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A marching band of 240 musicians are to march in a [#permalink]
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A marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each. There can be no less than 8 musicians per row and no more than 30 musicians per row. How many different rectangular formations are possible? A. 3 B. 4 C. 5 D. 6 E. 8
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Originally posted by TeHCM on 20 Nov 2005, 21:36.
Last edited by Bunuel on 03 Apr 2015, 04:30, edited 1 time in total.
Edited the question and added the OA.



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Re: A marching band of 240 musicians are to march in a [#permalink]
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20 Nov 2005, 21:57
TeHCM wrote: A marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each. There can be no less than 8 musicians per row and no more than 30 musicians per row. How many different rectangular formations are possible?
(1) 3 (2) 4 (3) 5 (4) 6 (5) 8
8<=t<=30 and t must be a factor of 240. There're 8 possible values of t: 8,10,12,15,16,20,24,30 > 8 possible rectangular formations.



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Re: A marching band of 240 musicians are to march in a [#permalink]
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20 Nov 2005, 22:15
TeHCM wrote: A marching band of 240 musicians are to march in a rectangular formation with s rows of exactly t musicians each. There can be no less than 8 musicians per row and no more than 30 musicians per row. How many different rectangular formations are possible?
(1) 3 (2) 4 (3) 5 (4) 6 (5) 8
find the exact multiple of integers greater than 8 but smaller than 30.
240 = 8x30
240 = 10x24
240 = 12x20
240 = 15x16
240 = just revers the above multiples ony by one = 16x15
240 = 20x12
240 = 24x10
240 = 30x8
altogather 8.



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Re: A marching band of 240 musicians are to march in a [#permalink]
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21 Nov 2005, 01:38
The combinations could be {(1,240),(2,120),(3,80),(4,60),(5,48),(6,40),(8,30),(10,24),(12,20),)15,16),(16,15),(20,12),(24,10),(30,8),(40,6),(48,5),(60,4),(80,3),(120,2),(240,1)}
Of these we are told 8<=t<=30 So we can remove these pairs, and we are left only with.
{(8,30,(10,24),(12,20),(15,16),(16,15),(20,12),(24,10),(30,8)}
Hence 8.



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Re: A marching band of 240 musicians are to march in a [#permalink]
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21 Nov 2005, 11:14
yup got 8 as well
8*30, 10*24, 12*20, 15*16 and viec versa.
so E is the answer.
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Re: A marching band of 240 musicians are to march in a [#permalink]
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23 Nov 2005, 06:22
krisrini wrote: The combinations could be {(1,240),(2,120),(3,80),(4,60),(5,48),(6,40),(8,30),(10,24),(12,20),)15,16),(16,15),(20,12),(24,10),(30,8),(40,6),(48,5),(60,4),(80,3),(120,2),(240,1)} Of these we are told 8<=t<=30 So we can remove these pairs, and we are left only with. {(8,30,(10,24),(12,20),(15,16),(16,15),(20,12),(24,10),(30,8)} Hence 8.
Kristrini> you get an A+ for the extra effort!
Agreed that the total # combinations are 8: (8X30) (10X24) (12X20) (15X16) and their four inverses.



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Re: A marching band of 240 musicians are to march in a [#permalink]
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25 Oct 2015, 06:03
I also did factor approach and got right.
However, OA also writes this :
Straightforward Math is the best route to take to solve this question. Since s × t must equal 240, s and t must be factors of 240. Find the prime factors that make up 240:
240 = 2×2×2×2×3×5.
Now find the combinations of these factors that give you s × t = 240, where 8≤t ≤30: 8×30, 10×24, 12×20, 15×16, 16×15, 20×12, 24×10 and 30×8, a total of 8 combinations.
Note: if you are still wondering how we found these factors using the prime factors, think about it this way. 8×30 = (2×2×2) × (2×3×5). 10×24 = (2×5) × (2×2×2×3). Thus, all of these combinations use all six prime factors. It's just a matter of how you can arrange them.
SO, can it be done via some permutation and combination approach. I need to select some or none from all these factors I guess. Any clue ?



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Re: A marching band of 240 musicians are to march in a [#permalink]
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17 Jan 2016, 18:32
Here's another approach to solving this:
Attachments
SmartSelectImage_20160117203013.png.png [ 569.75 KiB  Viewed 1882 times ]



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Re: A marching band of 240 musicians are to march in a [#permalink]
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17 Jan 2016, 21:19
rukna wrote: I also did factor approach and got right.
However, OA also writes this :
Straightforward Math is the best route to take to solve this question. Since s × t must equal 240, s and t must be factors of 240. Find the prime factors that make up 240:
240 = 2×2×2×2×3×5.
Now find the combinations of these factors that give you s × t = 240, where 8≤t ≤30: 8×30, 10×24, 12×20, 15×16, 16×15, 20×12, 24×10 and 30×8, a total of 8 combinations.
Note: if you are still wondering how we found these factors using the prime factors, think about it this way. 8×30 = (2×2×2) × (2×3×5). 10×24 = (2×5) × (2×2×2×3). Thus, all of these combinations use all six prime factors. It's just a matter of how you can arrange them.
SO, can it be done via some permutation and combination approach. I need to select some or none from all these factors I guess. Any clue ? They use the terms "combinations" and "arrange" in a very generic manner. You cannot use any permutation and combination formulas as such. Since t has to be between 8 and 30, some brute force will be required.
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