ShankSouljaBoi wrote:
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.
This a very elegant way of solving this question, tanks ShankSouljaBoi. I'm just reframing your sentences:
we have 2 numbers wx and cx and their product results in a 3 digit number
Solution:
1.lets rewrite wx and cx as (10w+x) and (10c+x) respectively
2. lets multiply the two numbers, wx * cx => (10w+x) * (10c+x)=> 100(wc) + 10x(c+w) + x^2 -> equation 1
3. for 3 digit number abc, its of the form => 100a+10b+c -> equation 2
4. we can compare equation 1 and equation 2
5. coming back to the question and more specifically the first statement, it says that if all the digits in the third equation are equal, then a=b=c => wc=w(c+x)=x2 -> equation 3
6. lets look at the second part in equation 3, x(c+w)=x^2 => w+c=x => w+c-x=0, so we can arrive at the solution
7. for the second statement I agree with what ShankSouljaBoi said, but IMO since we are sure that statement 1 is sufficient to answer the question, I didn't bother looking at statement 2, hence the answer to the solution is A