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In multiplication of the two digit numbers wx and cx, where w, x, and

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In multiplication of the two digit numbers wx and cx, where w, x, and  [#permalink]

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New post 11 Apr 2018, 09:09
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A
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D
E

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  95% (hard)

Question Stats:

45% (02:47) correct 55% (02:40) wrong based on 174 sessions

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In multiplication of the two digit positive numbers wx and cx, where w, x, and c are unique non-zero digits, the product is a three digit number. What is w + c - x?

(1) The three digits of the product are all the same and different from w, c and x.
(2) x and w + c are odd numbers.

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Re: In multiplication of the two digit numbers wx and cx, where w, x, and  [#permalink]

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New post 11 Apr 2018, 09:49
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1
Given : wx*cx = _ _ _ . Asked is w+c-x.

(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 re-written 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+c-x = 2+3-7 = -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+c-x = 2+3-1 = 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+c-x = 1+2-3 = 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.
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Re: In multiplication of the two digit numbers wx and cx, where w, x, and  [#permalink]

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New post 22 Apr 2018, 18:33
I) Number could be 111,222,....999
111 has factors as 37*3
222 has 37*6
...
...
999=37*27 only 999 has same digit in its unit place.
Sufficient

II) can have multiple combination

So answer is A

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Re: In multiplication of the two digit numbers wx and cx, where w, x, and  [#permalink]

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New post 13 Sep 2018, 09:04
3
Stem: wx and cx product to be commented upon.
(10w+x)(10c +x) = 100(wc) + 10wx + x^2 so , this is some number with three digits as wc (w+c)x and x^2 ----> hundreds tens and units digit.
we are asked w+c-x
1. all three digits are same. this means (w+c)x = x^2 ----> x= w+c. Hence, w+c-x =0. Sufficient
2. x is odd means, in our number x^2 can only be 1 or 9 ----> x can be 1 or 3. Ok Units digit done, lets move to tens digit.
w+c is odd means, tens digit---> odd*1 or odd*3. The relation between w+c and x cannot be determined hence insufficient.
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Re: In multiplication of the two digit numbers wx and cx, where w, x, and  [#permalink]

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New post 12 Oct 2018, 04:52
Wait, I found that I can calculate from statement 2.
1). X^2 is an 1d interger and x is odd => x = 1 or x = 3.
2) If x=1 => w+c=1 => impossible bc w is not equal to c.
3). If x=3 => w+c=3. It’s possible bs w and c may be 1 and 2 and vice versa

=> So, x=3 and w+c=3
Then w+c-x= 0.

So the answer should be D, right?

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Re: In multiplication of the two digit numbers wx and cx, where w, x, and &nbs [#permalink] 12 Oct 2018, 04:52
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In multiplication of the two digit numbers wx and cx, where w, x, and

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