Let See How Amazing is Mathematics (1=2)

[Bushy]

Final Answer: 25

Breakup:

Friend A------Friend B-----Friend C-----DOG
8---------------8------------8-----------1 ------- Remaining8+8) 16
5---------------5------------5-----------1 ------- Remaining5+5) 10
3---------------3------------3-----------1 ------- Remaining3+3) 6
2---------------2------------2

That way it can have infinite number of solutions, where the number of rotis is always 1 more than the LCM.

 

[FlyingEagle]

Final Answer: 25

Breakup:

Friend A------Friend B-----Friend C-----DOG
8---------------8------------8-----------1 ------- Remaining8+8) 16
5---------------5------------5-----------1 ------- Remaining5+5) 10
3---------------3------------3-----------1 ------- Remaining3+3) 6
2---------------2------------2

Good Work but where is the roti of the dog in the last?

 

[FlyingEagle]

Other possible answers:


106
35, 35, 35, 1
23, 23, 23, 1
15, 15, 15, 1
10, 10, 10,

187
62, 62, 62, 1
41, 41, 41, 1
27, 27, 27, 1
18, 18, 18

268
89, 89, 89, 1
59, 59, 59, 1
39, 39, 39, 1
26, 26, 26,

Impressive but the same mistake Sorry Dude its nt correct

 

[Bas_tum_Pak]

tell me One thing, When each friend wake up and divided to equal portion of rotis then he put to dog from his part or he saved one extra roti for dog.

 

[FlyingEagle]

tell me One thing, When each friend wake up and divided to equal portion of rotis then he put to dog from his part or he saved one extra roti for dog.

He saved one extra roti for dog and then divide the rest into 3 parts.

 

[Skeptic]

Good Work but where is the roti of the dog in the last?

If that is the condition then only one answer possible: infinity.

No other solution possible...

 

[FlyingEagle]

If that is the condition then only one answer possible: infinity.

No other solution possible...

No there is a solution dude. Tomorrow I'll explain it to u ppl.

 

[BlackenTheSky]

Seems like I am wrong I learned it from my grand mom.. And among the christian community in my village Friday 13 th is a very bad day for any thing and also they don't start a new venture on 13 th ..

No you are right,but its not number 13,its date 13 and the day should be friday..like we call friday 13,and that friday is called bloody friday..

 

[makikirkiri]

i dont believe mathematical enthusiasts like you ppl can't find sol of such simple problem.

One of the answers is 78.
first guy eats 1/3(78) and gives 1 to dog therefore 51 remain.
second guy eats 1/3(51) and gives 1 to dog therefore 33 remain
third guy eats 1/3(33) and gives 1 to dog therefore 21 remain.
at last all three make equal parts ie 21/3=7 .


There however could be many more solutions to this problem.But the least number is 78

 

[makikirkiri]

However with the newly mentioned conditions that
1.divisions into 3 parts after alloting 1 roti to the dog
2.the dog gets 1 roti even at the end,
condtions which are not present or are implied by the original problem statement, the solution would be of the form{(81n+53)/8}where n is an odd natural number.I have tried upto n=19 but haven't found the sol.


It would be better if questions which require pure logic are posted instead of such ones which require blunt substitutions & calculations

srysrysry... the sol is of the form {81n+65}/8
and we obtain the sol for n=7 which is 79. Forgive my silly mistake

 

[Zee-Gen]

Its 79…..!!!!!!

Total 79 - 1 (for dog) = 78

78/3 = 26 (first person eat)

Remaining 52 (78-26)

Now 52 - 1 (for dog) = 51

51/3 = 17 (second person eat)

Remaining 34 (51-17)

34 - 1 (for dog) = 33

33/3 = 11 (third person eat)

Remaining 22 (33-11)

At the end 1 for dog and remaining divided between three friends…….21/3 = 7

Or

26 + 26 + 26 +1
17 + 17 + 17 + 1
11 + 11 + 11 + 1
7 + 7 + 7 + 1

 

[Bas_tum_Pak]

Its 79…..!!!!!!

Total 79 - 1 (for dog) = 78

78/3 = 26 (first person eat)

Remaining 52 (78-26)

Now 52 - 1 (for dog) = 51

51/3 = 17 (second person eat)

Remaining 34 (51-17)

34 - 1 (for dog) = 33

33/3 = 11 (third person eat)

Remaining 22 (33-11)

At the end 1 for dog and remaining divided between three friends…….21/3 = 7

Or

26 + 26 + 26 +1
17 + 17 + 17 + 1
11 + 11 + 11 + 1
7 + 7 + 7 + 1

OMG

i thought for 52 .. but it became wrong when all friends wake up.

Good Work.

Let wait for FlyingEagle.

 

[Comet]

Its 79…..!!!!!!

Total 79 - 1 (for dog) = 78

78/3 = 26 (first person eat)

Remaining 52 (78-26)

Now 52 - 1 (for dog) = 51

51/3 = 17 (second person eat)

Remaining 34 (51-17)

34 - 1 (for dog) = 33

33/3 = 11 (third person eat)

Remaining 22 (33-11)

At the end 1 for dog and remaining divided between three friends…….21/3 = 7

Or

26 + 26 + 26 +1
17 + 17 + 17 + 1
11 + 11 + 11 + 1
7 + 7 + 7 + 1

take a look at my earlier post. I came to the same conclusion. Let us wait for Flying Eagle to comment on it

 

[Bushy]

I tried solving it, but one variable always remained undefined, leaving a scope for infinite number of solutions. I wonder if any of you guys got the same result.


On the other hand, none of the solutions would be divisible by 3, leaving us with no solution at all. I am very anxious to see if there is any solution.

 

[FlyingEagle]

Well the exact answer is "79". Umaip answered it but I equested him to wait for someother members. Good Going.................

Todays question is very easy:

A cow gives 5 kg milk, A buffallo gives 3 kg milk and a goat gives 1/4 kg milk. Collect 19 kg of milk from maximum 7 animals.

Kindly explain the number of cows, buffalos and goats?
Note: They shouldn't be more than 7 in total?