A group of 630 children is arranged in rows for a group photograph ses

Question

A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it. What number of rows is not possible?

A. 3

B. 4

C. 5

D. 6

E. 7

Kinshook · Accepted Answer

Given: A group of 630 children is arranged in rows for a group photograph session. Each row contains three fewer children than the row in front of it.
Asked: What number of rows is not possible?

Let the last row contain n children
Total children = n + (n+3)+....

A. 3
n + (n+3) + (n+6) = 630
3n + 9 = 630
3n = 639 
n = 213
Possible

B. 4
n + (n+3) + (n+6) + (n+9) = 630
4n + 18 = 630
4n = 612
n = 153
Possible

C. 5
n + (n+3) + (n+6) + (n+9) + (n+12) = 630
5n + 30 = 630
5n = 600
n = 120
Possible

D. 6
n + (n+3) + (n+6) + (n+9) + (n+12) + (n+15) = 630
6n + 45 = 630
6n = 585
n = 585/6
n is not an integer : Not possible

E. 7
n + (n+3) + (n+6) + (n+9) + (n+12) + (n+15) + (n+18) = 630
7n + 63 = 630
7n = 567
n = 81
Possible

IMO D

Jitu20 · Answer

4 B - 630 is not divisible by 4

Posted from my mobile device

ananya3 · Answer

IMO D

630 = total
let total number of rows = x
every rows contains 3 less than the previous so if first row contains n
=> subsequent rows - (n-3), (n-6), ....
=> total kids = n+ (n-3) + (n-6) +...... (n-3(x-1))
=> 630 = xn - 3*(1+2....+(x-1))
=> 630 = xn - 3*(x(x-1))/2 
=> 1260 = x (2n - 3(x-1))

All the options are divisible by 1260
therefore we will need to check the second term (2n - 3(x-1))
 case 1 - x = even  
=> 1260(even) = x(even) * (2n - 3(x-1))
=> (2n - 3(x-1)) can be even or odd
 but (2n - 3(x-1)) = even - odd = odd 
=>  x=even and (2n - 3(x-1))=even - not possible

case 2 - x = odd 
=> 1260(even) = x(odd) * (2n - 3(x-1))
=> (2n - 3(x-1)) = even
 and (2n - 3(x-1)) = even - even = even

So we mainly need to check the even options B and D
B. x = 4 (even)
1260 = 4 * 315 (315 - odd)
possible

D. x = 6 (even)
1260 = 6 * 210 (210 - even)
  not possible

----------------------------------------
checking other options

A. x=3 (odd)
1260 = 3 * 420 (420 - even)
possible

C. x = 5 (odd)
1260 = 5 * 252 (252 - even)
possible

E. x=7 (odd)
1260 = 7 * 180 (180-even)
possible

ScottTargetTestPrep · Answer

Solution:
 
Since each row contains three fewer children than the row in front of it, the number of children form an evenly spaced set. In other words, if there are an odd number of rows, the middle row has the median, or average, number of children. Of course, that number has to be an integer because the number of children has to be a whole number. Since 630 is divisible by 3, 5, and 7, we see the middle row will indeed have an integer number of children. If any of these numbers of rows is not possible, the number of children in the middle row will not be an integer. However, since they are, they can’t be the correct answer and we are only left with either choice B or D as the correct answer.

Let’s examine choice B first. We can let x = the number of children in the first row. Therefore, the number of children in the second, third and fourth rows are x - 3, x - 6, and x - 9, respectively. We can create the equation:

x + x - 3 + x - 6 + x - 9 = 630

4x - 18 = 630

4x = 648

x = 162

Since x is an integer, we see that the number of rows can be 4. Since we are looking for the number of rows that is not possible, we are left with choice D as the correct answer.

Answer: D

bumpbot · Answer

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A group of 630 children is arranged in rows for a group photograph ses

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