# Problem of the Week Problem D Five Digits

A sequence starts out with one $$5$$, followed by two $$6$$s, then three $$7$$s, four $$8$$s, five $$9$$s, six $$5$$s, seven $$6$$s, eight $$7$$s, nine $$8$$s, ten $$9$$s, eleven $$5$$s, twelve $$6$$s, and so on. (You should notice that only the five digits from $$5$$ to $$9$$ are used.)

The first $$29$$ terms of the sequence appear below.

$$5$$, $$6$$, $$6$$, $$7$$, $$7$$, $$7$$, $$8$$, $$8$$, $$8$$, $$8$$, $$9$$, $$9$$, $$9$$, $$9$$, $$9$$, $$5$$, $$5$$, $$5$$, $$5$$, $$5$$, $$5$$, $$6$$, $$6$$, $$6$$, $$6$$, $$6$$, $$6$$, $$6$$, $$7$$, $$…$$

Determine the $$2022^{\text{nd}}$$ digit in the sequence.

Note:
In solving the above problem, it may be helpful to use the fact that the sum of the first $$n$$ positive integers is equal to $$\dfrac{n(n+1)}{2}$$. That is,

$1 + 2 + 3 + … + n = \frac{n(n+1)}{2}$ For example, $$1 + 2 + 3 + 4 + 5 = 15$$, and $$\dfrac{5(6)}{2} = 15$$.

Also, $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = 36$$, and $$\dfrac{8(9)}{2} = 36$$.