Suppose there is a two digit number , whose tens digit is three times its ones digit. If its digits are reversed and this new number is added to 36 , then we get the original number. Find the original number.

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Solution

according to question
let, the unit digit = x
then ,the tens digit = 3x
because according to question tens digit is 3 times of unit digit.
now the number = 10(3x) + x = 31x
put it eq'n 1 now ,
reversing the original number
10(3x) + x = 10x + 3x
10x + 3x = 31x -36
36 = 31x -13x
next 36 = 18x
now x = 2 .
put the value of x in eq'n 1
, 31(2)= 62
the original number obtained is 62

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