Advertisements
Advertisements
प्रश्न
A two digit positive number is such that the product of its digits is 6. If 9 is added to the number, the digits interchange their places. Find the number.
उत्तर
Let the digit at the tens place be ‘a’ and at units place be ‘b’.
The two-digit so formed will be 10a + b.
According to the first condition, the product of its digits is 6.
⇒ a x b =6
`=> x = 6/b` ...(1)
According to second condition
10a + b + 9 = 10b + a
`⇒ 9a - 9b = 9
`=> a - b = 1`
`=> a - 6/a= 1` From 1
`=> a^2 - a - 6 = 0`
`=> (a - 3)(a + 2) = 0`
`=> a= -3 or 2`
Since a digit cannot be negative, a = 2.
`=> b = 6/a = 6/2 = 3`
Thus, the required number = 10a + b = 10(2) + 3 = 23
APPEARS IN
संबंधित प्रश्न
Using the Remainder Theorem, factorise each of the following completely.
3x3 + 2x2 – 23x – 30
Using Remainder Theorem, factorise : x3 + 10x2 – 37x + 26 completely.
If (x + 1) and (x – 2) are factors of x3 + (a + 1)x2 – (b – 2)x – 6, find the values of a and b. And then, factorise the given expression completely.
Find the value of a and b so that the polynomial x3 - ax2 - 13x + b has (x - 1) (x + 3) as factor.
In the following two polynomials. Find the value of ‘a’ if x + a is a factor of each of the two:
x3 + ax2 - 2x + a + 4
If x – 2 is a factor of each of the following three polynomials. Find the value of ‘a’ in each case:
x3 + 2ax2 + ax - 1
If (x + 3) and (x – 4) are factors of x3 + ax2 – bx + 24, find the values of a and b: With these values of a and b, factorise the given expression.
If (2x + 1) is a factor of both the expressions 2x2 – 5x + p and 2x2 + 5x + q, find the value of p and q. Hence find the other factors of both the polynomials.
Factorize completely using factor theorem:
2x3 – x2 – 13x – 6
If (x – a) is a factor of x3 – ax2 + x + 5; the value of a is ______.
