A number of two digits is five times the sum of its digit and it is 5 times more than the double digit of its digits . find the numbers.
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let the tens digit be x and ones digit be y
number=10x+y
sum of digits =x+y
product of digits=xy
axcording to question: 10x+y=5 (x+y)
10x+y=5x+5y
5x=4y
x=4/5y
then 2nd equation is: 10x+y=2xy+5
putting the value of x in the above equation:
10×4/5y+y=2×4/5y×y +5
8y+y=8/5y^2+5
45y=8y^2+25
8y^2-45y+25=0
solving quadratic equation:
8y^2-40y-5y+25=0
8y(y-5)-5 (y-5)=0
(8y-5)(y-5)=0
y=5 as digit cannot be 5/8, i.e in fraction
x=4/5y=4/5*5=4
therefore number=45
number=10x+y
sum of digits =x+y
product of digits=xy
axcording to question: 10x+y=5 (x+y)
10x+y=5x+5y
5x=4y
x=4/5y
then 2nd equation is: 10x+y=2xy+5
putting the value of x in the above equation:
10×4/5y+y=2×4/5y×y +5
8y+y=8/5y^2+5
45y=8y^2+25
8y^2-45y+25=0
solving quadratic equation:
8y^2-40y-5y+25=0
8y(y-5)-5 (y-5)=0
(8y-5)(y-5)=0
y=5 as digit cannot be 5/8, i.e in fraction
x=4/5y=4/5*5=4
therefore number=45
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