Slightly less rigorous, but you don't need the first part which sets the upper bound at 21. Since the product of digits is positive, we can set a lower bound by solving n^2 - 10n - 22 > 0 as you did, and we get n>=12. Since the product of two digits is 81 i.e. up to n=17 and we find n-12 is the only two digit solution. Since the product of three digits is 729, but n=100 already exceeds that bound, so there are no three-digit solutions possible. It should quickly become apparent that the expression will now always increase more rapidly than the product of the digits. Of course, you can show that the product increases as log n while the expression increases as n^2 if you aren't happy with the "obvious" result.
Let n = aN * 1O ^( N-1 ) + aN -1 * 10^ (N-2) + ...a1 be a N digit number. Its product of digits =P (n)= aN *aN-1 *... *a1 < aN * (10 * 10*...10) N-1 times ie P (n) < aN * 10 ^ (N-1) (the first term in the expansion of n) Using this result, We can see that n - P (n) > n - aN * 10 ^ (N-1) The first term in the expansion of n cancels out leaving one or more terms all of which are positive. Therefore n - P (n) is always greater than zero for any n of value 2 digits or more n - P (n) > 0 ⇒ n² - 10n - 22 < n Or n² - 11n - 22 < 0 So n = 0, we get n² - 10n - 22 >=0. Or n >= 12 So n = 12 n = 12 IS THE ONLY possibility and it satisfies the condition.
N has no decimal digits if it is an integer and natural numbers are positive integers so no decimals..so this question doesn't make any sense..hello? Trick question maybe?
@@leif1075 Decimal digits are just normal digits of any number. The word 'decimal' refers to the fact that it's base 10.. it doesn't really mean the digits following a point (' . ') Here it refers to digits of a base 10 natural number.
(before watching the video) Define P(n)=product of digits of n •Obviously, for *0=10* , *n>P(n)>=0* . The quadratic *n²-10n-22* is concave and has one root slightly smaller than -1 and the other bigger than 11, which means it's negative for all natural numbers up to eleven, but since we know *P(n)>=0* , we can rule them out. After that, since n can only be at least 12: n²-10n-22 = P(n) < n => n²-10n-22 n²-11n-22
Here is another approach. Probably not rigorous enough. 1) let P= product of digits d =number of digits E =n(n-10)-22 Solution at P=E P must be positive so n>=12 AND of course at n=12 P=2 which checks, so n=12 is a solution. 2) show it doesn't work for any number with more than 2 digits expressing log_10(P) d if d>2 3) write n as 10q+r P=qr E=nn-10n-22 E= 100qq +20rq +rr -100q-10r -22=P=qr rr+r(19q-10) -22 +100q(q-1). We how have an expression. for the second digit (r) in terms of the first digit (q). But note that the discriminant of the quadratic is negative for q>1 so q=1 and the quadratic becomes (r-2)(r+11)=0 giving r=2 hence n=12 is the only solution.
The 9 to the power of n+1 needs justification. In particular for very large values of n. What is saying is that 9^log(n+1) is always bigger than 10 ^log(n) . Besides that , very good problem and nice solution
Can you suggest some number theory resource for problems (preferably topic wise so that I can master each concept)I am a beginner I don't want to do tons of problems rather I want to do less but quality problems
Well a long way is to show that only 1nes and 2es could be contained as digits in n, as 2 can be only 1 time and when n is even. Case with only 1 drops with non rational D for n and with 2s we get 12 as a solution.
I have another solution: (1) n^2-10n-22=n^2-4n+2=(n-2)^2-2(mod3) is not a multiple of 3 because 3k+2 is never a square. (2) n^2-10n-22=n^2-2n-2=(n-1)^2-3(mod 4) is not a multiple of 4 because 4k+3 is never a square. (3) n^2-10n-22=n^2-2(mod 5) is not a multiple of 5 because 5k+2 is never a square. (4) n^2-10n-22=(n-5)^2+2(mod 7) is not a multiple of 7 because 7k+5 is never a square. Therefore digits 0,3,4,5,6,7,8,9 are all impossible so the only possible digits are 1 and 2. Furthermore since n^2-10n-22 is not a multiple of 4 and it is not 1 for obvious reason we know there are exactly one 2 and the remaining digits are all ones. Therefore n^2-10n-22=2 and (n-12)(n+2)=0.
The questions wording doesnt make any sense..if n is a natural number that means it is a positive INTEGER and therefore has no decimal digits..why has nonone else pointed this out?
Common Tricks involving IMO problems involving digits: 😉
Sum of digits: (mod 9 - inequalities - decimal representation)
Product of digits: (inequalities - decimal representation)
Sorry, but can you please explain what is the inequalities and decimal representation you're talking about? Perhaps give an example? Thanks!
Slightly less rigorous, but you don't need the first part which sets the upper bound at 21.
Since the product of digits is positive, we can set a lower bound by solving n^2 - 10n - 22 > 0 as you did, and we get n>=12.
Since the product of two digits is 81 i.e. up to n=17 and we find n-12 is the only two digit solution.
Since the product of three digits is 729, but n=100 already exceeds that bound, so there are no three-digit solutions possible.
It should quickly become apparent that the expression will now always increase more rapidly than the product of the digits. Of course, you can show that the product increases as log n while the expression increases as n^2 if you aren't happy with the "obvious" result.
I have done exactly the same reasonig.
Let n = aN * 1O ^( N-1 ) + aN -1 * 10^ (N-2) + ...a1 be a N digit number.
Its product of digits =P (n)= aN *aN-1 *... *a1 < aN * (10 * 10*...10) N-1 times
ie P (n) < aN * 10 ^ (N-1) (the first term in the expansion of n)
Using this result, We can see that n - P (n) > n - aN * 10 ^ (N-1)
The first term in the expansion of n cancels out leaving one or more terms all of which are positive.
Therefore n - P (n) is always greater than zero for any n of value 2 digits or more
n - P (n) > 0
⇒ n² - 10n - 22 < n
Or n² - 11n - 22 < 0
So n = 0, we get n² - 10n - 22 >=0. Or n >= 12
So n = 12
n = 12 IS THE ONLY possibility and it satisfies the condition.
brilliant!
N has no decimal digits if it is an integer and natural numbers are positive integers so no decimals..so this question doesn't make any sense..hello? Trick question maybe?
@@leif1075 Decimal digits are just normal digits of any number. The word 'decimal' refers to the fact that it's base 10.. it doesn't really mean the digits following a point (' . ') Here it refers to digits of a base 10 natural number.
@@sharathpr42 oh thanks for clarifying ive never heard the term decimal used that way before? Is it common?
@@leif1075 yeah.. it's as common as saying binary digits.
4:13, much faster would have been just noting 9^(log n + 1) < 10^(log n + 1) = 10n, instead of taking logs of both sides and doing algebra.
(before watching the video)
Define P(n)=product of digits of n
•Obviously, for *0=10* , *n>P(n)>=0* .
The quadratic *n²-10n-22* is concave and has one root slightly smaller than -1 and the other bigger than 11, which means it's negative for all natural numbers up to eleven, but since we know *P(n)>=0* , we can rule them out.
After that, since n can only be at least 12:
n²-10n-22 = P(n) < n => n²-10n-22 n²-11n-22
Here is another approach. Probably not rigorous enough.
1) let P= product of digits
d =number of digits
E =n(n-10)-22
Solution at P=E
P must be positive so n>=12
AND of course at n=12 P=2 which checks, so n=12 is a solution.
2) show it doesn't work for any number with more than 2 digits expressing log_10(P) d if d>2
3) write n as 10q+r
P=qr E=nn-10n-22
E= 100qq +20rq +rr -100q-10r -22=P=qr
rr+r(19q-10) -22 +100q(q-1).
We how have an expression. for the second digit (r) in terms of the first digit (q). But note that the discriminant of the quadratic is negative for q>1 so q=1 and the quadratic becomes
(r-2)(r+11)=0 giving r=2 hence n=12 is the only solution.
The 9 to the power of n+1 needs justification. In particular for very large values of n. What is saying is that 9^log(n+1) is always bigger than 10 ^log(n) . Besides that , very good problem and nice solution
Can you suggest some number theory resource for problems (preferably topic wise so that I can master each concept)I am a beginner I don't want to do tons of problems rather I want to do less but quality problems
I second this
Check out "1220 Number Theory Problems".
Noticed that
n >= product of digits of n and the rest became quite simple
You need only proof it for all n, but it's enough simple too. Good idea👍
Good!
What's odd to me is that the bound in the video of
Yup, like I did the same thing
Well a long way is to show that only 1nes and 2es could be contained as digits in n, as 2 can be only 1 time and when n is even. Case with only 1 drops with non rational D for n and with 2s we get 12 as a solution.
understand your logic, well done
12 is the answer , the only answer
The answer is only 12.... It was easy, but interesting! 👍
is 12 the only answer?
Yes
I have another solution: (1) n^2-10n-22=n^2-4n+2=(n-2)^2-2(mod3) is not a multiple of 3 because 3k+2 is never a square. (2) n^2-10n-22=n^2-2n-2=(n-1)^2-3(mod 4) is not a multiple of 4 because 4k+3 is never a square. (3) n^2-10n-22=n^2-2(mod 5) is not a multiple of 5 because 5k+2 is never a square. (4) n^2-10n-22=(n-5)^2+2(mod 7) is not a multiple of 7 because 7k+5 is never a square. Therefore digits 0,3,4,5,6,7,8,9 are all impossible so the only possible digits are 1 and 2. Furthermore since n^2-10n-22 is not a multiple of 4 and it is not 1 for obvious reason we know there are exactly one 2 and the remaining digits are all ones. Therefore n^2-10n-22=2 and (n-12)(n+2)=0.
The questions wording doesnt make any sense..if n is a natural number that means it is a positive INTEGER and therefore has no decimal digits..why has nonone else pointed this out?
CAN'T BELIEVE I SOLVED THIS YAY!!
12
12