Question

A two digit number is such that the product of its digits is 35. When 18 is added to the number, the digits interchange their places. Find the number

Answer 1

Answer:  Numbers are 57 and 75.

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 = 35

So, it becomes,

$xy=35-----------(1)$

If we add 18 to the number the new number obtained is number formed by interchange of digits.

$10x+y+18=10y+x\\\\10x-x+18=10y-y\\\\9x+18=9y\\\\18=9y-9x\\\\18=9(y-x)\\\\\frac{18}{9}=y-x\\\\2=y-x\\\\y=2+x-------------------(2)$

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

$xy=21\\\\x(2+x)=35\\\\x^2+2x=35\\\\x^2+2x-35=0\\\\x^2+7x-5x-35=0\\\\x(x+7)-5(x+7)=0\\\\(x+7)(x-5)=0\\\\x+7=0,x-5=0\\\\x=-7,x=5$

x=-7 will be rejected.

So, x=5 and

$xy=35\\\\5y=35\\\\y=\frac{35}{5}\\\\y=7$

Hence, original number will be

$10x+y=10\times 5+7=30+7=57$

New number will be

$10y+x=10\times 7+5=70+5=75$

Therefore, Numbers are 57 and 75.

Answer 2

→Given←

A two-digit number is such that the product of its digits is 35. If 18  is added to the number, the digits interchange their places.

$\rule{180}3$

→To Find←

The number by elimination method.

$\rule{180}3$

→Solution←

Let one of the digits be x, and the other digit is y.

A/q, product of the digits is 35.

⇒ xy = 35             . . .(i)

$\rule{100}2$

Now, when 18 is added, the number gets interchanged.

⇒ 10x + y + 18 = 10y + x

⇒ 10x - x + 18 = 10y - y

⇒ 9x + 18 = 9y

⇒ 9x - 9y = -18

⇒ 9(x - y) = -18

⇒ x - y = -2           . . . (ii)

$\rule{100}2$

Solving eq. (i) and eq. (ii) :-

In (ii),

x + 2 = y

Putting this value in eq. (i) :-

xy = 35

⇒ x(x + 2) = 35

⇒ x² + 2x - 35 = 0

⇒ x² -5x + 7x - 35 = 0

⇒ x(x - 5) + 7(x - 5) = 0

⇒ (x - 5)(x + 7) = 0

⇒ (x - 5) = 0 or (x + 7) = 0

⇒x = 5 or x = -7

The negative value is rejected. So, x = 5.

Putting x = 5 in eq.(i) :-

5y = 35

⇒ y = 35/5

⇒ y = 7

Therefore, the number is 10x + y = 10(5) + 7 = 57.