.

The periodicity of the repetition of the power of 7 is 4. This means that every 1st, 5th, 9th, and so on time, the unit digit will be 7. The 2nd, 6th, and subsequent times will have a unit digit of 9, while the 3rd, 7th, and subsequent times will have a unit digit of 3.

The probability of obtaining each of these unit digits is 12 (approximated as 50/4) out of 50.

Therefore, the probability for 7^a is 12/50, and the probability for 7^b is also 12/50.

Since there are a total of 4 combinations mentioned, the combined probability is calculated as (12/50 * 12/50) * 4 (approximated).

Simplifying this expression, we get (1/4) * (1/4) * 4 = 1/4.

Hence, the approximate probability of the sum (7^a + 7^b) being divisible by 5 is 1/4.

We can simplify the expression (73! - 72!) as follows:
73! - 72! = 72! * (73 - 1) = 72! * 72

Since we are looking for the highest power of 6 that can divide the expression, we need to determine how many factors of 6 are present in 72!.

Let's find the highest power of 6 in 72!. We can do this by finding the highest power of 6 in the prime factorization of 72.

Prime factorization of 72: 2³ * 3²

To find the highest power of 6, we need to consider the highest power of 2 and the highest power of 3.

Highest power of 2: 3 (2³)
Highest power of 3: 2 (3²)

Therefore, the highest power of 6 that can divide 72! is 6², which is 36.

Thus, the highest integer power of 6 that can divide (73! - 72!) is 36.

The correct answer is D: 36.