একটি সহজ প্রশ্ন
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- Posts:188
- Joined:Mon Jan 09, 2012 6:52 pm
- Location:24.4333°N 90.7833°E
একটি আয়তাকার বাগানে সারি ও কলামের ছেদবিন্দুতে একটি করে গোলাপের চারা লাগানো হয়েছে। বাগানের পরিসীমাতে যতগুলো চারা আছে বাগানের ভেতরেও ঠিক ততগুলো চারা রয়েছে। প্রতিটি সারিতে ১ মিটার দূরে এবং প্রতিটি কলামে ২ মিটার দূরে দূরে চারা থাকলে বাগানের সর্বনিম্ন ক্ষেত্রফল বের কর।
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
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- Posts:188
- Joined:Mon Jan 09, 2012 6:52 pm
- Location:24.4333°N 90.7833°E
Re: একটি সহজ প্রশ্ন
No...It is 88 square metres..By the way what was your approach??
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
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- Posts:188
- Joined:Mon Jan 09, 2012 6:52 pm
- Location:24.4333°N 90.7833°E
Re: একটি সহজ প্রশ্ন
The garden with those properties and least area is like this,take a look
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
- Fahim Shahriar
- Posts:138
- Joined:Sun Dec 18, 2011 12:53 pm
Re: একটি সহজ প্রশ্ন
Yap. It’s 88. I wrote wrong.
Suppose, the rectangle’s length is ‘a’ and width is ‘b’. Here ‘b’ is an even number and (a, b/2 ) > 3 or (2a,b)>6.
The number of plants over perimeter is 2a+b and the number of plants inside the rectangle is (b/2 -1)*(a-1).
2a+b = (b/2 -1)*(a-1)
4a+2b = (b-2)*(a-1)
4a+2b = ab-2a-b+2
3(2a+b) = ab+2
Here ab≡1 (mod 3). Now there are two cases-
(1) a,b≡1 (mod 3) or
(2) a,b ≡ -1 (mod 3)
b is an even number greater than 6 having b ≡ -1,1 (mod 3). The lowest number that
satisfies these conditions is 8. If b=8, then a=11. Hence the lowest possible area is 88 square meters.
Suppose, the rectangle’s length is ‘a’ and width is ‘b’. Here ‘b’ is an even number and (a, b/2 ) > 3 or (2a,b)>6.
The number of plants over perimeter is 2a+b and the number of plants inside the rectangle is (b/2 -1)*(a-1).
2a+b = (b/2 -1)*(a-1)
4a+2b = (b-2)*(a-1)
4a+2b = ab-2a-b+2
3(2a+b) = ab+2
Here ab≡1 (mod 3). Now there are two cases-
(1) a,b≡1 (mod 3) or
(2) a,b ≡ -1 (mod 3)
b is an even number greater than 6 having b ≡ -1,1 (mod 3). The lowest number that
satisfies these conditions is 8. If b=8, then a=11. Hence the lowest possible area is 88 square meters.
Name: Fahim Shahriar Shakkhor
Notre Dame College
Notre Dame College