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Re: In multiplication of the two digit numbers wx and cx, where w, x, and [#permalink]
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11 Apr 2018, 09:49
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Given : wx*cx = _ _ _ . Asked is w+cx.
(1). If three digits of the product are all the same, then the product has only 9 possibilities which are : (111, 222, 333, 444, .....................,888,999). Now, we already know that 111= 37*3, hence all the possibilities can be rewritten as follows: (3*37 , 6*37 , 9*37 ,.....................,24*37,27*37) Only the last possibility matches the given sets of conditions. Hence 27*37=999, because in this case ten's place is the same digit for both 27 as well as 37. So, w+cx = 2+37 = 4 (Easily deduced). Hence SUFFICIENT !
(2). This mentions that x and w + c are both odd numbers. There are many possibilities. We know only ODD+EVEN=ODD. So,either of (w,c) are (odd,even) or (even,odd). And x =ODD. Case1. Let's say for one possibility, x=1; w=2; c=3 . This also satisfies the inital given condition that multiplication of the two digit numbers gives 3 digited product. Hence, So, w+cx = 2+31 = 4. Case2. Now for another possibility, say x=3; w=1; c=2. This also satisfies the inital given condition that multiplication of the two digit numbers gives 3 digited product. Hence, So, w+cx = 1+23 = 0. (There could be many other cases too but we can conclude from 2 cases only.) So both the case gives a different answer, hence we cannot arrive at a unique solution from (2). Hence INSUFFICIENT !
So, the final answer to the question should be option A.
