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25 Feb 2004, 18:02
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If (3^4n) +1 is divided by 10, can the remainder be 0?
a) n=2m+1, m is a positive integer
b) n> 4
Answer with reason please!
a) n=2m+1, m is a positive integer
b) n> 4
Answer with reason please!
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25 Feb 2004, 18:10
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3^4n = 81^n
for Any value of n if it is integer be it even of odd (81^n+1) will not yield 0 as the last digit of the number will be 2
B is not a sufficient condition because n could be any number not necessarily integer.
For example n = 4.5 you have 81^4 * 81^0.5 + 1 which is divisible by 10
A is sufficient.
for Any value of n if it is integer be it even of odd (81^n+1) will not yield 0 as the last digit of the number will be 2
B is not a sufficient condition because n could be any number not necessarily integer.
For example n = 4.5 you have 81^4 * 81^0.5 + 1 which is divisible by 10
A is sufficient.
26 Feb 2004, 07:04
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Hmmm.
I came up with B.
3 raised to anything will have the unit digit end in 3,9,7 or 1 and starts after every 4th exponent. Example 3^4 = 81, 3^8 = 6561.
From (a):
n is an integer and n > 1 or n = 1
4n will always end in 1. so ((3^4n) + 1)/10 will never have a remainder of 0.
From (b):
n > 4 and n may not be an integer so 3^4n may have unit digit = 9 and in this instance, (3^4n) + 1 will have unit digit = 0 which is divisible by 10.
I came up with B.
3 raised to anything will have the unit digit end in 3,9,7 or 1 and starts after every 4th exponent. Example 3^4 = 81, 3^8 = 6561.
From (a):
n is an integer and n > 1 or n = 1
4n will always end in 1. so ((3^4n) + 1)/10 will never have a remainder of 0.
From (b):
n > 4 and n may not be an integer so 3^4n may have unit digit = 9 and in this instance, (3^4n) + 1 will have unit digit = 0 which is divisible by 10.
