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The largest number which divides 615 and 963 leaving remainder 6 in each case is

  • a)
    82

  • b)
    95

  • c)
    87

  • d)
    93

Correct answer is option 'C'. Can you explain this answer?
Most Upvoted Answer
The largest number which divides 615 and 963 leaving remainder 6 in ea...
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=29×3×7


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.
Community Answer
The largest number which divides 615 and 963 leaving remainder 6 in ea...
According to the question the new numbers are--
615-6=609,
and 963-6=957,
since , 957>609,
by euclid's division Lemma-->
957=609×1+348,
the reminder is not zero we apply the process again,
609=348×1+261,
is not equal to zero so, by applying the process again
348=261×1+87,
261=87×3+0,
since r=0 list of the process and get
the no.=87
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The largest number which divides 615 and 963 leaving remainder 6 in each case isa)82b)95c)87d)93Correct answer is option 'C'. Can you explain this answer?
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