Answer:
I assume you know Arithmetic Progression .
so, we have to find the first and last 4-digit number divisible by 5
first = 1000 , last = 9990
we have a formula, 
= a + (n-1)d
here, is the last 4-digit number divisible by 5.
n is the number of 4-digit even numbers divisible by 5
d is the common difference between the numbers, which is 10 in this case
a is the first 4-digit number divisible by 5
9990 = 1000 + (n-1)*10
899 = n-1
n = 900
Hence, there are 900 4-digit even numbers divisible by 5
Answer:
To run one lap Lin takes 1.6 minutes.
Step-by-step explanation:
To find this answer you would do 8 divided by 5. Always remember when dividing time goes first unless the instructions / directions / your teacher says differently. So, when you do 8 divided by 5 the answer you get is 1.6, and there is your answer. Also to check your answer you can do 1.6 times 5, and you get 8, so you know that 1.6 is the correct answer.
Well, we could try adding up odd numbers, and look to see when we reach 400. But I'm hoping to find an easier way.
First of all ... I'm not sure this will help, but let's stop and notice it anyway ...
An odd number of odd numbers (like 1, 3, 5) add up to an odd number, but
an even number of odd numbers (like 1,3,5,7) add up to an even number.
So if the sum is going to be exactly 400, then there will have to be an even
number of items in the set.
Now, let's put down an even number of odd numbers to work with,and see
what we can notice about them:
1, 3, 5, 7, 9, 11, 13, 15 .
Number of items in the set . . . 8
Sum of all the items in the set . . . 64
Hmmm. That's interesting. 64 happens to be the square of 8 .
Do you think that might be all there is to it ?
Let's check it out:
Even-numbered lists of odd numbers:
1, 3 Items = 2, Sum = 4
1, 3, 5, 7 Items = 4, Sum = 16
1, 3, 5, 7, 9, 11 Items = 6, Sum = 36
1, 3, 5, 7, 9, 11, 13, 15 . . Items = 8, Sum = 64 .
Amazing ! The sum is always the square of the number of items in the set !
For a sum of 400 ... which just happens to be the square of 20,
we just need the <em><u>first 20 consecutive odd numbers</u></em>.
I slogged through it on my calculator, and it's true.
I never knew this before. It seems to be something valuable
to keep in my tool-box (and cherish always).
