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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)`.

A

12

B

13

C

124

D

2612

Text Solution

Verified by Experts

The correct Answer is:
B
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.
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