Remainder is 'k-2' when 26 is divided by k. k=?

1. k > 5. we can have k =7 which gives remainder of 5 which is k-2. however, we can also have k = 28 which gives remainder of 26, which is k-2. hence k could be 7 or 28. so Insuff. throw A, D out of window.

2. k < 10. we know k = 7 satisfies the given constraint. But k = 4 also satisfies the given constraints. two values so insuff. cross off B.

Together we know, 5 < k < 10 only k = 7 satisfies given constraints so suff. Answer C



I think answer is C What is OA and OE?


Hope this helps and please post OE, thanks.

Hi Sumg

Though the OA given is C, I think the answer should be A

Since from the first stmt, the only possible value of K that satisfies the condidition is K=7
From Stmt 2: there are two possible values of K<10 which can be either K=2 or K=4.

Could you please confirm whether my reasoning is correct. Thanks.

Hi,
26 when divided by k leaves remainder(k-2)
So, 26 - (k-2) = k*p, where p is the quotient
So, 28-k = kp =>k(p+1) = 28 where k>2
So, k can be 4,7,14,28

From(1): k>5
So, k can be 7,14,28
Not sufficient

From(2): k<10
So, k can be 4 or 7
Not sufficient

Both(1) and (2):
k=7
Sufficient

Hence, C

love the algebraic solution by Frank :

Its important to note a lame example - >12/10 = 1 + 2/10

Mapping above to this question :

26/k = P + (k-2)/k
solving above gives :
K(P+1) = 28 where p can be {0,1,3,6...etc- since K is an integer}

Agree with Frank -- We need both the statements to answer this question.

k>5 gives -- 7, 14, 28
& k<10 gives -- 4, 7

combine both and we get our answer --- 7.

IMO:C