I dismissed that as we have decided, to aid calculation, that 0 divided by anything is exactly divisible and = 0 but it's not real :-)
If it were then there is an argument for:
0/1 =0/2=0/3=0/78=0/1023 etc hence 1=2=3=78=1023...since if a/b = a/c then b=c (I know, unless a is zero, but 'we' have defined that)
Going back to my wondering then, would it have helped answer the OP? Given the lowest number possible from the digits is 1-2+3-4+5-6+7-8+9 = 5, would you have found it easy to work out the changes needed so that it added up to a multiple of 11 leaving the smallest number?
I think in fairness FF
as a maff freek - or maths master as we used to call them
you should tell us how you did it
zebo SAYS that numberacy ability to add, is different to maff
I thought of you because you have pretty impressively Done This Before - so how dja do it?
List of tests for divisibility numbers that I can be bothered to work out. All numbers in base ten are written ....dcba (eg 1345 has d = 1, c = 3 etc):
1: automatic
2: a
3: sum of digits
4: 2b + a
5: a = 0 or 5
6: passes test for 2 and 3
7: a + 3b + 2c - d - 3e - 2f + g... or (abc) - (def) + (ghi) - ...
8: 4c + 2b + a
9: sum of digits
10: a=0
11: alternating sum of digits
12: passes test for 3 and 4
13: (abc) - (def) + (ghi) - ...
OK THE CORRECT ANSWER IS: 123475968
This is the smallest possible nine digit number which is divisible by 11 with no repeated digits. And i forgot to mention zero is not one of the digits.
THE METHOD:
We should use digits 1-9. And to be divisible by 11, the sum of alternate digits - other digits should be eleven.
1 2 3 4 7 5 9 6 8
1+3+7+9+8=28
2+4+5+6=17
28-17=11
You have to just find the combination in which the final number u get is the smallest. As u can see this the correct combination.
Sorry, we can't find any related questions. Try using the search bar at the top of the page to search for some keywords, or choose a topic and submit your own question.