Test: Number Properties - GMAT MCQ

To determine the numbers between 1 and 1000 that have an odd number of factors, we need to understand the concept of factors. A factor of a number is a whole number that divides the given number without leaving a remainder.

Let's analyze the factors of a few numbers:

• The factors of 1 are {1}. It has only one factor, which is itself.
• The factors of 2 are {1, 2}. It has two factors.
• The factors of 3 are {1, 3}. It has two factors.
• The factors of 4 are {1, 2, 4}. It has three factors.
• The factors of 5 are {1, 5}. It has two factors.
• The factors of 6 are {1, 2, 3, 6}. It has four factors.

From these examples, we can observe that if a number has an odd number of factors, it must be a perfect square. This is because factors come in pairs, except for the square root of the number. The square root of a number is a factor of the number that is not paired with another factor.

To find the perfect squares between 1 and 1000, we can take the square root of the smallest perfect square (1) and the square root of the largest perfect square (961, which is 31^2). So the perfect squares between 1 and 1000 are {1, 4, 9, 16, 25, 36, ..., 961}.

Now we need to count how many perfect squares are between 1 and 1000. Since 31^2 is the largest perfect square less than or equal to 1000, we have 31 perfect squares in this range.

Therefore, there are 31 numbers between 1 and 1000 (inclusive) that have an odd number of factors. Thus, the correct answer is C.

Few concepts..
R=1!+2!+3!+4!+5!+6!+7!+8!+9!+10!+.....
1) units digit...
All terms after 4! contain one 5 and one 2 atleast, so units digit of all terms after 4! Will be 0..
So units digit will depend on 1!+2!+3!+4! Only
2) last two digits..
All terms after 9! contain two 5 and two 2 atleast, so last 2- digit of all terms after 9! Will be 00.
So last two digits will depend on 1!+2!+...+9! Only

Units digit here =1+2+3*2+4*3*2=1+2+6+24=33
Hence 3..

To determine the greatest value of n such that 30!/6^n is an integer, we need to analyze the prime factorization of 30! (30 factorial) and the prime factorization of 6.

The prime factorization of 30! can be determined by decomposing all the numbers from 1 to 30 into their prime factors and multiplying them together.

30! = 1 × 2 × 3 × 4 × 5 × ... × 30

To simplify this process, let's focus on the prime factorization of 6:

6 = 2 × 3

Now, we can rewrite 30! in terms of its prime factors:

30! = 1 × 2 × 3 × 2 × 2 × 5 × 2 × 3 × ... × 2 × 2 × 3 × 5

As we can see, there are multiple factors of 2 and 3 in the prime factorization of 30!. We are interested in the factors of 6 since 6^n is present in the denominator.

To find the largest value of n, we need to determine the highest power of 6 that can be divided from 30!.

Let's count the number of factors of 2 and 3:

Factors of 2: 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2 (total 14 factors of 2)
Factors of 3: 3, 3, 3, 3 (total 4 factors of 3)

To form a factor of 6, we need both a factor of 2 and a factor of 3. Therefore, we can only create as many factors of 6 as the minimum number of factors of 2 and 3. In this case, we have 4 factors of 3, which means we can create a maximum of 4 factors of 6.

Hence, the greatest value of n is 4 (from 6^4).

Therefore, the correct answer is option D, 14.

To determine the smallest value of n such that n/420 is a terminating decimal, we need to find the greatest common divisor (GCD) of n and 420. If the GCD is 1, then n/420 will be a terminating decimal.

Let's analyze each option:

A. 18: The GCD of 18 and 420 is 6, not 1. Therefore, n/420 is not a terminating decimal for n = 18.

B. 21: The GCD of 21 and 420 is 21. In this case, n/420 simplifies to (21/21) = 1, which is a terminating decimal. Therefore, n = 21 satisfies the condition.

C. 24: The GCD of 24 and 420 is 12, not 1. Therefore, n/420 is not a terminating decimal for n = 24.

D. 30: The GCD of 30 and 420 is 30. In this case, n/420 simplifies to (30/30) = 1, which is a terminating decimal. However, we are looking for the smallest value of n, and n = 30 is not the smallest.

E. 42: The GCD of 42 and 420 is 42. In this case, n/420 simplifies to (42/42) = 1, which is a terminating decimal. However, we are looking for the smallest value of n, and n = 42 is not the smallest.

Therefore, the smallest value of n such that n/420 is a terminating decimal is n = 21. Hence, the correct answer is option B.

Since we are talking about consecutive integers, what is good for one set should be good for every set of consecutive integers. Let's take 1,2,3,4,5.

I. The largest of the integers is even. (Incorrect)
II. The sum of the integers is odd. (Correct)
III. The difference between the largest and smallest of the integers is an even number. (Correct)

Since II and III are correct, answer should be E.

This can be solved mathematically be considering a series of integers n-2, n-1, n, n+1, n+2
Since the average of an odd number of consecutive terms is the middle number, n must be odd.

I. The largest of the integers is even. (incorrect)
II. The sum of the integers is odd. (Sum of integers is 5n; odd integer * odd integer = odd integer, therefore correct)
III. The difference between the largest and smallest of the integers is an even number. (difference between largest and smallest number will be 4. therefore, correct)

Let's break down the information given in the question step by step to understand it better.

• On the scale, a reading of n+1 corresponds to an intensity that is 10 times the intensity corresponding to a reading of n.
  This means that if we increase the reading on the scale by 1 unit, the intensity value will increase by a factor of 10.
• We are given that the intensity corresponding to a reading of 8 is being compared to the intensity corresponding to a reading of 3.

To find the intensity corresponding to a reading of 8, we need to calculate how many times greater it is compared to the intensity corresponding to a reading of 3.

Starting with a reading of 3, we can increase the reading by 1 five times to reach a reading of 8:
3 → 4 → 5 → 6 → 7 → 8

According to the given information, each increase of 1 unit on the scale corresponds to a 10-fold increase in intensity. Therefore, to go from a reading of 3 to 8, we need to multiply the intensity by 10 five times:

Intensity at reading 3 * 10 * 10 * 10 * 10 * 10 = Intensity at reading 8

Simplifying the expression:

Intensity at reading 3 * (10⁵) = Intensity at reading 8

This means that the intensity at reading 8 is 10⁵ times greater than the intensity at reading 3. Therefore, the answer is (C) 10⁵.

To find the smallest value of N with exactly four different prime factors, we need to consider the given conditions.

We are given that 14N/60 is an integer. Let's simplify this expression:

14N/60 = N/5

Since N/5 is an integer, it means that N must be a multiple of 5. Let's substitute N = 5M, where M is an integer, into the expression:

N/5 = (5M)/5 = M

So, we have reduced the problem to finding the smallest value of M that has exactly four different prime factors.

Now, let's analyze the answer choices:

A) 30 = 5 × 2 × 3, has three prime factors.
B) 60 = 5 × 2 × 2 × 3, has three prime factors.
C) 180 = 5 × 2 × 2 × 3 × 3, has four prime factors.
D) 210 = 5 × 2 × 3 × 7, has four prime factors.

Option C and D both have four prime factors, but we are looking for the smallest value. Hence, the correct answer is option D) 210.

Thus, the smallest value of N that satisfies the conditions and has exactly four different prime factors is 210.

To determine which statement must be true, let's analyze the given information and the answer choices.

We are given that 3x + 2 is divisible by 5. In other words, there exists an integer k such that 3x + 2 = 5k.

Let's examine each answer choice:

(A) x is divisible by 3.
If x is divisible by 3, then we can write x as 3m for some integer m. Substituting this into our equation, we get 3(3m) + 2 = 5k, which simplifies to 9m + 2 = 5k. This equation does not provide any information about the divisibility of k by 5, so we cannot conclude that x is divisible by 3 based on the given information.

(B) 3x is divisible by 10.
If 3x is divisible by 10, then we can write 3x as 10n for some integer n. Substituting this into our equation, we get 10n + 2 = 5k, which simplifies to 2n + 2 = 5k. This equation does not provide any information about the divisibility of k by 5, so we cannot conclude that 3x is divisible by 10 based on the given information.

(C) x - 1 is divisible by 5.
Let's analyze this statement. If x - 1 is divisible by 5, then we can write x - 1 as 5p for some integer p. Adding 1 to both sides of the equation, we get x = 5p + 1. Substituting this into our original equation, we have 3(5p + 1) + 2 = 5k, which simplifies to 15p + 5 + 2 = 5k and further simplifies to 15p + 7 = 5k. Rearranging this equation, we have 5(3p + 1) + 2 = 5k. Since the left-hand side is divisible by 5, it follows that the right-hand side (5k) must also be divisible by 5. Therefore, x - 1 being divisible by 5 is a valid conclusion based on the given information.

(D) x is odd.
We cannot conclude that x is odd based on the given information. For example, if x = 2, then 3x + 2 = 3(2) + 2 = 8, which is divisible by 5, but x is not odd.

(E) 3x is even.
We cannot conclude that 3x is even based on the given information. For example, if x = 1, then 3x + 2 = 3(1) + 2 = 5, which is divisible by 5, but 3x is not even.

Based on the analysis above, the only statement that must be true is (C) x - 1 is divisible by 5.

Given:

• A: B = 3: 4
• LCM of A and B = 120

To find:
The GCD of A and B

Approach and Working Out:

• We can say that A = 3x and B = 4x
• This implies, the GCD of A and B = x, since 3 and 4 are co-primes.

We also know that, LCM (A, B) * GCD (A, B) = A * B

• 120 * x = 3x * 4x
• Therefore, x = 10

Hence, the correct answer is Option C.

To solve this problem, let's start by considering set X, which consists of eight consecutive integers. Let's assume the first integer in set X is "n."

So, set X can be written as follows:
X = {n, n+1, n+2, n+3, n+4, n+5, n+6, n+7}

Now, we need to find set Y, which consists of all the integers obtained by adding 4 to each integer in set X and subtracting 4 from each integer in set X.

When we add 4 to each integer in set X, we get the following set:
X_add4 = {n+4, n+5, n+6, n+7, n+8, n+9, n+10, n+11}

Similarly, when we subtract 4 from each integer in set X, we get the following set:
X_subtract4 = {n-4, n-3, n-2, n-1, n, n+1, n+2, n+3}

Now, set Y consists of all the integers in X_add4 and X_subtract4. To find the total number of integers in set Y, we can combine these two sets and remove any duplicates.

Combining X_add4 and X_subtract4, we get:
Y = {n-4, n-3, n-2, n-1, n, n+1, n+2, n+3, n+4, n+5, n+6, n+7, n+8, n+9, n+10, n+11}

To count the number of integers in set Y, we subtract the smallest integer in set Y from the largest integer and add 1. In this case, the smallest integer is (n-4), and the largest integer is (n+11).

Therefore, the number of integers in set Y is (n+11) - (n-4) + 1 = n + 11 - n + 4 + 1 = 16.

Hence, there are 16 more integers in set Y than in set X.

Therefore, the correct answer is C.