Modular Circle.......

Number Theory Level 4

The integers from I to 1000 are written in order around a circle. Starting at I, every fifteenth number is marked (that is, I, 16, 31, etc.). This process is continued until a number is reached which has already been marked. How many unmarked numbers remain?

The answer is 800.

2 solutions

U Z
Nov 1, 2014

we see a sequence - $1 , 16 , 31 ............$

$n^{th} term = 15n - 14$

Now

$(15n_{1} - 14) - (15n_{2} -14)$ ( can you answer why?)

should be a multiple of 1000 ,

then only this is possible it is equal to 3000.

$n_{2} = 1 , n_{1} = 201$

so 201th number will repeat , hence only 200 will be marked . thus unmarked is 800

Pavan Tiwari
Oct 3, 2014

number which leaves remainder 1 , 6 , 11 upon division by 15 are marked , so these number leaves remainder 1 when divided by 5.

total marked number are 1000/5=200. thus unmarked number are 800.

numbers of the form 15k+1 (case 1), 15k+6 (case 2) and 15k+11 (case 3) are marked. So, the last number marked in each case will be 991, 996 and 986 respectively. total numbers marked in the 1st and 2nd case will be 66+66=132. No. of numbers marked in the 3rd case will be 65. So, according to me, the answer should be

                      1000-(66+66+65)=803

PUSHPESH KUMAR - 6 years, 8 months ago

There are actually 67 numbers in the sequence $1, 16, 31, \ldots 991$ . While you may solve that $k = 66$ , you forget the case when $k = 0$ , which gives you $15 k + 1 = 1$ . Remember that in an AP, the number of terms is $\frac { \text{ Last - First } } { \text{common difference } } +1$ . Many people forget the +1.

As such, your answer is off by 3 (1 each for each sequence), and $803 - 3 = 800$ is the correct answer.

Calvin Lin Staff - 6 years, 6 months ago

Even I did the same way.........the answer should be 803

Vighnesh Raut - 6 years, 6 months ago

Modular Circle.......

The answer is 800.

2 solutions

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