Question

Question :
Find number of positive integers smaller than 1000 such that the sum of the first digir and the last digits is 10.!

Answer :

Answer 1

Untuk bilangan 2 digit :
19,28,37,46,55,64,73,82,91
Sn = 9/2 (19+91)
     = 9 x 110 / 2
     = 990/2
     = 495

Untuk 3 digit :

109,119,...,199 + 208,218,...,298 + 307,317,...,397 + 406,416,...,496 + 505,515,..,595 + 604,614,..,694 + 703,713,...,793 + 802,812,..,892, + 901,911,..,991
= 10/2 (109+199+208+298+307+397+406+496+505+595+604+694+703+793+802+892+901+991)
= 5 x 9900
= 49500

Jadi jumlahnya 49500+495 = 49995