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(52879)   (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(3817)  (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
|
|
|
