Question

find the sum of all three digits numbers divisible by 5 ​

Answer 1

Answer:

you do the math, I will outline how

Step-by-step explanation:

we know that since the numbers must be divisible by 5 they have to end on either 5 or 0, so they will be:

100, 105, 110, 115, 120,125...995

this can be sorted into two lists, the ones ending in 0 and the ones ending in 5:

100, 110, 120, 130...970, 980, 990    and     105, 115, 125... 975, 985, 995

the total sum of each list will be:

100 + 110 + 120+130...970 + 980 +990        

and  

105 + 115 + 125...+975 + 985 +995

now here comes the trick: Reanging the first list gives:

(100 +990) + (110 +980) + (120 +970) ... +

noticed something special about this list?

from each parathesis to the next the first number gets exactly 10 bigger and the second number gets exactly 10 smaller, so calculating the amount in the list gives:

(1090) + (1090) + (1090)...

so now you just have to figure out how many paranthesis there will be in the list (lets call that number n), and the list will sum up to:

n*(1090)

notice how the first two digits in the lists are 10,11,12,13...99, so figuring out how many numbers will be in that line will be quite easy (but be carefull, and make a smaller example and count on your fingers to make sure you got the principle correct)

The same trick can be done for the list which ends on the fives, and the sum of both lists can be added to get your result.

Answer 2

I'm not satisfya with this answer