The New Sudoku Players' Forum

[rjamil]

A group of 630 children is standing in rows for a group photo session. Each row contains three more children than the row in front of it. Which one of the following number of rows is NOT possible?

1. 3
2. 4
3. 5
4. 6

R. Jamil

[Maxito_Bahiense]

Let r be the number of rows, and n the number of children assigned to the first row. The sum of children would be

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C=n+(n+3)+(n+6)+...=r n + sum_{i=1}^{r-1} 3(r-i)

Making S(x)=sum_{i=1}^{r-1} x, we have

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C=r n+3S(r-i)=r n+3[(r-1)r-S(i)]
=r n+3[(r-1)r-r^2/2+r/2]


Since C==630, multiplying both sides of the equation by 2 gives

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3r^2+2n r-3r=1260

This equation can be solved in terms of n as

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n=[3(21-r)(20+r)]/[2r]

r=3 is admissible, since (21-r) would cancel out with 2r to give an integer number of children on each row. r=4 also cancels 20+r with 2r. r=5 has (21-r)(20+r) evenly dividing 2r. Finally, r=6 fails to give an integer number of children, since (20+r) cancels with 2 in the denominator, but 21-r is not divisible by r.

Edit: I should have written "3(21-r) is not divisible by r" for clarity.