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

 Mar 4, 2023
 #1
avatar+118623 
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

 Mar 4, 2023
edited by Melody  Mar 4, 2023
 #3
avatar+118623 
0

I have left many out,

I was thinking that they had to add to 10 , not to a multiple of 10.

Melody  Mar 4, 2023
 #2
avatar
+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

 Mar 4, 2023

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