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11 May 2019, 02:56
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
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Question Stats:
23% (03:01) correct
77%
(03:03)
wrong
based on 60
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'abc' is a 3 digit natural number such that 'abc' = \(a^3 + b^3 + c^3\) and 300 < 'abc' < 400. What is the value of sum of digits of 'abc' ?
1. 'abc' is an even number
2. 'abc' is a multiple of 5
Note- 'abc' represents 3 digit number, not the multiplication of a,b and c.
1. 'abc' is an even number
2. 'abc' is a multiple of 5
Note- 'abc' represents 3 digit number, not the multiplication of a,b and c.
11 May 2019, 19:42
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nick1816
a is clearly 3..
So 3bc=\(3^3+b^3+c^3\)...
Now b and c can be any of the digits..
Let us check cubes...
\(2^3=8.......3^3=27.........4^3=64.......5^3=125......6^3=216.......7^3=343\)
So b and c are one of digits from 0 to 7...
Now \(300<3^3+b^3+c^3<400..........273<b^3+c^3<373\)
Let us check statements..
I. abc is even.
So c can be one of 0, 2, 4, 6
Now \(273<b^3+0^3<373....\) ..b could only be 7.... So abc=370 and \(a^3+b^3+c^3=3^3+7^3+0^3=27+343=370\)... Possible.
None other will fit in.
Sufficient
II. abc is divisible by 5.
So c is 0 or 5..
If C is 0, we have seen above the number is possible and is 370.
If C is 5, then abc=3b5 and \(300<3^3+b^3+5^3<400.....148<b^3<248\)
b could be 5 or 6..
If 5, then \(3^3+5^3+5^3\), this will not equal abc as the last digit will not be 5...27+125+125
If 6, then \(3^3+6^3+5^3\), this will not equal abc as the last digit will not be 5...27+216+125
Only possibility is 370
Sufficient
D
11 May 2019, 22:49
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a is clearly 3..
So 3bc=\(3^3+b^3+c^3\)...
Now, 5< max(b,c) <8 must be true, because if max(b,c) is 8 or 9, a^3 + b^3 + c^3 will exceed 400. If max(b,c) is less than 6, a^3 + b^3 + c^3 will be smaller than 300
Case 1. When b or c equals to 6
Let b=6,
300 < a^3 + b^3 + c^3 < 400
300< 27+216+c^3< 400
57< c^3< 157
c can take only 4 or 5
When c is 4, a^3 + b^3 + c^3= 307 not equal to abc
When c is 5, a^3 + b^3 + c^3= 368 not equal to abc
Similarly When c=6, b can take 4 or 5...in both cases a^3 + b^3 + c^3 is not equal to abc
Case 2. When b or c equals to 7
Let b=7
300 < a^3 + b^3 + c^3 < 400
300< 27+343+c^3<400
or c^3< 30
c can take 0,1,2 or3
only c=0 or 1 satisfies a^3 + b^3 + c^3=abc
When c=7, b can take 0,1,2 or 3. None will satisfies our condition.
Hence 370 and 371 are only two 3-digits number between interval 300 and 400, when a^3 + b^3 + c^3=abc
Statement 1. When abc is even
abc=370
Sufficient
Statement2 When abc is multiple of 5
abc=370
Sufficient
Answer is D
So 3bc=\(3^3+b^3+c^3\)...
Now, 5< max(b,c) <8 must be true, because if max(b,c) is 8 or 9, a^3 + b^3 + c^3 will exceed 400. If max(b,c) is less than 6, a^3 + b^3 + c^3 will be smaller than 300
Case 1. When b or c equals to 6
Let b=6,
300 < a^3 + b^3 + c^3 < 400
300< 27+216+c^3< 400
57< c^3< 157
c can take only 4 or 5
When c is 4, a^3 + b^3 + c^3= 307 not equal to abc
When c is 5, a^3 + b^3 + c^3= 368 not equal to abc
Similarly When c=6, b can take 4 or 5...in both cases a^3 + b^3 + c^3 is not equal to abc
Case 2. When b or c equals to 7
Let b=7
300 < a^3 + b^3 + c^3 < 400
300< 27+343+c^3<400
or c^3< 30
c can take 0,1,2 or3
only c=0 or 1 satisfies a^3 + b^3 + c^3=abc
When c=7, b can take 0,1,2 or 3. None will satisfies our condition.
Hence 370 and 371 are only two 3-digits number between interval 300 and 400, when a^3 + b^3 + c^3=abc
Statement 1. When abc is even
abc=370
Sufficient
Statement2 When abc is multiple of 5
abc=370
Sufficient
Answer is D
