-1

A student has seven pieces of paper. She chooses some of them and cuts each of them into seven pieces. In the sequel, he chooses some of the pieces and cuts each of them into seven pieces. She continues this procedure many times with the pieces she has in hand every time. Is it possible to have $2016$ pieces at any given time? Justify your answer.

Asaf Karagila
  • 393,674
Temidire Adesiji
  • 29

2 Answers2

1

Note that every time you cut one piece of paper into $7$, the total number of pieces of paper reduces by $1$ $piece$ $of$ $paper$ increases by $7$. Thus, the student will have $7, 13 ,19, ...$ Which will give rise to a sequence of odd integers, Hence, Impossible to get $2016$ pieces of paper.

Temidire Adesiji
  • 29
  • You answered this 5 minutes after you asked it. Nothing particularly wrong with that, but it causes me to wonder if you truly had this question yourself? – Wildcard Oct 11 '17 at 01:38
0

Since whenever a piece is cut into $7$ pieces, you gain $6$ pieces ($7$ new pieces minus the replaced old one), it follows that at any stage, the number of pieces remains the same mod $6$.

Since at the start, you have $7$ pieces, it follows that at any stage, the number of pieces is congruent to $1$ mod $6$.

But $2016$ is congruent to $0$ mod 6.

quasi
  • 58,772
  • It is not correct. You are getting it wrong. She has $7$ pieces of paper. So $1$ is taken away, so we have $6$ papers left and that, that is taken is cut into $7$ and brought back. So that's $(7-1)+7=13$ and that follows the same pattern over and over again, so its therefore impossible to get $2016$, because it continues as a sequence of Odd numbers $7, 13, 19, 25,...$ – Temidire Adesiji Oct 14 '17 at 13:12
  • So after every cut, the net gain is $6$ pieces. Hence, the total number of pieces stays the same, mod $6$. The final number of pieces must be of the form $6k+1$, since the initial number of pieces is $7$, and $7 \equiv 1 (\text{mod} 6)$. Agreed, it must be odd (since $6k+1$ is odd), but $6k+1$ tells you more. For example, the number of pieces can't be $2015$, since $2015 \not\equiv 1 (\text{mod} 6)$. – quasi Oct 14 '17 at 13:35