greatest number that divides 510 and 760 leaving remainder 6 and 4 respectively
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we have to find the largest number which divide 615 and 963 leaving remainder 6 in each case.
so let us subtract 6 from 615 and 963
⇒ 615−6=609
⇒ 963−6=957
now lets find the HCF
⇒ prime factorisation of 609=29×3×3
prime factorisation of 957=29×3×11
now lets take out the common factors from both the cases
⇒ 29 and 3
x=29×3=87
∴ 87 is the number which will divide 615 and 963 leaving remainder 6 in each case.
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