I stumbled upon this equation when I was reading:
"Let S(n) be the sum of the digits of the natural number n. Find all natural numbers n that satisfies the equation: 2(S(n))^3+ 7(S(n))^2 + 16 = 3n. "
What I only know after trying this problem many times is that n can't be a one-digit number. Can anybody give me some clues on how I should approach this? ( plz don't spoil me the answers, I just want to know how to solve this)
S(n)=n (mod 9) is a good hint. Then it will be easier.
Thanks.
Sir i am not able to answer but please make a video on this question
Until then I will try to figure out what can be the solution of this
@sonimadhav332
yes sir
Assume "n" is a k digit number , and let s be the sum of it's digits.
Thus we know that :
1 <= s <= 9*k , because we know n has k digits , the greatest the sum of the digits can be is when all the digits are 9 , and are k of them
and 10^(k-1) <=n , n has k digits
we begin with : 1<=s<=9*k
<=> 2<=2*s^3<=1458*k^3
<=> 7<=7*s^2<=567*k^2
--------------------------- Adding the 2 of them together and with 16
<=> 25<=3*n<=1458*k^3+567*k^2+16 , Appling our given equation .
We divide everting by 3 and remembering that n is a natural number we have
9<= n < 486*k^3 + 189 * k^2 + 6 (1)
But we know that n , being a k digit number , also follows :
10^(k-1) <= n (2)
In the following let's consider the following function :
f(x) = 10^(x-1) - 486 * x^3 - 189 * x^2 - 6
And it's first , second , third and 4-th derivative.
we have
1st f'(x) = ln(10)* (10^(x-1)) - 1458 * x^2 - 378 * x
2nd f''(x) = ln(10)^2 * (10^(x-1)) - 2916* x - 378
3rd f'''(x) = ln(10)^3 * (10 ^(x-1)) - 2916
4th f''''(x) = ln(10) ^ 4 * 10^(x-1)
Notice that the 4-th derivative is always positive , it is a positive constant multiplied with an exponential . That means that the 3rd derivative is increasing on (-inf , inf )
Notice that f'''(4) = ln(10)^ 4 * 10^3 - 2916
Now we know e < 3 then e^2 < 9 < 10
then 2 < ln(10)
16 < ln(10)^4
0 < 16000 - 2916 < ln(10)^4 * 10 ^ 3 - 2916
Since we know the 3rd derivative is always increasing and at x = 4 is bigger then 0 , then we know that f'''(x) > 0 for x >=4
Than means the second derivative is always increasing , when x >=4 .
Notice f''(5) = ln(10)^2 * 10^4- 2916*5 - 378
Applying the same logic as above we have :
2 < ln(10)
0< 40000 - 14580 - 378 < f''(5)
Since we know the second derivative is increasing when x >= 4 and at 5 is bigger then 0 , we know that f''(x) > 0 when x >= 5
Following we have that the first derivative is always increasing when x >= 5
Notice that f'(6) = ln(10) * 10 ^5 - 1458 * 36 - 378 * 6
Again 2 < ln(10)
2 * 100000 - 52488 - 2268 < f'(6)
0 < 145244 < f'(6)
That means f'(x) is increasing when x >= 5 and bigger then 0 at 6 that means that f'(x) is bigger than 0 for all x >=6
That means f is always increasing when x >=6
Notice that f(7) = 10^6 - 486 * 343 - 189 * 49 - 6
f(7) = 1000000 - 166698 - 9261 - 6 = 824035 >0
Since f is increasing when x >= 6 and f(7) > 0 when know that
10^(x-1) - 486 * x^3 - 189 * x^2 - 6 > 0 when x >=7
<=> 10^(x-1) > 486 * x ^ 3 - 189 * x^2 - 6
Returning to our previous topic , we deduced that in order for the equation to hold we have that
9<= n < 486*k^3 + 189 * k^2 + 6 from (1)
and
10^(k-1) <= n from (2)
but for such a number n to exist we have that
10^(k-1) < 486 * k ^ 3 + 189 * k^2 + 6
But we proved above that for any number k , bigger than 7 :
10^(k-1) > 486 * x ^ 3 - 189 * x^2 - 6
9 <= n < 486 * k^3 + 189*k^2 + 6 <=> k <7
so we know our number n has at most 6 digits . In this case : 1 <= s <= 9* 6
1 <= s <= 54
Now , we know that for any number n and it's sum of digits s = S(n) we have :
s ≡ n ( mod 3 ) (3)
This is easily provable noticing that:
n = a0 + a1 * 10^1 + a2 *10 ^2 + ... + am * 10^m
where a0 , a1 , ... , am are the digits of this number
10 ≡ 1 ( mod 3 )
then