Habits of Mind Problem #2
The Sneaky Sheep
Due Sept 6, 2011
Eric the sheep is at the end of a line of 50 sheep all waiting to be shorn. But being an impatient sort of
sheep, Eric sneaks up the line two places every time the shearer takes a sheep from the front to be
shorn. So, for example, while the first sheep is being shorn, Eric moves ahead so that there are two
sheep behind him in line. If at some point it is possible for Eric to move only one place, he does that
instead of moving ahead two places.
i) Under these conditions, what will be Eric’s “number” when he is shorn? Explain why your
answer is correct.
ii) Eric gets more (and more) impatient! Explore how the answer will change if Eric sneaks past 3
at a time or 4 at a time.
iii) Suppose there were (m) sheep ahead of Eric (m = 49 in the first part of this problem) and Eric
sneaks past (n) at a time, (m is greater than n), what number will Eric be when he is shorn?
Explain.
(i) One way to solve this carefully and correctly is to make a table or chart of the position of Eric and all the
other sheep:
Stage =
Sheep
Sheared
0
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18 (Eric)
Sheep
Behind Eric
Eric's
Total Sheep
Sheep
Position in
yet in Line
Ahead of Eric Unshorn line (not shorn)
0
49
50
50
2
46
47
49
4
43
44
48
6
40
41
47
8
37
38
46
10
34
35
45
12
31
32
44
14
28
29
43
16
25
26
42
18
22
23
41
20
19
20
40
22
16
17
39
24
13
14
38
26
10
11
37
28
7
8
36
30
4
5
35
32
1
2
34
32
0
1
33
This shows that Eric's number (of shearing) is 18th.
(ii)
Stage =
Sheep
Sheared
Sheep
Behind Eric
0
1
2
3
4
5
6
7
8
9
10
11
12
13
Eric (14)
Eric's
Total Sheep
Sheep
Position in
yet in Line
Ahead of Eric Unshorn line (not shorn)
0
49
50
50
3
45
46
49
6
41
42
48
9
37
38
47
12
33
34
46
15
29
30
45
18
25
26
44
21
21
22
43
24
17
18
42
27
13
14
41
30
9
10
40
33
5
6
39
36
1
2
38
36
0
1
37
This shows that in this case, Eric's number (of shearing) is 14th.
In the case that Eric sneaks past 4, one has to be more careful:
Stage =
Sheep
Sheared
0
1
2
3
4
5
6
7
8
9
10
11
Eric( 12)
Sheep
Behind Eric
Eric's
Total Sheep
Sheep
Position in
yet in Line
Ahead of Eric Unshorn line (not shorn)
0
49
50
50
4
44
45
49
8
39
40
48
12
34
35
47
16
29
30
46
20
24
25
45
24
19
20
44
28
14
15
43
32
9
10
42
36
4
5
41
37
2
3
40
38
0
1
39
This shows that Eric's number (of shearing) is 12th.
One has to be more careful, because when Eric reaches the 5th place in line with 4 sheep ahead of him then the
next sheep to be shorn will leave 3 sheep in front of Eric and he cannot sneak past 4 sheep. Now one has to
make a decision which is not precisely detailed in the problem. In this case, the choice is that Eric will sneak
froward by 1 sheep in line while a sheep is being shorn. These actions lead to the italicized numbers in the table,
which follow a different pattern than the previous numbers.
(iii) In general, the number of sheep in line ahead of Eric decreases by (n+1) at each shearing. The number
of sheep behind Eric will be a multiple of (n+1). Now m can be written as m = p(n+1) + q, where 0 ≤ q
< (n+1) is the remainder after dividing m by (n+1). If q = 1 (as it is with both cases (i) and (ii)(a) ,
because 48 is a multiple of both 3 and 4 , namely 48 = 16*3 = 16*(2+1) and 48 = 16*4= 16*(3+1) then
49 = 16*3 + 1, and so Eric's number will be 16 + 1 + 1 = 18 and
49 = 12*4 + 1 and so Eric's number will be 12 + 1 + 1 = 14.
If however, q > 1, as it is for the case in (ii)(b) where 49 = 9*5 + 4, then we must begin the analysis again with
the 4 sheep in front of Eric and skipping 1 sheep at a time. Now the division is 4 = 2*2 + 0, and finally, Eric will
be first in line. Then the total number of sheep shorn to get to Eric is 9 + 2 + 1 = 12.