Question

The product of ab=1200. a is an integer and b is an odd integer. What is the largest possible value of b

Answer 1

The product of a and b is 1200. This means both a and b are the factors of 1200.

So in order to solve this question, we need to list all the possible factors of 1200 which satisfy the given conditions, to find the answer.

The factors of 1200 are:
1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 16, 20, 24, 25, 30, 40, 48, 50, 60, 75, 80, 100, 120, 150, 200, 240, 300, 400, 600, 1200

The factors pairs whose product is 1200 are:
(1, 1200)    (2, 600)    (3, 400)    (4, 300)    (5, 240)    (6, 200)    (8, 150)   
(10, 120)   (12, 100)    (15, 80)    (16, 75)    (20, 60)    (24, 50)    (25, 48)    (30, 40)

The factor pairs with odd values are:
(1200, 1)
(400, 3)
(240, 5)
(150, 8)
(80, 15)
(48, 25)

Thus the largest possible value of b can be 25, and corresponding value of a will be 48.

48 x 25 = 1200

Answer 2

Answer:

Largest possible value of b is:

75

Step-by-step explanation:

we are given that:

ab=1200

where a is an integer and b is an odd integer.

we know that product of odd numbers is always odd.

Now, 1200= 12×100

                = 2×6×10×10

               =2×2×3×2×5×2×5

It cannot be factorized further

Now, considering the product of all odd numbers

=3×5×5

=75

Now, if we multiply any other remaining number from the factorization of 1200, we will get a even number

Hence, The largest possible value of b is:

75