Find the number of six-digit numbers, where the sum of the digits is divisible by 10
+118609 i just noted down all the possibilities for the aditions.
These are not the final numbers but the digits that could be contained within the numbers
910000
820000
811000
730000
721000
711100
640000
631000
622000
621100
611110
550000
541000
532000
531100
522100
521110
511111
442000
441100
433000
432100
431110
422200
422110
421111
333100
332110
331111
322210
322111
and that is it, now you need to work out the number of acceptable permutations for each one
eg
532000
3*5!/3! = 5!/2 = 60
etc
Check your answer from Melody's suggestion against the 6-digit line in the following:
10 - 99 : 9
100 - 999 : 90
1000 - 9999 : 900
10000 - 99999 : 9000
100,000 - 999,999 : 90,000
1000000 - 9999999 : 900000
10000000 - 99999999 : 9000000
100000000 - 999999999 : 90000000
1000000000 - 9999999999 : 900000000
Note: ALL 6-digit sums that are divisible by 10 are the same as ALL 6-digit multiples of 10. That is: [999,999 - 100,000 + 1] =900,000 / 10 = 90,000.
This is the case also with: 2, 3, 5, 9 - but NOT with: 4, 6, 7, 8
