We are looking for the remainder of n being divided by k.
1. It's useful to know that the formula for raising (a + b) to the third power:
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3.
This means that n will be k^3 + 3k^2 + 3k + 1. Now, you can clearly see that the remainder will be 1 here, since k^3, 3k^2 and 3k are all divisible by k.
2. is insufficient. Knowing k = 5 is not enough: we also need some info about n.
What is the remainder ...
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cramya
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Rules:
1) (a+1)^x / a will always give a remainder of 1 where a is an integer >1
2) a^x / a+1 will always give a remainder of 1 when n is even and will always give a remainder of a when n is odd where a is an integer
Applying rule 1 to stmt I the remainder is always 1 since a=k and n = (a+1)^x i.e (a+1)^3
The other way would be to substitute numbers
k=2
n=9
remainder 1
k=3
n=64
remainder 1
k=4
n=124
remainder 1
It follows a pattern....
suff
Choose A
1) (a+1)^x / a will always give a remainder of 1 where a is an integer >1
2) a^x / a+1 will always give a remainder of 1 when n is even and will always give a remainder of a when n is odd where a is an integer
Applying rule 1 to stmt I the remainder is always 1 since a=k and n = (a+1)^x i.e (a+1)^3
The other way would be to substitute numbers
k=2
n=9
remainder 1
k=3
n=64
remainder 1
k=4
n=124
remainder 1
It follows a pattern....
suff
Choose A
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logitech
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DanaJ, we are all very lucky to have you on board! Thanks for the great work you are doing on this forum. It is joy to read and learn from your posts. I am sure I talk for everbody on this forum.DanaJ wrote:We are looking for the remainder of n being divided by k.
1. It's useful to know that the formula for raising (a + b) to the third power:
(a + b)^3 = a^3 + 3a^2b + 3ab^2+ b^3.
This means that n will be k^3 + 3k^2 + 3k + 1. Now, you can clearly see that the remainder will be 1 here, since k^3, 3k^2 and 3k are all divisible by k.
2. is insufficient. Knowing k = 5 is not enough: we also need some info about n.
LGTCH
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