Question

The question is in the chapter playing with numbers
So please help me to do this sum
The sum of the digit of a two digit number is 9 the
number is 6 times the unit digit . Find the number

Answer 1

Hello ather!!

Let the numbers be a and b..
a+b=9(sum of the 2 digits is 9)
a=9-b ----------(i)
As the the number is 6 times the units digit
Tens unit should be ×10
a×10+b
=10a+b=6b
{from equation (i)}
10(9-b)+b=6b
=90-10b+b=6b
90-9b=6b
90=15b
b=6
as
a=9-b from equation (i)
a=9-6=3
a=3...
Hence the required number is 10a+b
10×3+6
=36..
Hence the required number is 36!!
cheers!!


VERIFICATION:-
number =36
3+6=9....

36=6×units digit
36=6×6
36=36...
Hence verified!!!

Answer 2

Hello...

Here is your answer...

Let t = the tens digit, u = the unitis digit, t + u = 9

Solve t by adding -u to each side.

t = 9 - u

value of number is 10t + u.

(10t + u) = 6u

Substitute (9 - u) for t

10(9 - u) + u = 6u

90 - 10u + u = 6u

90 - 9u = 6u

add 9u

90 = 15u

divide side by 15

6 = u

t+u = 9

t + 6 = 9

add -6

t = 3

t+u = 9, 3+u = 9 and u = 6

number is 36