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Given 1651m + 2032n
2032 > 1651
(1) 2032 = 1651 * 1 + 381
(2) 1651 = 381 * 4 + 127
(3) 381 = 127 * 3 + 0
Here, the remainder is equal to 0. Therefore the HCF of 1651 and 2032 = 127.
Now,
= > 127 = 1651 - 381 * 4
= 1651 - (2032 - 1651) * 4
= 1651 - 2032 * 4 + 1651 * 4
= 1651(5) - 2032(4)
= 1651(5) + 2032(-4).
Therefore m = 5 and n = -4.
Hope this helps!
2032 > 1651
(1) 2032 = 1651 * 1 + 381
(2) 1651 = 381 * 4 + 127
(3) 381 = 127 * 3 + 0
Here, the remainder is equal to 0. Therefore the HCF of 1651 and 2032 = 127.
Now,
= > 127 = 1651 - 381 * 4
= 1651 - (2032 - 1651) * 4
= 1651 - 2032 * 4 + 1651 * 4
= 1651(5) - 2032(4)
= 1651(5) + 2032(-4).
Therefore m = 5 and n = -4.
Hope this helps!
aishwary88:
please tell me
??
1651 * 4 + 1651 = 1651(5).
more detail
In the 3rd step, it is 1651 - 2032 * 4 + 1651 * 4 = > It can be written as (1651 + 4 * 1651) - 2032 * 4 = > 1651 * 5 - 2032 * 4. Example : 8 * 4 + 8 = 40. it can be written as 8 * 5 = 40
yes i understood
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