Question

Product of digits of two digit number is 21.if we add 36 to the number the new number obtained is number formed by interchange of digits find the number

Answer 1

Answer: Numbers are 37 and 73.

Step-by-step explanation:

Let the unit digit be 'x'

Let the ten's digit be 'y'.

Original number will be

$10\times \text{ten's digit}+one's\ digit\\\\=10x+y$

On interchanging of digits,

Let unit's digit be 'y'.

Let ten's digit be 'x'.

New number will be

$10\times \text{ten's digit}+one's\ digit\\\\=10y+x$

According to question,

Product of digit of two digit number = 21

So, it becomes,

$xy=21-----------------------(1)$

If we add 36 to the number the new number obtained is number formed by interchange of digits.

$10x+y+36=10y+x\\\\10x-x+36=10y-y\\\\9x+36=9y\\\\36=9y-9x\\\\36=9(y-x)\\\\\frac{36}{9}=y-x\\\\4=y-x\\\\y=4+x----------------------------(2)$

Put the value of eq(2) in Eq(1).

$xy=21\\\\x(4+x)=21\\\\x^2+4x=21\\\\x^2+4x-21=0\\\\x^2+7x-3x-21=0\\\\x(x+7)-3(x+7)=0\\\\(x+7)(x-3)=0\\\\x+7=0,x-3=0\\\\x=-7,x=3$

x=-7 will be rejected.

So, x=3 and

$xy=21\\\\3y=21\\\\y=\frac{21}{3}\\\\y=7$

Hence, original number will be

$10x+y=10\times 3+7=30+7=37$

New number will be

$10y+x=10\times 7+3=70+3=73$

Therefore, Numbers are 37 and 73.

Answer 2

Step-by-step explanation:

unit's digit=X

ten's digit=21/X

Original Number

= 10(21/X)+X

=210/X+X

36 added in original number

210/X+X+36

New Number

=10(X)+21/X

210/X+X+36=10X+21/X

210+x²+36x=10X²+21

210+x²+36x-10X²-21=0

-9x²+36x+210-21=0

-9(x²-4X-21)=0

X²-4x-21=0

a=1,b=-4,c=-21

D=b²-4ac

=(-4)²-4(1)(-21)

=16+84

D=100

D>0

-b-√D/2a

-(-4)-√100/2(1)

4-10/2

-6/2

-3

-b+√D/2a

-(-4)+√100/2(1)

4+10/2

14/2

7

original number

210/x+x

210/7+7

30+7

37