In a 3 digit number, sum of the digit is 9 and the unit digit is twice the the the tens digit. on adding 99 to the number, the digit are interchanged. Find the number.
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1. a+b+c=9
2.c= 2b
3. 100c+10b+ a = 100a+10b+c+99
Next substitute given 2 into given 3.
100(2b) + 10b+a =100a+10b+c+99
210b-12b+a = 100a+12b-12b+99
198b+a-a=100a-a+99
198b-99= 99a+99-99
(198b-99)/99=99a/99
a=2b-1
Now substitute the value for a, and for c into given 1.
(2b-1) + b + (2b) = 9
5b-1+1=9+1
5b/5=10/5
b = 2
a=2b -1 so a = 3
c = 2b, so c = 4
So the number is 234.
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