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NathanMasky
16 Apr 2025, 14:35
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
85%
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Question Stats:
47% (01:58) correct
53%
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wrong
based on 131
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What is the remainder when a six digit number N = aaabcb is divided by 111 where a, b and c are distinct natural numbers?
A. 10 * a
B. 10 * b + c
C. 10 * (c - b)
D. 10 * c
E. None of the above
A. 10 * a
B. 10 * b + c
C. 10 * (c - b)
D. 10 * c
E. None of the above
19 Apr 2025, 05:03
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NathanMasky
Adhering to the restrictions:
Let the # be \(111232\), thus be have \(111232 = 111 *(1002) +10\) i.e. Remainder is \(10\)
Substituiting the values of \(a,b,c\) in the options we can elimate all except options \(1\) and \(3\).
\(3\) is not eliminated because \(-ve\) remainder is not possible.
Let the # be \(555262,\) ( taking \(c >b\) , to avoid negative remainder)
\(555262 = 111* (5002) + 40 \)
Eliminate \(1\).
Ans C
Note: Once an option has been eliminated, no need to test it again.
Hope it helped.
26 Apr 2025, 07:02
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NathanMasky
Lets convert the number to work with its digit
N=AAABCB = A*100000+A*10000+A1000+B*100+C*10+B = A (111000) + B (101) + C 10
Lets divide now this by 111:
A (111000)/111 + B (101)/111 + C 10/111 = A*1000 (integer) + B(101)/111 + C 10/111 -> We can forget about A term as it will be an integer with reminder 0.
The second term; B *101/ 111 = B * (111-10) / 111 = B *111/111 (integer, we can forget about it)- B *10/111 -> reminder will be -10B as -10B will always be less than 111.
The last term; C 10/111 -> the reminder will always be C10 as regarless of C, it will always be less than 111.
The reminder will be the sum of the two reminders = -10B + C10 = 10 (C-B)
Answer: C
