At this point, we see that 174, 350, 702, and 1406 are possible 1
st
terms which give a Dlin
sequence that includes 1406.
We may continue this process of working backward (dividing by 2 and subtracting 1) to deter
mine all possible 1
st
terms which give a Dlin sequence that includes 1406.
1406
→
702
→
350
→
174
→
174
2

1 = 86
→
86
2

1 = 42
→
42
2

1 = 20
→
20
2

1 = 9
Attempting to continue the process beyond 9 gives
9
2

1 =
7
2
which is not possible since the
1
st
term in a Dlin sequence must be a positive integer (and so all terms are positive integers).
Thus, the possible 1
st
terms which give a Dlin sequence that includes 1406 are 9, 20, 42, 86,
174, 350, 702, and 1406.
(c) Each of the integers from 10 to 19 inclusive is a possible first term, and so we must determine
the ones digit of each term which follows each of these ten possible first terms.
If the 1
st
term is 10, then the 2
nd
term (10 + 1)
×
2 = 22 has ones digit 2, and the 3
rd
term
(22 + 1)
×
2 = 46 has ones digit 6.
If the 1
st
term is 11, then the ones digit of the 2
nd
term (11 + 1)
×
2 = 24 is 4, and the 3
rd
term
(24 + 1)
×
2 = 50 has ones digit 0.
Given each of the possible first terms, we list the ones digits of the 2
nd
and 3
rd
terms in the
table below.
1
st
term
10
11
12
13
14
15
16
17
18
19
Units digit of the 2
nd
term
2
4
6
8
0
2
4
6
8
0
Units digit of the 3
rd
term
6
0
4
8
2
6
0
4
8
2
From the table above, we see that the only ones digit which repeats itself is 8.
Thus, if the 1
st
term in the sequence is 18 (has ones digit 8), then the 2
nd
and 3
rd
terms in the
sequence have ones digit 8 and so all terms will have the same ones digit, 8.
Similarly, if the 1
st
term in the sequence is 13 (has ones digit 3), then the 2
nd
and 3
rd
terms in
the sequence have ones digit 8.
It then follows that all further terms after the first will have ones digit 8.
The 1
st
terms (from 10 to 19 inclusive) which produce a Dlin sequence in which all terms after
the 1
st
term have the same ones digit are 13 and 18.
(d) If the 1
st
term in a Dlin sequence is
x
, then the 2
nd
term is (
x
+ 1)
×
2 = 2
x
+ 2, and the 3
rd
term is (2
x
+ 2 + 1)
×
2 = (2
x
+ 3)
×
2 = 4
x
+ 6.
For example, if
x
= 1 (note that this is the smallest possible 1
st
term of a Dlin sequence), then
the 3
rd
term is 4
×
1 + 6 = 10, and if
x
= 2, the 3
rd
term is 4
×
2 + 6 = 14.
What is the largest possible value of
x
(the 1
st
term of the sequence) which makes 4
x
+ 6 (the
3
rd
term of the sequence) less than or equal to 2020?
Setting 4
x
+ 6 equal to 2020 and solving, we get 4
x
= 2014 and so
x
= 503
.
5.
Since the 1
st
term of the sequence must be a positive integer, the 3
rd
term cannot be 2020.
Similarly, solving 4
x
+ 6 = 2019, we get that
x
is not an integer and so the 3
rd
term of a Dlin
sequence cannot equal 2019.
When 4
x
+ 6 = 2018, we get 4
x
= 2012 and so
x
= 503.
Thus, if a Dlin sequence has 1
st
term equal to 503, then the 3
rd
term of the sequence is a
positive integer between 1 and 2020, namely 2018.
Further, 503 is the largest possible 1
st
term for which the 3
rd
term has this property.
Each 1
st
term
x
will give a different 3
rd
term, 4
x
+ 6.