Find the number of six-digit numbers, where the sum of the digits is divisible by 10

Guest Mar 4, 2023

#1**0 **

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

Melody Mar 4, 2023

#2**+1 **

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

Guest Mar 4, 2023