Find the sum of the first 12 positive 3 digit integers ending in 4.

Question

The answer is 1908 and i dont know how

Answer 1

"first 12 positive 3 digit integers ending in 4"  with running sums are:

 

Count  Num    Sum So Far

------  ------    ------

  1      104       104
  2      114       218
  3      124       342
  4      134       476
  5      144       620
  6      154       774
  7      164       938
  8      174     1112
  9      184     1296
10      194     1490
11      204     1694
12      214     1908

  Now, at the pre-calc level, you might be expected to know Gauss' formula for the sum of the numbers from 1 to N:

  S = N(N+1)/2

  "sum of the first 12 positive 3 digit integers ending in 4" is:

  S = 104+114+124+134+144+154+164+174+184+194+204+214

  S = 12*104 + 0 + 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 + 90 + 100 + 110 [note: 12 numbers]

  S = 12*104 + 10*(1+2+3+4+5+6+7+8+9+10+11)

  S = 12*104 + 10*11*12/2

  S = 1908