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  • How many ways can two numbers be chosen from the multiples of 4 up to 99, such that their product is a multiple of neither 5 nor 7 nor 3?  Option:

How many ways can two numbers be chosen from the multiples of 4 up to 99, such that their product is a multiple of neither 5 nor 7 nor 3?

 

Option: 1

100


Option: 2

80


Option: 3

120


Option: 4

60


Answers (1)

best_answer

To determine the number of ways two numbers can be chosen from the multiples of 4 up to 99 such that their product is not a multiple of 5, 7, or 3, we need to consider the factors of 5, 7, and 3.

The multiples of 4 up to 99 are:

4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96

Numbers divisible by 5: 20, 40, 60, 80

Numbers divisible by 7: 28, 56, 84

Numbers divisible by 3: 12, 24, 36, 48, 60, 72, 84, 96

To find the number of ways to choose two numbers such that their product is not a multiple of 5, 7, or 3, we need to exclude the numbers divisible by 5, 7, or 3 and consider the remaining multiples of 4.

Excluding the numbers divisible by 5, 7, or 3, we have the following set of multiples of 4:

4, 8, 16, 24, 32, 36, 44, 48, 52, 64, 68, 72, 76, 88, 92, 96

Now, let's count the number of multiples of 4 in this set. There are 16 multiples of 4.

To choose two 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

Rakesh

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