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05 Apr 2006, 09:50
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When 7^m+1 is divided by 5, is the remainder 0?
1). m is 4n+2
2). m is 3b+1
1). m is 4n+2
2). m is 3b+1
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Updated on: 05 Apr 2006, 10:01
Originally posted by jcgoodchild on 05 Apr 2006, 09:59.
Last edited by jcgoodchild on 05 Apr 2006, 10:01, edited 1 time in total.
Last edited by jcgoodchild on 05 Apr 2006, 10:01, edited 1 time in total.
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When 7^m+1 is divided by 5, is the remainder 0?
1). m is 4n+2
2). m is 3b+1
All of this is dependent on whether or not n and b are positive integers. If this is not the case, the answer is E.
Statement 1:
Using m=4n+2, let's rework the original equation.
7^(4n+2) + 1
(7^2)*(7^4n) + 1
(49)*(2401^n) + 1
Now, whatever n is, 2401 will always end in a 1, which means that 7^(4n+2) will always end in a 9. This plus 1 will result in a 0 at the end, which meaks the number divisible by 5 with no remainder. Sufficient.
Statement 2: Same logic.
7^(3b+1) +1
7*(7^3b) + 1
7*(343^b) + 1
Now, the statement is heavily devependent on what b is. But if we think about it, the final digit will either end in 3, 9, 7 or 1. If it ends in 7, then the final number will end in a 9 (which, plus 1 would make the number divisible by 5), but if it ends in 3, 9 or 1, the final number will end in either a 1, 3 or 7 respecitively, and adding 1 would not make the number divisible by 5. Therefore, Insufficient.
Answer is A
1). m is 4n+2
2). m is 3b+1
All of this is dependent on whether or not n and b are positive integers. If this is not the case, the answer is E.
Statement 1:
Using m=4n+2, let's rework the original equation.
7^(4n+2) + 1
(7^2)*(7^4n) + 1
(49)*(2401^n) + 1
Now, whatever n is, 2401 will always end in a 1, which means that 7^(4n+2) will always end in a 9. This plus 1 will result in a 0 at the end, which meaks the number divisible by 5 with no remainder. Sufficient.
Statement 2: Same logic.
7^(3b+1) +1
7*(7^3b) + 1
7*(343^b) + 1
Now, the statement is heavily devependent on what b is. But if we think about it, the final digit will either end in 3, 9, 7 or 1. If it ends in 7, then the final number will end in a 9 (which, plus 1 would make the number divisible by 5), but if it ends in 3, 9 or 1, the final number will end in either a 1, 3 or 7 respecitively, and adding 1 would not make the number divisible by 5. Therefore, Insufficient.
Answer is A
