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 there places. Find the number

Answer 1

Let the unit digit of the no. be x and tens place be y.

So ,

x y = 35

No. = 10 y + x

A/Q, 10 y + x + 18 = 10 x + y

⇒ 18 = 9 (x-y)

so ,

x-y = 2 .......(i)

We know that

x² + y² = (x-y)² +2 xy

         = 2 x 2  + 2 x 35 = 74

and also

x² + y² = ( x+y)² - 2xy

74 = (x+y)² - 2 x 35

so (x+y)² = 144

x + y = 12.............(ii)

Adding both equation , we get

2 x = 14

x = 7

so

y = 5

No.  = 10 y + x = 10 x 5 + 7 = 57

Answer 2

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GOOD EVENING!!☺️

↪️The number is 57.

Solution⤵️⤵️
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Consider a two digit number xy

Therefore the decimal expansion is 10x + y

The product of their digit is 35

Therefore, xy = 35

Thus, y = (35/x) ........ (1)

When 18 is added, the sum is (10x + y) + 18

Given that when 18 is added to the number, the digits are interchanged

When digits are interchanged, the decimal expansion of the number (yx) is 10y + x

Therefore, (10x + y) + 18 = 10y + x

9x - 9y + 18 = 0

x - y + 2 = 0

Substituting the value of y from equation (1) we have

x - (35/x) + 2 = 0

x - 35/x + 2 = 0

=> x² + 2x - 35 = 0

=> x² + 7x - 5x - 35 = 0

=> x(x +7) - 5(x + 7) = 0

=> (x - 5) (x + 7) = 0

=> x - 5 = 0 or x + 7 = 0

=> x = 5 or x = -7

Digits cannot be negative, therefore, the first digit is 5 and from the equation (1), we have, y = 7

Thus the required number is 57.

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