Bunuel wrote:

The three-digit integer kss is the sum of the two-digit integers ks and rs, where k, r, and s are the digits of the integers. Which of the following must be true?

I. k = 2

II. r = 9

III. s = 5

A. I only

B. II only

C. III only

D. I and II

E. II and III

Number kss =100k+10s +s

Number ks = 10k+s

Number rs = 10r+s

So, 100k + 10s + s = 10k + s +10r + s

-> 90 k + 9s = 10 r

-> r = 9 (10k + s)/10

To make r an integer s must be a multiple of 10, So s = 0 (to cancel 10 in denominator).

we can't take s as 5 since if s = 5 , r = 9*5(2k+1)/10 . So it won't result an integer r.

Now, if s = 0 , r = 9(10k+0)/10 = 9k

So r i a multiple of 9, r= 0 or 9 . But r can't be 0 as it is tens digit, also if r = 0, then k =0 (not possible)

So r = 9

-->9k = 9

--> k =1

So the values are k = 1 , r = 9 , s = 0

hence only II is correct .. Answer B.

We can't solve it by plugging values in first go as, we have 3 unknowns. So we need to simplify it by common logic and then we can put the values and check..

Anyway it was a tricky but nice question.

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