How many integers between 50 and 100, inclusive, are divisi

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[GMAT math practice question]

How many integers between 50 and 100, inclusive, are divisible by 2 or 3?

A. 35
B. 37
C. 42
D. 47
E. 52

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by GMATGuruNY » Thu Jun 07, 2018 2:20 am
Max@Math Revolution wrote:[GMAT math practice question]

How many integers between 50 and 100, inclusive, are divisible by 2 or 3?

A. 35
B. 37
C. 42
D. 47
E. 52
Total integers = (multiples of 2) + (multiples of 3) - (multiples of both 2 and 3).

For any evenly spaced set:
Count = (biggest - smallest)/increment + 1.
The INCREMENT is the difference between one value and the next.

Multiples of 2 between 50 and 100, inclusive:
Smallest = 50.
Biggest = 100.
Increment = 2.
Count = (100-50)/2 + 1 = 26.

Multiples of 3 between 50 and 100, inclusive:
Smallest = 51.
Biggest = 99.
Increment = 3.
Count = (99-51)/3 + 1 = 17.

Multiples of both 2 and 3 between 50 and 100, inclusive:
An integer divisible by both 2 and 3 is a MULTIPLE OF 6.
Thus:
Smallest = 54.
Biggest = 96.
Increment = 6.
Count = (96-54)/6 + 1 = 8.

Total integers = 26 + 17 - 8 = 35.

The correct answer is A.
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by Scott@TargetTestPrep » Fri Jun 08, 2018 10:22 am
Max@Math Revolution wrote:[GMAT math practice question]

How many integers between 50 and 100, inclusive, are divisible by 2 or 3?

A. 35
B. 37
C. 42
D. 47
E. 52
For this problem, we must recall that to calculate the number of integers in a range of values that are divisible by a certain number n, we use the formula: (largest multiple - smallest multiple)/n + 1. For example, to calculate the number of integers between 10 and 29 that are divisible by 3, we see that 27 is the largest multiple of 3 in that range of numbers, and 12 is the smallest multiple of 3. Thus, we have (27 - 12)/3 + 1 = 15/3 + 1 = 6.

The number of integers from 50 to 100, inclusive, that are divisible by 2 is:

(100 - 50)/2 + 1 = 26

The number of integers from 50 to 100, inclusive, that are divisible by 3 is:

(99 - 51)/3 + 1 = 17

Now we need to subtract the overlap, that is, we have to subtract the number of multiples of 6 because multiples of 6 were counted as both multiples of 2 and multiples of 3.

The number of integers from 50 to 100, inclusive, that are divisible by 6 is:

(96 - 54)/6 + 1 = 8

Thus, the number integers between 50 and 100, inclusive, that are divisible by 2 or 3 is 26 + 17 - 8 = 35.

Answer: A

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by Max@Math Revolution » Sun Jun 10, 2018 5:18 pm
=>

Let A be the set of integers between 50 and 100 that are divisible by 2.
Let B be the set of integers between 50 and 100 that are divisible by 3.
Let C be the set of integers between 50 and 100 that are divisible by both 2 and 3. This is the same as the set of integers between 50 and 100 that are divisible by 6. Then
A = { 50, 52, ..., 100 }
B = { 51, 54, ..., 96, 99 }
C = { 54, 60, ..., 96 }

The number of elements of the set A is |A| = ( 100 - 50 ) / 2 + 1 = 26.
The number of elements of the set B is |B| = ( 99 - 51 ) / 3 + 1 = 17.
The number of elements of the set C is |C| = ( 96 - 54 ) / 6 + 1 = 8.
Using a Venn diagram, we can see that we need to find |A| + |B| - |C| as the integers in the intersection of sets A and B are counted twice.


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|A| + |B| - |C| = 26 + 17 - 8 = 35.

Therefore, the answer is A.

Answer : A