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27 Apr 2016, 00:10
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B
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Difficulty:
35%
(medium)
Question Stats:
66% (01:14) correct
34%
(01:21)
wrong
based on 169
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If A and D are positive integers, what is the remainder when 7^(40A) + D is divided by 5?
(1) A = 5
(2) D = 4
(1) A = 5
(2) D = 4
Balajikarthick1990
27 Apr 2016, 01:58
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[quote="Bunuel"]If A and D are positive integers, what is the remainder when 7^(40A) + D is divided by 5?
(1) A = 5
(2) D = 4
7^200 ends in 1 ...so if D is 4 then rem=0 else reminder = something else
pattern of 7 ends in 7,9,3,1
7+4=11 reminder is different
9+4=13....
Combination yes possible C
(1) A = 5
(2) D = 4
7^200 ends in 1 ...so if D is 4 then rem=0 else reminder = something else
pattern of 7 ends in 7,9,3,1
7+4=11 reminder is different
9+4=13....
Combination yes possible C
27 Apr 2016, 02:02
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40A = 4*10*A --> Multiple of 4
7^40A will always have a units digit of 1 as cyclicity of 7^x is 7, 9, 3, 1.
A digit will be completely divisible by 5 if the units digit is 5 or 0.
St1: A = 5 --> Not Sufficient as we already know that 7^40A will have units digit as 1
St2: D = 4 --> Units digit(7^40A) + D = 1 + 4 = 5
The term is completely divisible by 5 and hence remainder is 0.
Sufficient
Answer: B
7^40A will always have a units digit of 1 as cyclicity of 7^x is 7, 9, 3, 1.
A digit will be completely divisible by 5 if the units digit is 5 or 0.
St1: A = 5 --> Not Sufficient as we already know that 7^40A will have units digit as 1
St2: D = 4 --> Units digit(7^40A) + D = 1 + 4 = 5
The term is completely divisible by 5 and hence remainder is 0.
Sufficient
Answer: B
