Let g : N ->N with g(n) being the produc...

Let `g : N ->N` with `g(n)` being the product of the digits of n.(a) Prove that `g(n)<= n` for all `n in N`.(b) Find all `n in N`, for which `n^2-12n + 36 = g(n)`.

124

2612

Text Solution

Verified by Experts

The correct Answer is:

Product of digits of natural number will be a non-negative integer
so, `n^(2)-10^(n)-36ge0`
`rArrnin(-oo,5-sqrt61]uu(5+sqrt61,oo)`
`"but "nin IN`
so `nge 13," where "ninN.`
Casw-1 for all 2 digit natural numbers max value of product of digits`=9xx9=81`
so `n^(2)-10n-36ge81`
`rarrnin[5-sqrt142,5+sqrt142]`
but n is taken as a 2 digit natural no., so `13genlt17,`
`rArr" product of digits "=3, 4, 5` "or 6 for 13, 14 ,15 and 16 respectively
checking n = 12
product of digits `=1xx3=3`
and `13^(2)-10xx13-36=3`
so 13 satisties the given condition
Hence it is a solution
chcking for n = 14
product of digits `=1xx4=4`
`142-10xx14-36=196-140-36=20gt6`
and `n^(2)-10n-36,` is increasing function for n gt 5, rest of the 2 digit integers won't satisfy the given condition
case-2 for all 3-digit integers max product `=9xx9xx9=729`
The smallest 3 digit no. is 100
`f(n)=n^(2)-10n-36,f(100)=100^(2)-10xx100-36=8964gt729`
and f(n) is increasing Hence no 3 digit Integers and similary any higher integer will not satisfy
`rArrn=13` is the only answer.

Topper's Solved these Questions

KVPY
KVPY PREVIOUS YEAR|Exercise Matematics|20 Videos
KVPY
KVPY PREVIOUS YEAR|Exercise Part 1 Mathematics|45 Videos
KVPY
KVPY PREVIOUS YEAR|Exercise PART-2 MATHMATICS|10 Videos
KVPY 2021
KVPY PREVIOUS YEAR|Exercise PART II MATHEMATICS|4 Videos

Let a_(n) = number of n-digit positive integers formed by the digits 0, 1 or both such that no consecutive digits in them is 0. Let b_(n) = number of such n digit integers ending with digit 1, and c_(n) = the numner of such n digit integers ending with digit 0, then for n ge3 ,

`a_(n)=b_(n-1)+c_(n-2)`

`a_(n)=b_(n-1)+c_(n-1)`

`a_(n)=b_(n-2)+c_(n-1)`

`a_(n)=b_(n)+c_(n)`

Let a_n denote the number of all n-digit positive integers formed by the digits 0, 1, or both such that no consecutive digits in them are 0. Let b_n= The number of such n-digit integers ending with digit 1 and c_n= Then number of such n-digit integers with digit 0. The value of b_6 is

If S be the sum, P the product and R the sum of the reciprocals of n terms of a G.P., then ((S)/(R))^(n) =

`P^(2)`

`P^(3)`

`sqrt(P)`