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27 Aug 2016, 16:14
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If K is a positive integer; What is the remainder when K^2 is divided by 8
A K is odd
B) K is the smallest odd prime
Kudos for an Algebraic Approach
A K is odd
B) K is the smallest odd prime
Kudos for an Algebraic Approach
27 Aug 2016, 18:37
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St1:
k is odd.
Let k be of type 2n + 1, where n is an even integer.
k^2 = (2n + 1)^2
= 4n^2 + 4n + 1
= 4n (n + 1) + 1
If n is the even integer 0, expression 4n (n + 1) will be 0 and we are left with 1. Dividing 1 by 8 leaves a remainder of 1.
If n is an even integer > 0, expression 4n (n + 1) will have an 8 in it and an extra 1 will be left over. Dividing by 8 will again leave a remainder of 1.
Sufficient.
St2:
k is the smallest odd prime.
Algebraic approach is same as that for st1 since k is again odd.
Alternatively, 3 is the smallest odd prime.
3^2 = 9
9 divided by 8 leaves a remainder of 1.
Sufficient.
Answer (D).
k is odd.
Let k be of type 2n + 1, where n is an even integer.
k^2 = (2n + 1)^2
= 4n^2 + 4n + 1
= 4n (n + 1) + 1
If n is the even integer 0, expression 4n (n + 1) will be 0 and we are left with 1. Dividing 1 by 8 leaves a remainder of 1.
If n is an even integer > 0, expression 4n (n + 1) will have an 8 in it and an extra 1 will be left over. Dividing by 8 will again leave a remainder of 1.
Sufficient.
St2:
k is the smallest odd prime.
Algebraic approach is same as that for st1 since k is again odd.
Alternatively, 3 is the smallest odd prime.
3^2 = 9
9 divided by 8 leaves a remainder of 1.
Sufficient.
Answer (D).
Answer: B
K is a positive integer: k can be 1, 2, 3, 4, 5, ...
1- K is odd: k can be 1, 3, 5, ...
a- let k=1 ⇒ K^2 = 1
dividing 1 by 8, the remainder will be 1 -- kkrrsshh you are right
b- let k=3 ⇒ K^2 = 9
dividing 9 by 8, the remainder will be 1
a ≠ b ⇒ insufficient
2- K is the smallest odd prime ⇒ k is only 1 number - independent of the result - it is sufficient
K is a positive integer: k can be 1, 2, 3, 4, 5, ...
1- K is odd: k can be 1, 3, 5, ...
a- let k=1 ⇒ K^2 = 1
dividing 1 by 8, the remainder will be 1 -- kkrrsshh you are right
b- let k=3 ⇒ K^2 = 9
dividing 9 by 8, the remainder will be 1
a ≠ b ⇒ insufficient
2- K is the smallest odd prime ⇒ k is only 1 number - independent of the result - it is sufficient
