k is the greatest number that divides 2996, 4752 and 7825 leaving the remainder in each case equal. Find the product of the marks of K.
Solution
Let a, b, c be three numbers, that is 2996, 4752 and 7825
So, H.C.F will be (b - a), (c - b), (c - a)
Desired H.C.F of
4752 - 2996, 7825 - 4752, 7825 - 2996
= 1756, 3073, 4829
Now, the factors of
1756 = 1, 2, 3, 4, 398, 1756
3073 = 1, 7, 439, 3073
4829 = 1, 11, 439
Therefore, the greatest number = Highest common factor = 439
So, k = 439 is a required greatest number.
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Let a, b, c be three numbers, that is 2996, 4752 and 7825
So, H.C.F will be (b - a), (c - b), (c - a)
Desired H.C.F of
4752 - 2996, 7825 - 4752, 7825 - 2996
= 1756, 3073, 4829
Now, the factors of
1756 = 1, 2, 3, 4, 398, 1756
3073 = 1, 7, 439, 3073
4829 = 1, 11, 439
Therefore, the greatest number = Highest common factor = 439
So, k = 439 is a required greatest number.
The part of question "find the product of marks of k" is not clear. Recheck and post it again with a correct query and our experts would be happy to help you.