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Re: ***Riddle Of The Day***
Quote:
Originally Posted by usman_latif_ch
agar mere pehlay answers galat hain tu unka answer question deny wale ko batana chaheya na....mein ny aesy questions kafi kiye or perhay hain mujhy tu wohi laga tu ? likha tha
mein ny isska answer jo perha ha wo kuch or ha lqin mujhay abhi khud sahi smjh nahi aa rahi k wo answer kaise sahi ha smjh raha hn abhi...hehe
ap waise bataein tu kaise nikala ha apny yeh
Mera answer 6 isli ay tha
kay 100 Pora krna tha na
94 already ho gay thay ^ add karnay say 100 pora ho jata hay na
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Class Roll Number: BME-01083029 (The University Of Lahore)
Re: ***Riddle Of The Day***
Quote:
Originally Posted by .BZU.
Mera answer 6 isli ay tha
kay 100 Pora krna tha na
94 already ho gay thay ^ add karnay say 100 pora ho jata hay na
hmmmm....waise apka answer bhi sahi ha
lqin aesy tu koi bhi no missing hota ap 100 poora kerny k liye baqi add kr k 100 sy minius sy nikal sakty hain or riddles itnay asan nahi hoty laqin phir bhi well done...achi logic nikali ha
Class Roll Number: BME-01083029 (The University Of Lahore)
Re: ***Riddle Of The Day***
pehlay isska answer tu deny do...
Quote:
Arrange the digits from 1 to 9 to make a 9-digit number ABCDEFGHI which satisfies the following conditions:
1) AB is divisible by 2;
2) ABC is divisible by 3;
3) ABCD is divisible by 4;
4) ABCDE is divisible by 5;
5) ABCDEF is divisible by 6;
6) ABCDEFG is divisible by 7;
7) ABCDEFGH is divisible by 8;
8) ABCDEFGHI is divisible by 9.
Class Roll Number: BME-01083029 (The University Of Lahore)
Re: ***Riddle Of The Day***
Answer:
From condition 4, we know that E equals 5.
From conditions 1, 3, 5 and 7, we know that B, D, F, H are even numbers, therefore A, C, G, I are 1, 3, 7, 9 in some order.
Furthermore, from conditions 3 and 7 we know that CD is divisible by 4 and GH is divisible by 8 (because FGH is divisible by 8 and F is even). Because C and G are odd, D and H must be 2 and 6 in some order.
From conditions 2 and 5, we know that A+B+C, D+E+F, G+H+I are all divisible by 3.
If D=2, then F=8, H=6, B=4. A+4+C is divisible by 3, therefore A and C must be 1 and 7 in some order, G and I must be 3 and 9 in some order. G6 is divisible by 8, therefore G=9. But neither 1472589 nor 7412589 is divisible by 7.
Therefore D=6, and F=4, H=2, B=8. G2 is divisible by 8, therefore G=3 or 7. A+8+C is divisible by 3, therefore one of A and C is chosen from 1 and 7, the other is chosen from 3 and 9.
If G=3, then one of A and C is 9, the other is chosen from 1 and 7. But none of 1896543, 7896543, 9816543 and 9876543 is divisible by 7.
Therefore G=7, one of A and C is 1, the other is chosen from 3 and 9. From 1836547, 1896547, 3816547 and 9816547, only 3816547 is divisible by 7 (the quotient is 545221).
Therefore, the number we are looking for is 381654729
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1- filled the big jar
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10- Moved water from bigger one to smaller ...
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