Author |
Topic: Math Trick II (Read 556 times) |
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender: 
Posts: 4863
|
 |
Math Trick II
« on: Jun 25th, 2006, 12:02pm » |
 Quote  Modify
|
Like the post that inspired it, this one is really too easy for the Putnam forum, but I am posting it here anyway because I think it is an interesting mathematical curiosity, and the other forums didn't fit all that well.
The majority of "Math tricks" I've seen rely on the easily proven fact that the sum of the digits of a number is congruent to the number itself modulo 9. Let S(x) be the sum of the digits of the integer x (in base 10). So 9 | (x - S(x)).
However, not all multiples of 9 can be so represented. For example, there is no value of x for which x - S(x) = 90.
Identify all positive integers n such that 9n != x - S(x) for any x.
|
|
 IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
Deedlit
Senior Riddler
Posts: 476
|
|
Re: Math Trick II
« Reply #1 on: Aug 1st, 2006, 9:08pm » |
Quote Modify
|
Let's make a number system where the ith digit from the right represents (10i - 1)/9. So
abcde = 11111a + 1111b + 111c + 11d + e.
Observe that
Sum [i = 1 to n] 9 * ((10i - 1)/9) = (10n+1 - 1)/9 - n - 1 < (10n+1 - 1)/9
so each number has at most one representation; you have to take the largest possible digit at each step, because otherwise you can't make up the difference at lower digits. The above says
100...0 = 99...9 + n + 1
so there are n numbers in between a number ending with exactly n 9's and the next number ending with n 0's. Since
x - S(x) = sum[i = 1 to n] xi10i - sum[i = 1 to n] xi
= 9 sum[i = 1 to n] xi(10i - 1) / 9
the numbers that can be represented as x - S(x) are exactly the numbers represented by our number system.
A simple way to fill out the number system: note that
Sum [i = m+1 to n] 9 * (10i - 1)/9) + 10 * (10m - 1)/9) = (10n+1 - 1)/9 - n + m - 1
so we get the n missing numbers by letting m go from 1 to n. Letting a stand for 10, the numbers between, say, 32549999 and 32550000 are 3254999a, 325499a0, 32549a00, and 3254a000. So our number system includes the normal sequnces of digits 0-9, or sequnces of such digits followed by a and a string of 0's.
|
| « Last Edit: Aug 1st, 2006, 9:11pm by Deedlit » |
IP Logged |
|
|
|
Icarus
wu::riddles Moderator
Uberpuzzler
Boldly going where even angels fear to tread.
Gender:
Posts: 4863
|
|
Re: Math Trick II
« Reply #2 on: Aug 2nd, 2006, 6:44pm » |
Quote Modify
|
That is a unique approach! Well done.
|
|
IP Logged |
"Pi goes on and on and on ...
And e is just as cursed.
I wonder: Which is larger
When their digits are reversed? " - Anonymous
|
|
|
|
RIDDLES SITE
WRITE MATH!
Home 
Help 
Search 
Members 
Login 
Register
Reply
Notify of replies
Send Topic
Print