Let N - aaaaaa be a 6 digit number all digits repeated and N is divisible by 924- Let a- be the two roots of the equation x2-11x - - - 0- Then product of all possible values of - is

For N to be a divisible by 924, a should be 4 or 8. Given, a and β are roots of x^2 nbsp; – 11x + λ=0 a+β= 11 nbsp; So, possible pairs (a, β) are (4, 7) and ...

You visited us 6 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Parametric Differentiation

Let N = aaaaa...

Question

Let $N = a a a a a a$ be a $6$ digit number (all digits repeated) and $N$ is divisible by $924.$ Let $a, β$ be the two roots of the equation $x^{2} - 11 x + λ = 0.$ Then product of all possible values of $λ$ is

Open in App

Solution

For $N$ to be a divisible by $924,$ $a$ should be $4$ or $8.$
Given, $a$ and $β$ are roots of $x^{2} - 11 x + λ = 0$
$a + β = 11$
So, possible pairs $(a, β)$ are $(4, 7)$ and $(8, 3) .$
and $a \cdot β = λ$
$\Rightarrow$ Possible values of $λ = 28, 24$
$∴$ Product of $λ = 672.$

Suggest Corrections

Similar questions

Q. Let

N = α α α α α α

be a

6

digit number (all digits repeated) and

N

is divisible by

924

and let

α, β

be the roots of the equation

x^{2} - 11 x + λ = 0

. If the products of all possible values of

λ

168 k

, then value of

k

Q. Let

N = α α α α α α

be a

6

digit number (all digits repeated) and

N

is divisible by

924

and let

α, β

be the roots of the equation

x^{2} - 11 x + λ = 0

. If the products of all possible values of

λ

168 k

, then value of

k

Q. Let

N

be the number of six digit numbers that can be formed by using the digits

1, 2, 3, 4

such that all the digits should appear but exactly two digits can repeat. Then the number of non-trivial divisors of

N

N is a five-digit number such that all its digits are non-zero and even, and there is only a single repetition. If the repeated digits are adjacent, and N is divisible by 4, how many possible values can N take?

Q. For each positive real number

λ

, let

A_{λ}

be the set of all natural numbers n such that

| sin (\sqrt{n + 1}) - sin (\sqrt{n}) | < λ

. Let

A_{λ}^{c}

be the complement of

A_{λ}

in the set of all natural numbers. Then,

Join BYJU'S Learning Program

Let N=aaaaaa be a 6 digit number (all digits repeated) and N is divisible by 924. Let a,β be the two roots of the equation x2−11x+λ=0. Then product of all possible values of λ is

For N to be a divisible by 924, a should be 4 or 8. Given, a and β are roots of x2–11x+λ=0 a+β=11 So, possible pairs (a,β) are (4,7) and (8,3). and a⋅β=λ ⇒ Possible values of λ=28,24 ∴ Product of λ=672.

Let $N = a a a a a a$ be a $6$ digit number (all digits repeated) and $N$ is divisible by $924.$ Let $a, β$ be the two roots of the equation $x^{2} - 11 x + λ = 0.$ Then product of all possible values of $λ$ is

For $N$ to be a divisible by $924,$ $a$ should be $4$ or $8.$
Given, $a$ and $β$ are roots of $x^{2} - 11 x + λ = 0$
$a + β = 11$
So, possible pairs $(a, β)$ are $(4, 7)$ and $(8, 3) .$
and $a \cdot β = λ$
$\Rightarrow$ Possible values of $λ = 28, 24$
$∴$ Product of $λ = 672.$