IIT JEE , AIEEE Forum - Mathematics , Algebra

little_genius

prove that there does not exist a natural no. n whose product of digits is ........

srujana

n^3-25n^2+151n

n(n^2-25n+151)

Let n_1 and n_2 be any two natural numbers whose product is

Hence they satisfy the equation,

Sum of the roots

Product of the roots 151

151 is a prime no with its only factors being 151 and , hence the above two conditions are not satisfied by any $n \in N$

Hence there does not exist a natural number whose product of digits is n^3-25n^2+151n

hsbhatt

One query with the above solution: n₁ and n₂ are natural numbers such that n₁n₂ = n^3-25n^2+151n

Further it is stated that n^3-25n^2+151n = 0. This means that n₁n₂= 0 which is not possible in natural numbers.

Instead:

If you denote the product of the digits of a number n by P(n), we can prove that P(n) n

Let $n = 10^k a_1+10^{k-1}a_2+....+a_{k+1}$

$a_1 a_2 a_3 ....a_{k+1} = a_1 (a_2 a_3.....a_{k+1}) < a_1\underbrace {(10 \times 10 \times 10 \times ... \times 10)}_{ k \ \text{terms}} = 10^k a_1 < n$

Equality occurs for single digit numbers.

So, we must have $n^3 - 25n^2 + 151n \le n$

$\Rightarrow n(n^2 - 25n + 151) \le n \Rightarrow (n-10)(n-15)+1 \le 1$

$\Rightarrow (n-10)(n-15)+1 = 1 \ \text{(as it cannot be zero)}$ and also it should be positive

This gives n = 10 or n= 15, for which we get the given function to be equal to 10 and 15, which is obviously greater than the product of roots.

hsbhatt

yaar, is it right or wrong?

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