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  • The two numbers are chosen from a range of 2, 4, 6, 8,... 120, 128, and 130 and multiplied together in every possible way. The number of ways that the product can be multiplied by 8 is<div clas

The two numbers are chosen from a range of 2, 4, 6, 8,... 120, 128, and 130 and multiplied together in every possible way. The number of ways that the product can be multiplied by 8 is

Option: 1

904


Option: 2

304


Option: 3

604

 


Option: 4

804


Answers (1)

best_answer

The given number series is 2, 4, 6, 8,....,120, and 128.

From the given series, the multiple of 8 is 8, 16, 24,...., 120, and 128, which is an arithmetic sequence.

Using the formula,

\begin{aligned} & T_n=a+(n-1) \times D \\ & 128=8+(n-1) \times 8 \\ & n=16 \end{aligned}

If the total number o terms in the given sequence is m, then

130=2+(m-1)\times 2

m=65

Thus, the number of ways is given by,

The product is a multiple of 8 = (both numbers from 8, 16, 24,...120, and 128) or (one number from 8, 16, 24,...120, and 128 and the remaining from other numbers).

{ }^{16} C_2+{ }^{16} C_1 \times{ }^{(65-16)} C_1=\frac{16 !}{2 ! 14 !}+\frac{16 !}{1 !} \times \frac{49 !}{1 !}\\ { }^{16} C_2+{ }^{16} C_1 \times{ }^{(65-16)} C_1=120+16 \times 49

{ }^{16} C_2+{ }^{16} C_1 \times{ }^{(65-16)} C_1=904

Therefore, the total number of ways of selecting two numbers is 904 ways. 

 

 

Posted by

Kshitij

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