3 digit number trick
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Found this on the net. Can anyone proof this? Looks chim but I am sure there is a maths theory behind this.
A 3-digit number trick.
Everybody must have heard of the following number trick that involves 'reverse & add'-ing
the digits of a random 3-digit number. If not here your chance to catch up.
I'm grateful to Mitch Beck for making me aware of this property of 1089, a truly funny number.
Take any three integers from zero to nine, then subtract its reversal.
Then, if the difference is positive, add its reversal.
If the difference is negative, then subtract its reversal.
NO MATTER WHAT 3-DIGIT INTEGERS YOU BEGIN WITH, THE FINAL ANSWER IS ALWAYS 1089 !
856 159 872
- 658 - 951 - 278
------ ------ ------
198 - 792 594
+ 891 - 297 + 495
------ ------ ------
1089 -1089 1089
"Curious and Interesting Numbers", by David Wells, page 163, about 1089 :
If a 3-digit number is reversed and the result subtracted,
and that answer added to its reversal, the answer is always 1089:
623 - 326 = 297 and 297 + 792 = 1089.
"Recreations in the Theory of Numbers", by Albert H. Beiler, page 63, says :
A common trick with an almost infinite number of variations is to
have someone write a three-digit number, then write the number
with the digits in reverse order, subtract the smaller from the larger,
reverse the digits again and add this time; you will then be able to
give the answer without having received any information about the
number. The answer is 1089. Suppose any number, say 173, were
taken. Subtract from 371, leaving 198, reverse, giving 891, and the
sum of the last two numbers is 1089. -
I dun know how to prove using the number theory by algbera. However, I can use divisibility tests to prove to a certain stage.Originally posted by dibilo:Found this on the net. Can anyone proof this? Looks chim but I am sure there is a maths theory behind this.
A 3-digit number trick.
Everybody must have heard of the following number trick that involves 'reverse & add'-ing
the digits of a random 3-digit number. If not here your chance to catch up.
I'm grateful to Mitch Beck for making me aware of this property of 1089, a truly funny number.
Take any three integers from zero to nine, then subtract its reversal.
Then, if the difference is positive, add its reversal.
If the difference is negative, then subtract its reversal.
NO MATTER WHAT 3-DIGIT INTEGERS YOU BEGIN WITH, THE FINAL ANSWER IS ALWAYS 1089 !
856 159 872
- 658 - 951 - 278
------ ------ ------
198 - 792 594
+ 891 - 297 + 495
------ ------ ------
1089 -1089 1089
"Curious and Interesting Numbers", by David Wells, page 163, about 1089 :
If a 3-digit number is reversed and the result subtracted,
and that answer added to its reversal, the answer is always 1089:
623 - 326 = 297 and 297 + 792 = 1089.
"Recreations in the Theory of Numbers", by Albert H. Beiler, page 63, says :
A common trick with an almost infinite number of variations is to
have someone write a three-digit number, then write the number
with the digits in reverse order, subtract the smaller from the larger,
reverse the digits again and add this time; you will then be able to
give the answer without having received any information about the
number. The answer is 1089. Suppose any number, say 173, were
taken. Subtract from 371, leaving 198, reverse, giving 891, and the
sum of the last two numbers is 1089.
Take two numbers 856 and 658 for example or abc and cba in general.
Divisibility test of 9 states that if the respectative digits of the number add up together and is divisible by 9, the whole number is divisible by 9. If not, that number is a remainder.
Sum of abc = a + b + c
Sum of cba = c + b + a
Difference between abc and cba = 0
Therefore, the result of abc - cba is always divisible by 9.
Same example but using divisibility test of 11. Divisibility test of 11 states that if the difference of the sum of alternate digits of the number is divisible by 11, the whole number is divisible by 11. If not, that number is the remainder.
Using abc,
a + c - b
Using cba
c + a - b
Using difference between abc and cba,
a + c - b - (c + a - b) = 0
Difference between abc and cba is always divisible by 11.
This also implies that difference between abc and cba is divisible by the LCM of 9 and 11 which is 99.
After that, you will notice that it is the same case as of 9.
09, 90 = 99
18, 81 = 99
27, 72 = 99
36, 63 = 99
45, 54 = 99
54, 45 = 99
63, 36 = 99
72, 27 = 99
81, 18 = 99
90, 09 = 99
However, I am not sure of how to explain this above occurrance but notice that 1089 = 99 x 11.
Hence, I am certain that it is related to the above occurrance. -
oot abit.. u muz have alot of white hair by coming up with this formula..Originally posted by tanjun:I dun know how to prove using the number theory by algbera. However, I can use divisibility tests to prove to a certain stage.
Take two numbers 856 and 658 for example or abc and cba in general.
Divisibility test of 9 states that if the respectative digits of the number add up together and is divisible by 9, the whole number is divisible by 9. If not, that number is a remainder.
Sum of abc = a + b + c
Sum of cba = c + b + a
Difference between abc and cba = 0
Therefore, the result of abc - cba is always divisible by 9.
Same example but using divisibility test of 11. Divisibility test of 11 states that if the difference of the sum of alternate digits of the number is divisible by 11, the whole number is divisible by 11. If not, that number is the remainder.
Using abc,
a + c - b
Using cba
c + a - b
Using difference between abc and cba,
a + c - b - (c + a - b) = 0
Difference between abc and cba is always divisible by 11.
This also implies that difference between abc and cba is divisible by the LCM of 9 and 11 which is 99.
After that, you will notice that it is the same case as of 9.
09, 90 = 99
18, 81 = 99
27, 72 = 99
36, 63 = 99
45, 54 = 99
54, 45 = 99
63, 36 = 99
72, 27 = 99
81, 18 = 99
90, 09 = 99
However, I am not sure of how to explain this above occurrance but notice that 1089 = 99 x 11.
Hence, I am certain that it is related to the above occurrance.
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Such tricks are almost universally related to the magic number 9.
I actually have the book "Recreations in the Theory of Numbers", collecting dust. One of the chapters is indeed about the magic number. Other topics include recurring numbers, prime numbers, pi etc. Read liao sure headache.
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9 is something like a magic number in math. I learnt from some Vedic Maths book on multiplying numbers with 99, 999, 9999, etc in seconds.
3456 x 9999 = 34556544
1234569 x 9999999 = 12345688765431
25489785412562 x 99999999999999 = 2548978541256274510214587437
Actually, its instantaneous. :p -
Ever like this number: 1234567890
There are some ways to play with this number, pretty nice one. And there's one other number that I can't recall. A small number less than 2000. -
Numbers are very interesting.
Try this.
The result of 12345679 x 9. Sweet number formed.
Divisibility Tests are very interesting especially for 9 and 11.
By the way, I get this interest from buses as I try to look the regos of buses.
Initially, I identify patterns of bus regos with the type of buses.
Notice that for SBST buses, the last letter of the below regos stated are the same.
SBS 242U and SBS 9542U.
SBS 289K and SBS 9589K
The last letter of the regos of x and x + 9300 are the same. -
Perhaps bus forum pple can shed more lightOriginally posted by tanjun:Numbers are very interesting.
Try this.
The result of 12345679 x 9. Sweet number formed.
Divisibility Tests are very interesting especially for 9 and 11.
By the way, I get this interest from buses as I try to look the regos of buses.
Initially, I identify patterns of bus regos with the type of buses.
Notice that for SBST buses, the last letter of the below regos stated are the same.
SBS 242U and SBS 9542U.
SBS 289K and SBS 9589K
The last letter of the regos of x and x + 9300 are the same. -
111 x 111 =12321
1111 x 1111 =1234321
11111 x 11111 =123454321
111111 x 111111=12345654321 -
interesting sia. didnt kno it b4.
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vehicle suffixes are calculating from a checksum produced by this equation:Originally posted by eagle:Perhaps bus forum pple can shed more light
for vehicle PRE abcd,
X = 12a+2b+11c+d mod 19
In this case, adding 9 to a and 3 to b produces an additional of 114 to the checksum, which changes nothing since it is mod 19. -
Thanks for the info. By the way, does a remainder of 1 means that the suffix is A?Originally posted by ^tamago^:vehicle suffixes are calculating from a checksum produced by this equation:
for vehicle PRE abcd,
X = 12a+2b+11c+d mod 19
In this case, adding 9 to a and 3 to b produces an additional of 114 to the checksum, which changes nothing since it is mod 19.
By the way, does anyone know the divisiblity tests of other prime numbers like 7, 13 and so on? Thanks -
No. It also depends on the suffix. If not, each number will generate the same suffix regardless of the prefix.Originally posted by tanjun:Thanks for the info. By the way, does a remainder of 1 means that the suffix is A?
By the way, does anyone know the divisiblity tests of other prime numbers like 7, 13 and so on? Thanks
But within the same prefix you can quickly infer if you know one or more suffixes near what you are looking for.
http://en.wikipedia.org/wiki/Divisibility_test
7
1. Form the alternating sum of blocks of three from right to left.
2. Double the number with the last two digits removed and add the last two digits.
3. Add 5 times the last digit to the rest.
4. Subtract twice the last digit from the rest.
5. You should get a multiple of 7.
13
1. Add the digits in alternate blocks of three from right to left
2. Subtract the two sums.
3. Add 4 times the last digit to the rest.
4. You should get a multiple of 13. -
chim la
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Originally posted by ^tamago^:
Thanks for the info Tamago. Anyway, I realise that for most methods to work, the number tested must be divisible by the prime number that one is testing except one method which is the rather complicated method which involves finding the mod of 10^1 to 10^6 for 7 and then multiply by the value of the mod according to the number of digits that the number has. NUmbers are interesting.
No. It also depends on the suffix. If not, each number will generate the same suffix regardless of the prefix.
But within the same suffix you can quickly infer if you know one or more suffixes near what you are looking for.
http://en.wikipedia.org/wiki/Divisibility_test
[b]7
1. Form the alternating sum of blocks of three from right to left.
2. Double the number with the last two digits removed and add the last two digits.
3. Add 5 times the last digit to the rest.
4. Subtract twice the last digit from the rest.
5. You should get a multiple of 7.
13
1. Add the digits in alternate blocks of three from right to left
2. Subtract the two sums.
3. Add 4 times the last digit to the rest.
4. You should get a multiple of 13.[/b] -
It's ok. This method is described in Wiki.Originally posted by tanjun:Thanks for the info Tamago. Anyway, I realise that for most methods to work, the number tested must be divisible by the prime number that one is testing except one method which is the rather complicated method which involves finding the mod of 10^1 to 10^6 for 7 and then multiply by the value of the mod according to the number of digits that the number has. NUmbers are interesting.
10^1 mod 7 = 3
10^2 mod 7 = 2
10^3 mod 7 = 6
10^4 mod 7 = 4
10^5 mod 7 = 5
10^6 mod 7 = 1
1. Reverse the order of the digits.
2. Multiply each digit successively by 1, 3, 2, 6, 4, 5.
3. Add the products.
4. The result should be divisible by 7 if the original number is.
1050 ---> 0501 (reverse) ---> 0x1 + 5x3 + 0x2 + 1x6 = 0 + 15 + 0 + 6 = 21. So, 1050 is divisible by 7.
1454978 ---> 8794541 (reverse) ---> 8x1 + 7x3 + 9x2 + 4x6 + 5x4+ 4x5+ 1x1= 8 + 21 + 18 + 24 + 20 + 20 + 1 = 42. So, 1454978 is divisible by 7. -
Originally posted by ^tamago^:
Thanks for putting the detail.
It's ok. This method is described in Wiki.
10^1 mod 7 = 3
10^2 mod 7 = 2
10^3 mod 7 = 6
10^4 mod 7 = 4
10^5 mod 7 = 5
10^6 mod 7 = 1
1. Reverse the order of the digits.
2. Multiply each digit successively by 1, 3, 2, 6, 4, 5.
3. Add the products.
4. The result should be divisible by 7 if the original number is.
[b]1050 ---> 0501 (reverse) ---> 0x1 + 5x3 + 0x2 + 1x6 = 0 + 15 + 0 + 6 = 21. So, 1050 is divisible by 7.
1454978 ---> 8794541 (reverse) ---> 8x1 + 7x3 + 9x2 + 4x6 + 5x4+ 4x5+ 1x1= 8 + 21 + 18 + 24 + 20 + 20 + 1 = 42. So, 1454978 is divisible by 7.[/b]
Another way or trick to test the divisibility of 7 similar to the method above. I think of this method quite a while ago but no concrete algebra proof to prove my theory is correct. I have tested a couple of numbers and it seems to tally with my theory.
Instead of using 1, 3, 2, 6, 4, 5. Since, 1001 is a factor of 7, 10010 is also a factor of 7, 100100 is also a facotr of 7.
1. For the ones digit, it must be multiplied by 1 since an increase in 1 will result in an increase in the value of the remainder.
2. For the tens digit, follow the method above.
3. For the hundred digit, follow the method above.
4. For the thousand digit, follow the method above.
5. For the 10^4 digit, instead of using 6, notice that 1001 is divisible by 7,
Using the method above,
1 x 1 + 0 x 3 + 0 x 2 + 1 x a = 0 (since remainder = 0)
1 + a = 0
a = -1
6. For the 10^5 digit, instead of using 5, notice that 10010 is divisible by 7,
Using the method above and the answer above,
0 x 1 + 1 x 3 + 0 x 2 - 0 x 1 + 1 x b = 0
1 x 3 + 1 x b = 0
b = -3
7. So when u do it for the 10^6 digit, it will be -2.
Take note of the pattern, 1, 3, 2, -1, -3, -2. This method is not only the divisibility test of 7, it is also the remainder test of 7.
Using some numbers as tests.
14 -> 4 x 1 + 1 x 3 = 7
58 -> 8 x 1 + 5 x 3 = 23 mod 7 = 2 (58 = 7 x 8 + 2 ) Therefore remainder is 2.
237 -> 7 x 1 + 3 x 3 + 2 x 2 = 20 mod 7 = 6 (237 = 33 x 7 + 6) Therefore remainder is 6.
Let try the bigger numbers.
123456 -> 6 x 1 + 5 x 3 + 4 x 2 - 3 x 1 - 2 x 3 - 1 x 2 = 18 mod 7 = 4
4680123 -> 3 x 1 + 2 x 3 + 1 x 2 - 0 x 1 - 8 x 3 - 6 x 2 + 4 x 1 = -21 mod 7 = 0
Try this.