SOLUTION: How many 3-digit numbers are multiple of 21? I can solve it by counting there is another way to solve this type of problem Thanks

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Question 1029343: How many 3-digit numbers are multiple of 21?
I can solve it by counting there is another way to solve this type of problem
Thanks

Found 3 solutions by robertb, ikleyn, Edwin McCravy:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
The highest three-digit number divisible by 21 is 987, while the smallest is 105.
Use the formula for the nth term of an AP
a%5Bn%5D+=+a%5B1%5D+%2B+%28n-1%29d, where a%5Bn%5D+=+987 and a%5B1%5D+=+105, and d = 21.
==> 987 = 105 + (n-1)21 ==> 882 = 21(n-1) ==> 42 = n-1 ==> n = 43.
Thus there are 43 such numbers.

Note: I know there is another, much shorter (or more elegant) method in doing this, and expect another tutor to post it.

Answer by ikleyn(52788) About Me  (Show Source):
You can put this solution on YOUR website!
.
How many 3-digit numbers are multiple of 21?
I can solve it by counting there is another way to solve this type of problem
Thanks
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

First such number is  105.
Last  such number is  987.

How many intervals by 21 integer numbers are between these two numbers?
It is

%28987-105%29%2F21 = 42.

So, there are 42 intervals.
Hence, the number of multiples 21 from 105 to 987 inclusively is 42 + 1 = 43.

Answer. There are 43 multiples of 21 that are 3-digit integers.


Answer by Edwin McCravy(20056) About Me  (Show Source):
You can put this solution on YOUR website!
The smallest 3-digit number is 100 and the largest
3-digit number is 999.

Dividing 100 by 21 gives 4.76, which rounds up to 5,
so the smallest 3-digit multiple of 21 will be 5*21 or 105.

Dividing 999 by 21 gives 47.57, which rounds down to
47, so the largest 3-digit multiple of 21 will be 47*21 or 987.

So the 3-digit multiples of 21 form the arithmetic sequence 

5*21,6*21,...,46*21,47*21

Dividing them all by 21 gives the sequence

5,6,...,46,47, which has the same number of terms.

Subtracting 4 from each gives the sequence

1,2...,42,43, which also has the same number of terms.

Answer: 43

Edwin