Debashis Roy wrote:
Bunuel VeritasKarishma chetan2uFor Statement(2): (2) The tens’ digit of x is 2 more than the units digit, and the tens digit of y is 2 less than the units digit.
If we approach analytically, Lets assume units digit of X is x...SO tens digit of X is x+2...X=10(x+2)+x=11x+20...(1)
Lets assume units digit of Y is y...SO tens digit of Y is y-2... Y=10(y-2)+y =11y-20...(2)
From (1) and (2)...X-Y=11(x-y)+40...
so (X-Y)/9= [11(x-y)+40]/9....
40/9 gives a remainder of 4.....So if 11(x-y)/9 gives a remainder of 5 or -4.....only then will the entire expression be divisible by 9...
(x-y) will range from 0 to 9 (+/-)....so 11(x-y) can have values 0,11,22,33,..99.....which will never give a remainder of 5 when divided by 9....
So this entire expression will never be divisible by 9....
Plz explain where did I go wrong..
Thanks
Considering x and y to be single digits as you defined them,
x and (x+2) need to be single digits so the value of x can vary from 0 to 7.
y and (y-2) need to be single digits such that (y-2) is not 0 so the value of y can vary from 3 to 9
Value of (x -y) can go from -9, -8, ... -3, -2, -1, 0, 1, 2, 3, 4
When (x-y) = -2, 11*(x - y) = -22 which gives remainder 5.
In other words, when say x = 1 and y = 3, the two two-digit numbers X and Y are 31 and 13.
(31 - 13)/9 = 18/9 = 2 is an integer.
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Karishma
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