Two consecutive positive integers, each greater than 9, are divided by

Two consecutive positive integers, each greater than 9, are divided by 5. What is the sum of the remainders?

Let two consecutive integers are x and x+1.
\(Re[\frac{x}{5}] + Re[\frac{x+1}{5}] = ?\)

Since the two integers would be odd and even or even or odd, the remainder would be odd and even or even and odd. Thus sum would always be odd except when the 1st integer is an even integer and 2nd is a multiple of 5.
Example: For 10 and 11,
\(Re[\frac{10}{5}] + Re[\frac{11}{5}] = 0 + 1 = 1\)

For 11 and 12,
\(Re[\frac{11}{5}] + Re[\frac{12}{5}] = 1+2 = 3\)

....

For 14 and 15,
\(Re[\frac{14}{5}] + Re[\frac{15}{5}] = 4 + 0 = 4\)

(1) The sum of the remainders is even.
For 14 and 15,
\(Re[\frac{14}{5}] + Re[\frac{15}{5}] = 4 + 0 = 4\)

SUFFICIENT.

(2) The sum of the units digits of the two original integers is 9.
Let x = 10a + b, where and b positive integer. a>0 & b≥0
x + 1 = 10a + b + 1
So units digit of integers = b + b + 1 = 9
b = 4
Hence possible integers are 14 and 15, 24 and 25, 34 and 35 ......

Therefore,
\(Re[\frac{14}{5}] + Re[\frac{15}{5}] = 4 + 0 = 4\) always.

SUFFICIENT.

Answer D.