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reminders

by hpgmat » Tue Dec 08, 2009 7:54 pm
if n and m are positive integers , what is the reminder when 3 ^ (4n + 2) + M is divided by 10

n= 2
m=1
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by papgust » Wed Dec 09, 2009 12:03 am
IMO C

I assume that this is the expression - [3 ^ (4n + 2)] + M.

(1) and (2) are clearly insufficient alone. Because, the value of m or n can be any no and when divided by 10, you get different possible remainders.

Take n=2, 3 ^ (8+2) + M = 3^10 + M = 59049 + M. M could take any value from 0-9 and when added with the 59049 and divided by 10, different remainder values turn up.

Combined, only then it is sufficient.
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by hpgmat » Wed Dec 09, 2009 12:23 am
papgust wrote:IMO C

I assume that this is the expression - [3 ^ (4n + 2)] + M.

(1) and (2) are clearly insufficient alone. Because, the value of m or n can be any no and when divided by 10, you get different possible remainders.

Take n=2, 3 ^ (8+2) + M = 3^10 + M = 59049 + M. M could take any value from 0-9 and when added with the 59049 and divided by 10, different remainder values turn up.

Combined, only then it is sufficient.
your assumption is correct
Answer is incorrect
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by Ian Stewart » Wed Dec 09, 2009 3:19 am
hpgmat wrote:if n and m are positive integers , what is the reminder when 3 ^ (4n + 2) + M is divided by 10

n= 2
m=1
Notice that 3^(4n + 2) = (3^4n)(3^2) = (3^4)^n * 9 = 81^n * 9. Since the units digit of 81^n will be 1 no matter what the value of n, the units digit of 81^n * 9 will always be 9. So we don't care what n is. We do need the value of m, however, so the answer is B.
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by hpgmat » Wed Dec 09, 2009 12:04 pm
Ian Stewart wrote:
hpgmat wrote:if n and m are positive integers , what is the reminder when 3 ^ (4n + 2) + M is divided by 10

n= 2
m=1
Notice that 3^(4n + 2) = (3^4n)(3^2) = (3^4)^n * 9 = 81^n * 9. Since the units digit of 81^n will be 1 no matter what the value of n, the units digit of 81^n * 9 will always be 9. So we don't care what n is. We do need the value of m, however, so the answer is B.
Perfect explanation.
Thank you so much.
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