Zapisz dziewięć kolejnych liczb naturalnych trzycyfrowych. wskaż liczbę podzielą przez 9. nie wykonując dzielenia, znajdź reszty z dzielenia tych liczb przez 9 oraz przez 3

Start with the smallest 3-digits number 100. Then 9 consecutive natural 3-digits numbers are:

100, 101, 102, 103, 104, 105, 106, 107, 108.

Rule: A number is divisible by 9 if the sum of its digits is divisible by 9.

The remainders after division by 9 are:

8, 7, 6, 5, 4, 3, 2, 1, 0 (number 108 have the sum of digits 1+0+8=9 that is divisible by 9 without remainder. Each previous number gives the remainder that is 1 more than the next one).

Rule: A number is divisible by 3 if the sum of its digits is divisible by 3.

The remainders after division by 3 are:

2, 1, 0, 2, 1, 0, 2, 1, 0 (numbers 102, 105, 108 have the sums of digits 3, 6, 9 that are divisible by 3).

no, because the sides are not congruent

the answer is somewhere there guys it really . there im just not gonna answer

step-by-step explanation:

but i'm guessing ur first option tho