Question

can anyone solve this question

Answer 1

Let tens number be x
Let units number = x+2
Then the number = 10x+x+2
=11x+2
Sum of digits = x+x+2
=2x+2
According to the question
"product of given number and sum of its digits is 144"
Therefore,
(11x+2)(2x+2) = 144
22x²+22x+4x+4 = 144
22x²+26x+4 = 144
11x²+13x+2 =72
11x²+13x+2-72 = 0
11x²+13x-70 = 0
11x²(-22x+35x)-70 = 0
11x(x-2)+35(x-2) = 0
(11x+35)(x-2) = 0
11x+35 = 0, x-2 = 0
X= - 35/11, x=2
Then 10s digit =x = 2
Units digit = 2+2 = 4
Therefore the number = 24

Answer 2

Given Information:
A two digit number is to be found, in which the units place exceeds the tens place by two and the product of the given digits, and their sum is together equal to 144.

Let x denote the ten's place for this two digit number.
Equation for Unit's digit is x + 2
Then we get the formation for the number in equational terms as,
 10x + (x+2) = 11 x+2
Now, the sum of the two digits can be represented using
x + (x+2) = 2x + 2
Now to find out the final number, we will multiply the formation of the number as well as the sum of the two digits to find our answer,
(11 x+2)(x+2) = 144 (As per the condition given to us about the product)
$22 x^{2} + 26x - 140 = 0$
$11 x^{2} + 13x - 70 = 0$
$(x - 2)(11x + 35) = 0$
$11x+35 = 0 or x-2 = 0$
x = -35/11 or x = 2
x = -35/11 is not satisfactory, as it does not satisfy any conditions.
So, x = 2 is the value.
Now put the value in the primary equation, which was
11x + 2
We get, 11(2) + 2 = 22 + 2 = 24.
Hence, 24 satisfies the condition and is a valid answer.