geno3141

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 #1
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+5
17 ene 2015
 #7
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These are the numbers 3, 7, 13, 17, 23, 27, ..., 93, 97, 103, 107, ...

I'm going to divide these into two groups. First, the ones that end in 3.

Now, I'm just going to add the first few:

          3 + 13 + 23 + 33 +  ... + 93.

If you use the formula for the sum of an arithmetic series:  Sum = N(F + L)/2

  N  =  number of terms = 10          F = first term = 3          L = last term = 93

--->   Sum  =  10(3 + 93) / 2  =  480

Now, I'm going to add these:  103, 113, 123, ..., 193

--->   Sum  =  10(103 + 193)/2  =  1480

     (Notice that this sum is 1000 greater than the previous sum because each of the ten numbers is 100                greater than the previous ones and 10 x 100 = 1000.)

Now, to add:  203, 213, 223, ..., 293:  their sum is 2480.  (Use either the formula or just add 1000 to the previous answer.

Adding 303, 313, 323, ..., 393:  their sum is 3480

Etc.

Adding 903, 913, 923, ..., 993: their sum is 9480.

We end with these sums:  480 + 1480 + 2480 + ... + 9480.

Add these together -- by calculator, or again use the formula.

This gets you halfway home -- you still need to find the sum of 7, 17, 27, 37, ..., 997.

You can do this in the same manner that was used to find the sum of 3, 13, 23, 33, ..., 993

   or you can realize that each of these numbers is 4 larger than the numbers of the first set   ---   if you can      figure out how many number end with 7, you can take that number, multiply it by 4 and add that to the          answer for the numbers ending with 3.

Finally, add the answer for the numbers ending with a three with the answer for the numbers ending with a seven.

17 ene 2015
 #1
avatar+23252 
+5
17 ene 2015
 #4
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0
17 ene 2015