Question

A number consists of two digits whose product is 18. If 27 is added to the number , the digits interchange their places. Find the number​

Answer 1

• Let one's digit be M and ten's digit be N.

• Original number = 10N + M

» A number consists of two digits whose product is 18.

→ MN = 18

→ M = 18/N ________ (eq 1)

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» If 27 is added to the number its digits get interchanged.

• Revered number = 10M + N

A.T.Q.

→ 10M + N = 10N + M + 27

→ 10M - M + N - 10N = 27

→ 9M - 9N = 27

→ M - N = 3

→ (18/N) - N = 3 [From (eq 1)]

→ (18 - N²)/N = 3

→ 18 - N² = 3N

→ N² + 3N - 18 = 0

→ N² + 6N - 3N - 18 = 0

→ N(N + 6) - 3(N + 6) = 0

→ (N - 3) (N + 6) = 0

→ N = -6 (neglected) and + 3

Put value of N i.e. +3 in (eq 1)

→ M = 18/3

→ M = 6

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Number = 10N + M => 10 × 3 + 6

→ 36

________ [ ANSWER ]

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Answer 2

Question : A number consists of two digits whose product is 18. If 27 is added to the number, the digits interchange their places. Find the number.

$\underline{\mathfrak {Solution}}$

Let the digit in tens place be x and the digit in unit place y.

$=> x \times y = 18$

$=> xy = 18$

$=> y = \frac{18}{x}\:\:\:\:\: .... (1)$

Original number = 10x+y

Reverse number = 10y+x

$=>10y+x=(10x+y)+27$

$=>10y+x=10x+y+27$

$=>(10y+x)-(10x+y)=27$

$=>10y+x-10x-y=27$

$=>9y-9x=27$

$=>9(y-x)=27$

$=>y-x=\frac {27}{9}$

$=>y-x=3\:\:\:\:\: ...... (2)$

Putting value of y in equation (2).

$=>y-x=3$

$=>\frac{18}{x}-x=3$

$=>\frac{18-x^2}{x} =3$

$=>18-x^2=3x$

$=>x^2+3x-18 =0$

$=>x^2+(6-3)x-18 =0$

$=>x^2+6x-3x-18 =0$

$=>x(x+6)-3(x+6)=0$

$=>(x+6)(x-3)=0$

x+6 = 0

=>x=0-6

=>x=-6

and

x-3=0

=>x=0+3

=>x=3

Value of x= 3 because we can't find digit of number in negative.

$=>x\times y=18$

$=>3\times y = 18$

$=>y=\frac{18}{3}$

$=>y=6$

Hence, required number

= 10x+y

=10×3+6

=30+6

=36