A two digit number is such that the product of the digits is 8. When 18 is added to the number they interchange their places. Determine the number.
Answers
Let the two-digit number = “ xy ”
Where x takes the “tens” place and y, the unit. Hence you can write the two-digit no. “ xy” as “(10x + y)”.
Product of digits equals 8; x(y) = 8 …….(1)
Adding 18 to the number reverses the digits;
xy + 18 = yx Or (10x + y) + 18 = (10y + x)
This simplifies to; 9x + 18 = 9y …….(2)
Multiplying eq/n (2) through by “y” gives; 9x(y) + 18(y)= 9y(y)……(3).
Substituting in x(y) = 8 from eq/n (1) into (3) gives; 9(8) + 18y = 9y^2, which reduces to 8 + 2y = y^2. This is a quadratic eq/n whose solution for y equals 4 & -2. For y = 4, x = 2 by substitution into eq/n (1).
Hence the two-digit no. xy = 24
Answer:
Required number = 24
Step-by-step explanation:
Let ten's place digit = x ,
Unit place digit = y
The number = 10x + y ----( 1 )
If interchange the places then
the new number so formed
= 10y + x ----( 2 )
Given ,
xy = 8 ---( 3 )
And
10x + y + 18 = 10y + x
=> 10x - x -10y + y + 18 = 0
=> 9x - 9y + 18 = 0
=> x - y + 2 = 0
=> x - y = -2 ---( 4 )
( x + y )² = ( x - y )² + 4xy
=> ( x + y )² = ( - 2 )² + 4 × 8
= 4 + 32
= 36
=> ( x + y ) = √36
=> x + y = 6 ---( 5 )
Add ( 4 ) and ( 5 ), we get
2x = 4
=> x = 4/2 = 2
Put x = 2 in equation ( 5 ), we get
y = 6 - 2 = 4
Therefore ,
Required number = 10x + y
= 10 × 2 + 4
= 20 + 4
= 24
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