Are two digit number is such that the product of its digit is 18 when 63 is subtracted from the number?

A two digit number is such that the product of its digits is 18 . When 63 is subtracted from the number, the digits interchange their places. Find the number.[4 MARKS]

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Text Solution

Solution : Let the tens and units digits of the required number be x and y respectively. Then, <br> `xy=18impliesy=(18)/(x)" "…(i)` <br> And, `(10x+y)-63=10y+x` <br> `implies" "9x-9y=63impliesx-y=7" "…(ii)` <br> Putting `y=(18)/(x)` from (i) into (ii), we get <br> `x-(18)/(x)=7` <br> `impliesx^(2)-18=7ximpliesx^(2)-7x-18=0` <br> `impliesx^(2)-9x+2x-18=0impliesx(x-9)+2(x-9)=0` <br> `implies(x-9)(x+2)=0impliesx-9=0" or "x+2=0` <br> `impliesx=9" or "x=-2` <br> `impliesx=9" "[because" a digit cannot be negative"].` <br> Putting x=9 in (i), we get y=2. <br> Thus, the tens digit is 9 and the units digit is 2. <br> Hence, the required number is 92.

Let the tens and the units digits of the required number be x and y, respectively.Then, we have:xy = 18                     …….(i)Required number = (10x + y)Number obtained on reversing its digits = (10y + x)∴(10x + y) - 63 = 10y + x⇒9x – 9y = 63⇒ 9(x – y) = 63⇒ x – y = 7                       ……..(ii)We know:`(x + y)^2 – (x – y)^2 = 4xy``⇒ (x + y) = ± sqrt((x−y)2+4xy)``⇒ (x + y) = ± sqrt(49+4 ×18)`                `= ± sqrt(49+72)`                 `= ± sqrt(121) = ±11`⇒ x + y = 11          ……..(iii)    (∵ x and y cannot be negative)

On adding (ii) and (iii), we get:

2x = 7 +11 = 18⇒x = 9On substituting x = 9in (ii) we get9 – y = 7⇒ y = (9 – 7) = 2∴ Number = (10x + y) = 10 × 9 + 2 = 90 + 2 = 92

Hence, the required number is 92.