Question

Easy math, but not sure what I did wrong.

Answer 1

Answer:

8

Step-by-step explanation:

In his 5th year, he took 3 times as many exams as the first year.  So the number of exams taken in the 5th year must be a multiple of 3.

If a₁ = 1, then a₅ = 3.  However, this isn't possible because we need 4 integers between them, and a sum of 31.

If a₁ = 2, then a₅ = 6.  Same problem as before.

If a₁ = 3, then a₅ = 9.  This is a possible solution.

If a₁ = 4, then a₅ = 12.  If we assume a₂ = 5, a₃ = 6, and a₄ = 7, then the sum is 34, so this is not a possible solution.

Therefore, Alex took 3 exams in his first year and 9 exams in his fifth year.  So he took 19 exams total in his second, third, and fourth years.

3 < a₂ < a₃ < a₄ < 9

If a₂ = 4, then a₃ = 7 and a₄ = 8.

If a₂ = 5, then a₃ = 6 and a₄ = 8.

If a₂ = 6, then there's no solution.

So Alex must have taken 8 exams in his fourth year.