All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars    
perplexus dot info

Home > Just Math
Chairs galore (Posted on 2014-10-31) Difficulty: 3 of 5
A famous furniture store staged a big sale of chairs which attracted numerous customers.
The 1st customer bought 2 chairs plus a half of the remaining chairs.
The 2nd customer bought 3 chairs plus a third of the remaining chairs.
The 3rd customer bought 4 chairs plus a quarter of the remaining chairs
... etc till it was impossible to proceed in the said manner.
How many customers could be accommodated and how many chairs were sold to them?

No Solution Yet Submitted by Ady TZIDON    
No Rating

Comments: ( Back to comment list | You must be logged in to post comments.)
Solution solution and discussion | Comment 3 of 4 |
It's to be assumed that indeed at least 3 customers were served.

Let n be the number of chairs in stock at the beginning.

After the first customer left, (n-2)/2 chairs were left.
After the second customer left, (n-8)/3 chairs were left.
After the third customer left, (n-20)/4 chairs were left.
After the fourth customer would have left, (n-40)/5 chairs would have been left.
After the fifth customer would have left, (n-70)/6 chairs would have been left.

The first formula above requires n to be even.
The second requires n to be 2 more than a multiple of 3.
The third requires n to be a multiple of 4.
The fourth requires n to be a multiple of 5.

The fifth would require n to be 4 more than a multiple of 6 and therefore 1 more than a multiple of 3, which is incompatible with the remaining chairs after customer 2.

Therefore only 3 or 4 customers were accommodated.

Depending on the exact numbers of chairs originally in stock, differing numbers of chairs had been sold:

 n     customers   chairs sold 
 20       3           20
 32       3           29
 44       3           38
 56       3           47
 68       3           56
 80       4           72
 92       3           74
104       3           83
116       3           92
128       3          101
140       4          120
152       3          119
164       3          128
176       3          137
188       3          146
200       4          168
212       3          164
224       3          173
236       3          182
248       3          191
260       4          216
272       3          209
284       3          218
296       3          227
308       3          236
320       4          264
332       3          254
344       3          263
356       3          272
368       3          281
380       4          312
392       3          299
404       3          308
416       3          317
428       3          326
440       4          360
452       3          344
464       3          353
476       3          362
488       3          371
500       4          408

 For n = 8 To 500 Step 12
   remain = n
   For cust = 1 To 10000
    saveremain = remain
    remain = remain - (cust + 1)
    If remain Mod (cust + 1) = 0 And remain >= 0 Then
     remain = remain - remain / (cust + 1)
    Else
     Exit For
    End If
   Next cust
   If cust - 1 >= 3 Then
    Text1.Text = Text1.Text & mform(n, "###") & mform(cust - 1, "#######0") & mform(n - saveremain, "############0") & crlf
    DoEvents
   End If
 Next


  Posted by Charlie on 2014-10-31 10:19:44
Please log in:
Login:
Password:
Remember me:
Sign up! | Forgot password


Search:
Search body:
Forums (0)
Newest Problems
Random Problem
FAQ | About This Site
Site Statistics
New Comments (12)
Unsolved Problems
Top Rated Problems
This month's top
Most Commented On

Chatterbox:
Copyright © 2002 - 2024 by Animus Pactum Consulting. All rights reserved. Privacy Information