Question

Consider the mathematical expression

(num1 + num2) × (num3 − num4).

We have been asked to replace num1, num2, num3 and num4 by numbers between 1 and 35, i.e. numbers from the set {1, . . . , 35}, so that after substituting these numbers in place of num1, num2, num3 and num4, the evaluation of the expression using PEDMAS leads to an odd positive integer. Find the number of ways in which this can be done. For example, one way of doing this is to use num1 = 2, num2 = 3, num3 = 5, and num4 = 2. If we substitute these values we will get the expression

(2 + 3) × (5 − 2) = 5 × 3 = 15.

On the other hand, num1 = 1, num2 = 3, num3 = 5, and num4 = 2 leads to an even integer, and num1 = 2, num2 = 3, num3 = 2, and num4 = 5 leads to a negative odd integer, both of which are not allowed.

Answer 1

Answer:-

Given That:-

(num1 + num2) × (num3 − num4).

We have been asked to replace num1, num2, num3 and num4 by numbers between 1 and 35, i.e. numbers from the set {1, . . . , 35}, so that after substituting these numbers in place of num1, num2, num3 and num4, the evaluation of the expression using PEDMAS leads to an odd positive integer. Find the number of ways in which this can be done.

For example, one way of doing this is to use num1 = 2, num2 = 3, num3 = 5, and num4 = 2. If we substitute these values we will get the expression

(2 + 3) × (5 − 2) = 5 × 3 = 15.

On the other hand, num1 = 1, num2 = 3, num3 = 5, and num4 = 2 leads to an even integer, and num1 = 2, num2 = 3, num3 = 2, and num4 = 5 leads to a negative odd integer, both of which are not allowed.

Given,

(num 1+ num 2)*(num 3 - num 4) should be odd positive integer.

Now, product of two integers is odd if and only if both the integers are odd.

=> (num 1 + num 2) is odd integer. It is greater than 0 since both num 1 and num 2 are positive.

=> (num 3 - num 4) is odd integer. It should also be positive since, then only the product (num 1 + num 2)*(num 3 - num 4) will be positive.

Now we know (num 1 + num 2) is odd positive integer. Sum of two integers is odd only if one is even while other is odd.

=> Either num 1 is even and num 2 is odd or vice versa.

Now, if num 1 is even, it can be selected in 17 ways (since, there are 18 odd numbers and 17 even numbers from 1 to 35)

Then, num 2 can be selected in 18 ways.

Hence total ways of selecting = 17 x 18

Also if num 2 is even and num 1 is odd, total ways of selecting = 18 x 17

=> Total ways of selecting (num 1 + num 2) = 2 x 17 x 18 = 612

Now, (num 3 - num 4) is odd positive integer. Difference of two integers is odd positive only and only if one is even and other is odd and the first number is greater than the second one.

=> Now suppose num 3 is 35. Then num 4 can take values 34, 32, 30.......2. i.e. total 17 ways.

When num 3 is 34, then num 4 can take values 33, 31, 29,......1. i.e. total 17 ways.

When num 3 is 33, then num 4 can take values 32, 30, 28.........,2. i.e. total 16 ways.

When num 3 is 32, then num 4 can take values 31, 29, 27,.........1. i.e. total 16 ways

Similarly the number of ways for each is taken

Total number of ways = 17 x 2 + 16 x 2 + 15 x 2 +................+ 1 x 2

= 2 x ( 17 + 16 + 15 +....+1) =

= 2 x 153 = 306 ways

Hence, (num 3 - num 4) can be selected in 306 ways.

Finally , (num 1 + num 2)*(num 3 - num 4) can be selected in a total of 612 x 306 = 187,272 ways

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