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Find the largest no. which divides 615 and 963 leaving remainder 6 in each case.

Posted by Vishal Pratap 7 years, 2 months ago

CBSE > Class 10 > Mathematics

2 answers

Anubhav Kumar Singh 7 years, 2 months ago

Since 6 is the remainder in each case Therefore, 615 - 6=609 And,963 - 6=957 Now required no.=hcf of 609 and 957 957=609×1+348 609=348×1+261 348=261×1+87 261=87×3+0 Therefore, required no.=87

2Thank You

Yogita Ingle 7 years, 2 months ago

To find the largest number which divides 615 and 963 leaving remainder 6 in each case i.e. HCF.

Consider HCF be x.

In order to make 615 and 963 completely divisible by x, we need to deduct the remainder 6 from both the cases.

609 = 3 x 3 x 29

957= 3 x 11 x 29

⇒ x = 3 x 29 = 87

∴ largest number which divides 615 and 963 leaving remainder 6 in each case is 87.

1Thank You

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