Thread: adding numbers !!!!

i need to know fast (by 5 oclock tommorow eastern time) a faster way to do this problem. We need an alternate way other than adding 200 times. Help would be appreciated, thank you!!!!!! :Find the sum of the digits of the first 200 even natural numbers.

Originally Posted by jarny

i need to know fast (by 5 oclock tommorow eastern time) a faster way to do this problem. We need an alternate way other than adding 200 times. Help would be appreciated, thank you!!!!!! :Find the sum of the digits of the first 200 even natural numbers.

(n^2 + n)/2 will come in handy; there are other ways to solve this too.

S_n = n(2a_1+(n-1)d)/2

a_1 is the first term; d is the difference between the two terms

Sn=200(2*2+(199)2)/2

= 40,200

OR:

(n^2 + 200)/2 = 20,100, but you want the even numbers;

Thus, multiply by 2 = 40,200

Hello, jarny!

Here's a rather "primitive" approach . . .

Find the sum of the digits of the first 200 even natural numbers.

In a set of any ten consecutive natural numbers,
. . the even ones end in:
Their sum is:

Among the first 200 natural numbers, there are 20 such sets-of-ten.
. . Hence, the unit-digits add up to:


Now consider the ten-digits.

In the first hundred numbers, they are:
. .
Their sum is:

Since this happens again in the second hundred numbers,
. . the ten-digits total:


Now, the hundred-digits.
. . They are all in the first 99 numbers.
. . They are all in the next 100 numbers.
. . Finally, the "200" has a .
Hence, the hundred-digits add up to:


Therefore, the total of all the digits is:

can somebody help? i dont know whos right, the first one i dont know but soroban, did you do the first 200 or 200 even ones and would it go up to 400 then?

Soroban, you went up to 200 but it would be up to four hundred right? so it would be 0 as the third digit in the first 49 a 0, next 50 a 1, third 50 a 2, fourth 50 a 3, then last one 4 right? could you reexplain the work then?

When you said you wanted the sum of the digits, do you mean the actual digits, such as 2, 4, 6, 8 ,0, 2, 4, 6, 8, ... and so forth, or the actual numbers? To me it seems more realistic you'd want to know the numbers, in which case 40,200 is correct.

That is, 2 + 4 + 6 + ... + 400 = 40,200

Hello, jarny!

You're absolutely right . . .

The first 200 even numbers:

I'll try again . . . *blush*


In a set of any ten consecutive natural numbers,
. . the even ones end in:
Their sum is:

Among the first 400 natural numbers, there are 40 such sets-of-ten.
. . Hence, the unit-digits add up to:


Now consider the ten-digits.

In the first hundred numbers, they are:
. .
Their sum is:

Since this happens again in the 100's, 200's and 300's,
. . the ten-digits total:


Now, the hundred-digits.
. . They are all in the first 99 numbers.
. . They are all in the next hundred numbers.
. . They are all in the next hundred numbers.
. . They are all in the next hundred numbers.
. . Finally, the "400" has a .
Hence, the hundred-digits add up to:


Therefore, the total of all the digits is:

Hope I got it right this time . . .

Originally Posted by jarny

Soroban, you went up to 200 but it would be up to four hundred right? so it would be 0 as the third digit in the first 49 a 0, next 50 a 1, third 50 a 2, fourth 50 a 3, then last one 4 right? could you reexplain the work then?

Hello,

you are right but I presume it's too late now, but nevertheless...:

Do the same method as Soroban has explained.
You'll get:
1. Sum of all one-digits: 40 * 20 = 800
2. Sum of all ten-digits: 4 * 225 = 900
3. Sum of all hundred-digits: 100 * (1+2+3) +4

Total: 2104

EB