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.
Advertisements
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
When divided by x – 3 the polynomials x3 – px2 + x + 6 and 2x3 – x2 – (p + 3) x – 6 leave the same remainder. Find the value of ‘p’.
The expression 4x3 – bx2 + x – c leaves remainders 0 and 30 when divided by x + 1 and 2x – 3 respectively. Calculate the values of b and c. Hence, factorise the expression completely.
Using remainder Theorem, factorise:
2x3 + 7x2 − 8x – 28 Completely
Find the values of a and b in the polynomial f(x) = 2x3 + ax2 + bx + 10, if it is exactly divisible by (x+2) and (2x-1).
In the following two polynomials, find the value of ‘a’ if x – a is a factor of each of the two:
x5 - a2x3 + 2x + a + 1.
Prove that (5x + 4) is a factor of 5x3 + 4x2 – 5x – 4. Hence factorize the given polynomial completely.
Use factor theorem to factorise the following polynomials completely: x3 – 19x – 30
If x3 – 2x2 + px + q has a factor (x + 2) and leaves a remainder 9, when divided by (x + 1), find the values of p and q. With these values of p and q, factorize the given polynomial completely.
f 2x3 + ax2 – 11x + b leaves remainder 0 and 42 when divided by (x – 2) and (x – 3) respectively, find the values of a and b. With these values of a and b, factorize the given expression.
The polynomial 3x3 + 8x2 – 15x + k has (x – 1) as a factor. Find the value of k. Hence factorize the resulting polynomial completely.
