+0  
 
0
70
4
avatar+169 

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: What is the 5000th integer in his list?

matth99  Aug 13, 2018
 #1
avatar+88899 
+1

Note that we  have a  cycle of  26 digits - 13 up and 13  down

 

So

 

5000 mod 26  =  8

 

So 8 will be the 5000th digit

 

Proof

 

192  * 26  +  8    =  5000

 

 

cool cool cool

CPhill  Aug 13, 2018
 #2
avatar
+1

Does he count like THIS?:

1  2  3  4  5  6  7  8  9  10  11 12  13  12  11 10  9  8  7  6  5  4  3  2  1  2  3  4  5  ......

 

or like THIS: ?

1 2  3  4  5  6  7  8  9 10 11 12  13   13  12  11 10  9  8  7  6  5  4  3  2  1  1  2  3  4  5  ....

 

 

It SOUNDS like the former.....

 

So you would START with 13 up the 12 DOWN, then 12 UP then 12 DOWN

Guest Aug 13, 2018
edited by Guest  Aug 13, 2018
edited by Guest  Aug 13, 2018
 #3
avatar+88899 
+1

Mmmm...I assumed the second way....but I see your point, Guest  !!!

 

Let's assume your first way is the  correct one...so we have

 

1 2 3 4 5 6 7 8 9 10 11 12 13

12 11 10  9  8  7  6  5 4 3 2 1

2 3 4 5 6 7 8 9 10 11 12 13

12 11 10  9  8  7  6  5  4 3  2  1

 

So...the first  row contains 13 total digits...and every othe row will contain 12 digits

Subtracting 13 from 5000 we get 4987  digits remaiining

 

So   4987 / 12 = 415 * 12 + 7

So....this will be the 7 digit on the 416th row after the first one

Which will be the row  2 3 4 5 6 7 8 9 10 11 12 13

 

So...the 5000th digit will still   = 8  

 

cool cool cool

CPhill  Aug 13, 2018
 #4
avatar+20009 
+1

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: What is the 5000th integer in his list?

 

Variation 1:

\([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13,][13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1] \ldots\)


\(\text{position $= n \pmod{26} \qquad$ position $0 =$ position $26$} \\ \text{$5000th \pmod {26} =$ position $8$}\)

 

Variation 2:

\([1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12,][ 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2 ] \ldots\)

 

\(\text{position $= n \pmod{24} \qquad$ position $0 =$ position $24$} \\ \text{$5000th \pmod {24} =$ position $8$}\)

 

laugh

heureka  Aug 14, 2018

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