Advertisements
Advertisements
Question
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.
Advertisements
Solution
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
RELATED QUESTIONS
Find the number that must be subtracted from the polynomial 3y3 + y2 – 22y + 15, so that the resulting polynomial is completely divisible by y + 3.
Factorise x3 + 6x2 + 11x + 6 completely using factor theorem.
Using remainder Theorem, factorise:
2x3 + 7x2 − 8x – 28 Completely
Using the Reminder Theorem, factorise of the following completely.
2x3 + x2 – 13x + 6
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
When 3x2 – 5x + p is divided by (x – 2), the remainder is 3. Find the value of p. Also factorise the polynomial 3x2 – 5x + p – 3.
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.
One factor of x3 – kx2 + 11x – 6 is x – 1. The value of k is ______.
If (x – a) is a factor of x3 – ax2 + x + 5; the value of a is ______.
(x – 2) is a factor of ______.
