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  • How many ways can three numbers be chosen from the set {1, 4, 7, 10, ..., 142, 145, 148} such that their product is a multiple of 5 and 7 but not a multiple of 3?<strong

How many ways can three numbers be chosen from the set {1, 4, 7, 10, ..., 142, 145, 148} such that their product is a multiple of 5 and 7 but not a multiple of 3?

Option: 1

16,844


Option: 2

18,424

 


Option: 3

12,444

 


Option: 4

10,800

 


Answers (1)

best_answer

To determine the number of ways three numbers can be chosen from the given set such that their product is a multiple of 5 and 7 but not a multiple of 3, we need to consider the factors of 5, 7, and 3.

The given set starts with 1 and has a common difference of 3, so it can be rewritten as:

1, 4, 7, 10, ..., 142, 145, 148

To find the numbers that are multiples of 5, 7, and not a multiple of 3, we can exclude the numbers that are multiples of 3 and consider the remaining numbers.

Numbers divisible by 3: 3, 6, 9, 12, ..., 147

Numbers divisible by 5: 5, 10, 15, 20, ..., 145

Numbers divisible by 7: 7, 14, 21, ..., 147

Now, let's find the numbers that are multiples of 5 and 7, but not a multiple of 3.

Numbers divisible by both 5 and 7: 35, 70, 105

Numbers divisible by 3, 5, and 7: 105

To choose three numbers from the given set such that their product is a multiple of 5 and 7 but not a multiple of 3, we need to exclude the numbers divisible by both 5 and 7 and consider the remaining numbers.

Excluding the numbers divisible by both 5 and 7, we have the following set of numbers:

1, 4, 7, 10, ..., 142, 145, 148

Now, let's count the number of numbers in this set. The first number is 1, and the last number is 148. The common difference is 3.

The number of terms in an arithmetic sequence can be found using the formula:

Number of terms = (last term - first term) / common difference + 1

Number of terms = (148 - 1) / 3 + 1 = 49

Therefore, there are 49 numbers in this set.

To choose three numbers from this set, we can use the formula for combinations:

\mathrm{n C r=n ! /(r !(n-r) !)}

Therefore, the number of ways to choose two numbers from the multiples of 4 up to 99 and multiply them together to obtain a product that is not a multiple of 5, 7, or 3 is (16 choose 2) = 120 ways.

Thus, there are 120 ways to choose two numbers from the multiples of 4 up to 99 and multiply them together to obtain a product that is a multiple of neither 5, 7, nor 3.


 

Posted by

Deependra Verma

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