find the greatest number which divide 281 and141 leaving remainder 5
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To find the largest number which divides 281 and 141 leaving remainder 6 in each case i.e. HCF.
Consider HCF be x.
In order to make 281and 141 completely divisible by x, we need to deduct the remainder 5 from both the cases.
i.e., 276 and 136
The factors of 136 are: 1, 2, 4, 8, 17, 34, 68, 136
The factors of 276 are: 1, 2, 3, 4, 6, 12, 23, 46, 69, 92, 138, 276
⇒ x = 1*2*4=8
∴ largest number which divides 141 and 281 leaving remainder 5 in each case is 8.
hope it helped.....
pls mark me as the brainliest....
Consider HCF be x.
In order to make 281and 141 completely divisible by x, we need to deduct the remainder 5 from both the cases.
i.e., 276 and 136
The factors of 136 are: 1, 2, 4, 8, 17, 34, 68, 136
The factors of 276 are: 1, 2, 3, 4, 6, 12, 23, 46, 69, 92, 138, 276
⇒ x = 1*2*4=8
∴ largest number which divides 141 and 281 leaving remainder 5 in each case is 8.
hope it helped.....
pls mark me as the brainliest....
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