LM wrote:
What is the value of length n<100 meter of wire?
1) When divided equal part of 10 meter each, a piece of 5 meter is left.
2) When divided equal of 6 meter each, a piece of 1 meter is left.
What is the value of length n<100 meter of wire?(1) When divided equal part of 10 meter each, a piece of 5 meter is left --> \(n=10q+5\): 5, 15,
25, ..., 95. Not sufficient.
(2) When divided equal of 6 meter each, a piece of 1 meter is left --> \(n=6p+1\): 1, 7, 13, 19,
25, ..., 97. Not sufficient.
(1)+(2) General formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\) would be \(n=30m+25\) --> \(n\) can be: 25, 55 or 85. Not sufficient.
Answer: E.
To elaborate more. How to derive general formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\): divisor will be the least common multiple of above two divisors 6 and 10, hence 30. Remainder will be the first common integer in above two patterns, hence 25. So, to satisfy both conditions, \(n\) must be of a type \(n=30m+25\): 25, 55 or 85.
For more about this concept see:
manhattan-remainder-problem-93752.html#p721341when-positive-integer-n-is-divided-by-5-the-remainder-is-90442.html#p722552when-the-positive-integer-a-is-divided-by-5-and-125591.html#p1028654Hope it helps.
To elaborate more. How to derive general formula of \(n\) based on \(n=10q+5\) and \(n=6p+1\): divisor will be the least common multiple of above two divisors 6 and 10, hence 30. Remainder will be the first common integer in above two patterns, hence 25. So, to satisfy both conditions, \(n\) must be of a type \(n=30m+25\): 25, 55 or 85.
Thanks very much. Above would have been very difficult to figure out in the real exam.