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02 Feb 2024, 02:30
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E
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Difficulty:
25%
(medium)
Question Stats:
76% (01:13) correct
24%
(01:29)
wrong
based on 171
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When a positive integer x is divided by 7, the remainder is 3. What is the remainder when x^3 is divided by 7?
A. 2
B. 3
C. 4
D. 5
E. 6
A. 2
B. 3
C. 4
D. 5
E. 6
When x is divided by 7, the remainder is 3. When x^3 is divided by 7, what is the remainder?
02 Feb 2024, 03:54
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x=7k+3
when k=0, x=3; xcube=27 and remainder is 6
Similarly when k=1, x=10; x^3=1000 i.e 142*7 + 6
IMO E
when k=0, x=3; xcube=27 and remainder is 6
Similarly when k=1, x=10; x^3=1000 i.e 142*7 + 6
IMO E
02 Feb 2024, 11:20
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Correct option is E
Said that when x is divided by 7 remainder is 3. So, we can assume the number x to be x = 7*k + 3
So, \(x^3\) = \((7k+3)^3\)
=\((7k)^3\)+3*\((7k)^2\)*3+3*7k*\(3^2\)+27
All the first 3 terms are divisible by 7, only the last term 27 is not divisible which gives a remainder of 6.
Hence, correct option is option E
Said that when x is divided by 7 remainder is 3. So, we can assume the number x to be x = 7*k + 3
So, \(x^3\) = \((7k+3)^3\)
=\((7k)^3\)+3*\((7k)^2\)*3+3*7k*\(3^2\)+27
All the first 3 terms are divisible by 7, only the last term 27 is not divisible which gives a remainder of 6.
Hence, correct option is option E
