SOLUTION: Find the number of 7-digit positive integers, where the sum of the digits is divisible by 3.

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Question 1208684: Find the number of 7-digit positive integers, where the sum of the digits is divisible by 3.
Found 2 solutions by ikleyn, math_tutor2020:
Answer by ikleyn(52778)   (Show Source): You can put this solution on YOUR website!
.
Find the number of 7-digit positive integers, where the sum of the digits is divisible by 3.
~~~~~~~~~~~~~~~~~~~ 

First 7-digit positive integer number is 1,000,000 (one million).

The last  7-digit positive integer number is 9,999,999.

The number of all 7-digit positive integer number is 9,999,999 - 999,999 = 9,000,000.


The 7-digit positive integer numbers, whose sum of digits is divisible by 3 
are those and only whose of this set that are divisible by 3.


The number of such numbers is one third of 9,000,000, i.e. 3,000,000.


ANSWER.   The number of 7-digit positive integers, where the sum of the digits is divisible by 3 is 3,000,000.

Solved.



Answer by math_tutor2020(3816)   (Show Source): You can put this solution on YOUR website!

I have edited out the dollar signs from your question.
It seems to be a strange copy/paste error.

The number 1,000,002 is the smallest 7-digit number where its digits add to 3.
As such, that large number is a multiple of 3.

Then 1,000,002+3 = 1,000,005 is the next multiple of 3 and 1,000,005+3 = 1,000,008 is the next, and so on.
This arithmetic sequence has common difference d = 3.

The nth term of this arithmetic sequence is
an = 1,000,002 + 3(n-1)

The largest 7-digit multiple of 3 is 9,999,999.
We don't have to add up the digits since 9 is a multiple of 3, sums of 9 will also get us a multiple of 3.

If you solved this equation
9,999,999 = 1,000,002 + 3(n-1)
then you would get the result n = 3,000,000 = 3 million
I'll let the student handle the scratch work.


Answer: 3,000,000 aka 3 million
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