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The square of two digit number(ab) divided by half of itself resultant added by 18 then divided by 2 then we get reverse of original number(ba). how many combination of ab exist

1. 9
2. 8
3. 7
4. 6
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$\frac{(ab)^{2}}{\frac{1}{2}* ab} =2* ab$

2*ab+18

$\frac{2*ab+18}{2}$

ab+9 =ba

ba-ab=9

their are 8 possibilities

21-12 =9

32-23=9

43-34=9

54-45=9

65-56=9

76-67=9

87-78=9

98-89=9
answered by (1.9k points) 8 22 42
edited
$(2(10a+b)^2/(10a+b) + 18) / 2 = 10b+a \Rightarrow 10a+b+9 = 10b +a \Rightarrow 9a-9b+9 = 0 \Rightarrow a - b + 1 = 0$

Therefore the possible values of a and b are:

a    b

1    0

2    1

3    2

.

.

.

9    8

Hence, possible combinations of a and b are 9.