Find the sum of all two-digit numbers that are modulo 4 when divided by 5.

Any two-digit number that, when divided by 5, gives a remainder of 4, can be written as 5n + 9, where n is some positive integer.

Consider the sequence аn = 5n + 9.

The first term in this sequence is 5 * 1 + 9 = 14.

Find the largest member of this sequence, which is a two-digit number.

To do this, we solve the inequality in integers:

5n + 9 <100;

5n <100 – 9;

5n <91;

n <91/5;

n <18 1/5.

Therefore, the largest member of this sequence, which is a two-digit number, is obtained with n = 18 and this term is 5 * 18 + 9 = 99.

Let us show that this sequence an is an arithmetic progression:

an + 1 – an = 5 * (n + 1) + 9 – (5n + 9) = 5n + 5 + 9 – 5n – 9 = 5.

Therefore, the given sequence an is an arithmetic progression with the difference d equal to 5.

We find the sum of the first 18 members of this progression:

S18 = (2 * a1 + d * (18 – 1)) * 18/2 = (2 * a1 + d * 17) * 9 = (2 * 14 + 5 * 17) * 9 = (28 + 85) * 9 = 113 * 9 = 1017.

Answer: the sum of all two-digit numbers that give, when divided by 5, in the remainder of 4 is 1017.