Question

a 2digit number is 5 times the sum of its digit .if 9 is added to the number its digits are reversed.find the no.

Answer 1

Let’s call the demanded (decimal) number “XY”
Then the to given conditions tranlate to
I.: X+Y=5 
I.’: X=5-Y

II.: X*10 + Y + 9 = Y*10 + X (i.e. “YX”)

10X + Y + 9 = 10Y + X | -X
9X + 9 + Y = 10Y |-Y

II.’: 9X + 9 =9Y

I.’ in II’.: 
9(5-Y) + 9 = 9Y
45 - 9Y + 9 = 9Y |+9Y 
54 = 18Y =>

II.’’: Y = 3, II’’. in I’. => X = 2

=> “XY” = 23 
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Answer 2

Hey

Let a and b the digits we can write the number as "ab" but the number is 10*a+b
(for example if a=3 and b=2 the number would be 32 and 32 =10*3+2) 

a two digit number is 5 times the sum of its digit then 10*a+b=5(a+b)
when 9 is added to the number the result is the original number with its reversed 10*a+b+9=10*b+a 
then we have the following equations: 
10*a+b=5(a+b)
10*a+b+9=10*b+a 
then 
10a+b=5a+5b
10a+b+9=10b+a 
then 
5a-4b=0
9a-9b=-9 
then 
5a-4b=0
a-b=-1 
then using 2nd eq a=b-1 
using 1st eq 5(b-1)-4b=0 
then 5b-5-4b=0
then b=5 and a=5-1=4 
so the original number is 45 


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