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.

[Leren]

Using the Fundamental Theorem of Arithmetic we have 630 = 2 * (3^2) * 5 * 7. So 630 = 0 Mod 2, 0 Mod 3, 2 Mod 4, 0 Mod 5 & 0 Mod 6.

Now suppose the number of rows (n) is 2, 3, 4, 5, or 6 and the first row has 1 child. The numbers of children in these row groups is 5 = 1 Mod 2, 12 = 0 Mod 3 & 0 Mod 6, 22 = 2 Mod 4, 35 = 0 Mod 5 and 51 = 3 Mod 6.

The trick is that increasing the number of children in Row 1 by k adds k * n to the total, which means that the Mod n value of each row group of n rows does not change.

For n = 3 the number of children in the first row will be odd so that the 0 Mod 6 case applies.

So for 2 rows or 6 rows the Mod value <> the Mod value of 630. For 3, 4 or 5 rows the Mod value = the Mod value of 630.

So 2 and 6 are not possible and 3, 4 & 5 are possible.

QED

[rjamil]

A simplified algebric equation denoted for each option are as follows:

For 3 rows:
x + (x+3) + (x+6) = 630
3x + 9 = 630
3x = 621
x = 207 (Possible)

For 4 rows:
x + (x+3) + (x+6) + (x+9) = 630
4x + 18 = 630
4x = 612
x = 153 (Possible)

For 5 rows:
x + (x+3) + (x+6) + (x+9) + (x+12) = 630
5x + 30 = 630
5x = 600
x = 120 (Possible)

For 6 rows:
x + (x+3) + (x+6) + (x+9) + (x+12) + (x+15) = 630
6x + 45 = 630
6x = 585
x = 97.5 (Not Possible)

Hence it is clear that 6 rows is the correct answer as a fraction of children in first row is impossible.

R. Jamil