Question

A two digit number is such that the sum of the digits is 6 the product of the number and the number obtained by interchanging its digits is 765.find the number

Answer 1

Hi there!
Here's the answer:

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In the two digit No.,

Let the Units digit be x
& the tens digit = 6 - x
{ °•° Given that, Sum of digits = 6
x + (6-x) = 6 }

¶¶ No. = (10 × face Value of Tens place) + (1 × face Value of Units place )

•°• Original No. = 10(6-x) + (x)

If digits are interchanged, New No. is obtained
New No. = 10(x) + (6-x)

Given,
Original No. × New No. = 765
=> [10(6-x) + x] [10x + (6-x)] = 765

=> [60x-10x+x] [10x+6-x] = 765
=> (60-9x) (9x+6)= 765
=> [3(20-3x)] [3(3x+2)] = 765
=> 9(20-3x)(3x+2) = 765
=> (20-3x)(3x+2) = 85
=> 20(3x)+20(2)-3x(3x)-3x(2)-85 = 0
=> 60x+40-9x²-6x-85 =0
=> -9x² + 54x -45 = 0
=> -9 (x² - 6x +5) = 0
=> x² - 6x + 5 = 0

Solve for x
=> x² - 5x - x + 5 = 0
=> x(x-5) -1(x-5) = 0
=> (x-5)(x-1) = 0
•°• x = 1 or x = 5

CASE- (1): If x = 1
•°• The Units digit = x = 1
& the tens digit = 6-x = 5
•°• The two digit No. = 51

CASE- (2): If x = 5
•°• The Units digit = x = 5
& the tens digit = 6-x = 1
•°• The two digit No. = 15

•°• The required No. = 15 or 51

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Hope it helps

Answer 2

Step-by-step explanation:

I believe this would help you