|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
11 May 2019, 02:56
Kudos
Bookmarks
D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
23% (03:01) correct
77%
(03:03)
wrong
based on 61
sessions
History
Date
Time
Result
Not Attempted Yet
'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
Kudos
Bookmarks
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
Kudos
Bookmarks
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
