फाइंड द लार्जेस्ट नंबर विच डिवाइड्स 615 एंड 963 लीविंग रीमेंदर इन ईच केस
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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.
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