+0  
 
0
95
5
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John counts up from 1 to 13, and then immediately counts down again to 1, and then back up to 13, and so on, alternately counting up and down: \begin{align*} &(1, 2, 3,4,5,6,7,8,9,10,11,12,13,\\ &\qquad\qquad12,11,10,9,8,7,6,5,4,3,2,1,2,3,4,\ldots ). \end{align*}What is the $5000^{\text{th}}$ integer in his list?

 May 5, 2022
 #1
avatar+124595 
+1

Here's a very similar problem :

 

https://web2.0calc.com/questions/counting-problem_27

 

cool cool cool

 May 5, 2022
 #2
avatar
0

CPhill, that post doesn't help.  Can't you just answer the question?

Guest May 5, 2022
 #3
avatar+124595 
+1

OK.....!!!!

 

We noted that the pattern  repeated every  120   digits

 

So

 

5000 mod 120   = 80

 

Looking  at what I posted before.....the 80th digit  in the pattern  =     9

 

So.....the 5000th digit  =   9  

 

 

 

cool cool cool

 May 5, 2022
 #4
avatar
0

The computer says the 5000th integer on list is 8:

 

1 1
2 2
3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10
11 11
12 12
13 13
14 12
15 11
16 10
17 9
18 8
19 7
20 6
21 5
22 4
23 3
24 2
25 1
26 2
27 3
28 4
29 5
30 6
31 7
32 8
33 9
34 10
35 11
36 12
37 13
38 12
39 11
40 10
41 9
42 8
43 7
44 6
45 5
46 4
47 3
48 2
49 1
50 2
51 3
52 4
53 5
54 6
55 7
56 8
57 9
58 10
59 11
60 12
61 13
62 12
63 11
64 10
65 9
66 8
67 7
68 6
69 5
70 4
71 3
72 2
73 1
74 2
75 3
76 4
77 5
78 6
79 7
80 8
81 9
82 10
83 11
84 12
85 13
86 12
87 11
88 10
89 9
90 8
91 7
92 6
93 5
94 4
95 3
96 2
97 1
98 2
99 3
100 4
101 5
102 6
103 7
104 8
105 9
106 10
107 11
108 12
109 13
110 12
111 11
112 10
113 9
114 8
115 7
116 6
117 5
118 4
119 3
120 2
121 1

.

.

.

4993 1
4994 2
4995 3
4996 4
4997 5
4998 6
4999 7
5000 8

 May 5, 2022
 #5
avatar
0

5000 mod (13 - 1 ) ==8

OR

5000 mod(25 - 1) == 8

 May 5, 2022

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