Question

a two digit number is 4 times the sum of its digits and twice the product of the digits find the number

Answer 1

Hi ,

Let ten's place digit = x

Units place digit = y

Required number = 10x + y ----( 1 )

According to the problem given ,

10x + y = 4( x + y )

10x + y = 4x + 4y

6x = 3y

6x/3 = y

2x = y ---( 2 )

And

10x + y = 2xy

10x + 2x = 2x × 2x [ from ( 2 ) ]

12x = 4x²

0 = 4x² - 12x

4x( x - 3 ) = 0

4x = 0 or x - 3 = 0

x = 0 or x = 3

Put x = 3 in equation ( 2 ) , we get

y = 2 × 3 = 6

Now ,

Required number = 10x + y

= 10 × 3 + 6

= 30 + 6

= 36

I hope this helps you.

: )

Answer 2

Hello dear user.....
Here is your answer
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$let \: \: the \: place \: \: of \: ten \: digit \: \: = x \\ and \: \: \: units \: \: place \: = y \\ \\ so \: the \: \: required \: \: number \: \: = 10x + y \: \: \: \: \: \: \: \: \: ......(1) \\ \\ given \: in \: question \: \: \\ the \: sum \: of \: digits \\ 10x + y = 4(x + y) \\ = > 10x + y = 4x + 4y \\ = > 10x - 4x = 4y - y \\ = > 6y = 3x \\ y = \frac{6x}{3} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: ........(2) \\ \\ and \: given \: \: twice \: product \: \: of \: the \: digits \\ \\ 10x + y = 2xy \\ = > 10x + 2x = 2x \times 2x \\ = > 12x = 4 {x}^{2} \\ = > 0 = 4 {x}^{2} - 12x \\ = > 4x(x - 3) = 0 \\ = > 4x = 0 \: \: or \: \: x - 3 = 0 \\ = > x = 3 \: \\ \\ putting \: \: x = 3 \: in \: equation \: \: (2) \: \\ we \: get \: \\ y = 2 \times 3 \\ y = 6 \\ \\ so \: \\ in \: question \: \: required \: \: number \: \\ 10x + y \\ = > 10 \times 3 + 6 \\ = > 30 + 6 \\ = > 36$
so, the required number is 36 .

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