bhramarraj
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Posts posted by bhramarraj
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For them who require hint:
We know from clue [18], that numeral ‘5’ appears once only, in the complete table. So numeral ‘5’ will appear only in column D.
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4100
Cut all three links of one chain, and join remaining three chains with the help of these thee links to make a loop.
Two other men will be requied, all will carry four day food with them. After one day journey each of the three men will consume one day food. So on helper will give one day food to theexplorer and the other helper, and taking one day food along with him, he will return. After further one day journey again one dy food will be consumed by the explorer and the helper each. Helper will give one day food to the explorer, so explorer will have now four day food and the helper will return with the balance two day food with him. Now explorer has four days to cross the desert and has four days food with him.
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For 50 min unreal time = real time is 60 min. So for 180 min unreal time = real time is 60/50 * 180 = 216 min. 216 min = 3hr.36min.
A cork generally comes free with the bottle. However if the bottle with cork costs a dollar and a nickel then cork should cost 1/2 nickeland the bottle without cork will cost a doolar and 1/2 nickel.
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I also agree to altuss. It will take 4 number tears to separate each stamp.... same applies for 3X3 size also.
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Of course no two boxes contain the same number.Clarifications.
No two Boxes contain the same number?
8. Once and only once?
15. One and only one column?
16. Do you mean exactly two times? To prohibit having 2 sixes and 3 eights?
17. The five column sums share one and only one prime factor?
18. Once and only once?
Then, if CL is 45 then 50-59 do not appear in col D?
19. Major diagonal?
Thanks.
Clue8: Yes, once and only once.
Clue15: Yes, one and only one column.
Clue16:In each column there is one and only one numeral that appears two times. other numerals could be one or three or more times. So the condition of this clue cannot prohibit double digit numbers, at any space in the table.
Clue17: there could be more than one prime factors for the sums of different individual columns, but there is one and only one prime factor which is common to all sums.
Clue18: you are absolutely right, if CL is 45 then 50-59 do not appear in col D?
Clue19: Yes, we are talking of the major diagonals.
I think I have clarified all the doubts...?
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You are right....!So we need 24 numbers because 'CL' is the only blank.
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Solution is not concerned to a bingo card, or stack. One is required to creat a 5X5 table containing numbers as per the clues given.Well that's easy (not)
Do bingo cards have a specific amount of numbers on them?
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You are right. The puzzle is to create a 5X5 array of numbers that satisfy the given clues.I'm guessing that the puzzle is to create a 5x5 array of numbers that satisfy the clues.
I'm not sure about the reference to a stack of cards.
Nowhere does it say that at least one crd in the stack has the required array of numbers.
It is a lenghthy math puzzle requires lot patience.
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There is a stack of Bingo Cards. It is required to find a specific card having the number arrangement as hinted in the following clues:
[1] On the cards, there are five columns (named hereafter as A, B, C, D, & E) and five rows (named hereafter as J, K, L, M, & N).
[2] Column 'A' contains numbers 1 through 15.
[3] Column 'B' contains numbers 16 through 30.
[4] Column 'C' contains numbers 31 through 45.
[5] Column 'D' contains numbers 46 through 60.
[6] Column 'E' contains numbers 61 through 75.
[7] Center space 'CL' (a place in column C and row L) is free space i.e. contains no number.
[8] In row 'J', each numeral (0 through 9) appears only once; e.g. if there is number 43 at space DJ, then at other spaces in this row, there would be no number containing numeral 3 & 4.
[9] In row M, sum of all numbers is a perfect square.
[10] Row N contains the smallest number in each column.
[11] Each row contains only one '2-digit Prime Number'.
[12] In column C, difference of lowest to highest number is 8.
[13] In column D, each number has digit at 10th place smaller than the digit at unit place.
[14] In column E each number is an odd number.
[15] In only one column, the numbers are in descending order from top to bottom.
[16] In each column, there is only one numeral that appears two times.
[17] In each column, the sum of numbers shares a single common prime factor.
[18] On the bingo card, numeral 5 appears only once.
[19] The sum of the numbers in each diagonal is an odd number.
[20]The product of numbers at spaces AL & EL has a unit digit 2.
[21] The product of numbers at spaces BL & DL has a unit digit 4.
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If one link is attached to another & it is linear chain, one link per day is to be paid then Minimum 11 cuts will be required.
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I have solved the case when the weights given are 1,3,&9; and we have to find max. moves of weights to find out exact value of an unknown weight known to be less than 14.
CASE-A
Move-1a: W=3. But if W<3, then
Move-2a: W+1=3, i.e. W=2. But If W<2, then W=1, since W cannot be 0 .
CASE-B
Move-1b: W>3, then
Move-2b: W=9+3 i.e. W=12. But if W>12, then W=13 (as from condition given in OP W<14). But if W<12, then
Move-3b: W=9 But if W>9, then
Move-4b: W=9+1 i.e. W=10. But if W>10, & since from Move-2b W<12, therefore W=11.
CASE-C
Move-1c: W>3,
Move-2c: W<9+3 i.e. W<12
Move-3c: W<9
Move-4c: W+3=9 i.e. W=6, But if W>6, then
Move-5c: W+3=9+1 i.e. W=7 But if W>7, & since from Move-3c W<9, therefore W=8.
CASE-D
Move-1d: W>3,
Move-2d: W<9+3 i.e. W<12
Move-3d: W<9 Move-4d: W+3<9 i.e. W<6,
Move-5d: W+3+1=9, i.e. W=5. But if W<5 & since from Move-1d W>3, therefore W=4.
Thus we need maximum 5 movements of given weights, in order to know exact value of unknown weight which is known to be less than 14.
Now for 1,3,9,27,81 & 243, I have to work out.
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Excelent....!!!! I also did not solve it with such simplicity. My original solution isBita ate the sweets.
I assumed that each friends would only eat two items.
(A = Anita, B = Bina, C = Chitra, D = Dimpy and E = Ekta)
From clue 7 either D or E ate an apple so A, B, C must have eaten an apple (four apples eaten)
From clue 6 either A and B, or C and D ate banana.
From clue 8 either B, C and a third person ate oranges, or A, D and E ate oranges.
If B and C ate oranges then both friends would have eaten two items, therefore making it impossible for clue 6 to be true.
Since clue six is assumed true, B and C could not have eaten oranges, leaving A, D and E.
Making D now eating two item, So D can't ate the other apple (clue 6) so it must be E who eats it.
So adding it all up we have:
A Apple & Orange
B Banana
C Apple & Banana
D Banana & Orange
E Apple & Orange
Leaving B to have the sweets.
From [1] & [7]: Four apple eater are (i) A, B, C, & D OR (ii) A, B, C, & E
Similarly from [2] & [6]: Three orange eater are either (iii) B, C, & A (iv) B, C & D (v) B, C, & E OR (vi) A, D, & E
From [3} & [5]: Two banana eaters are (vii) A & B OR (viii) C & D.
From above Different sets are as under:
SET-I: (i), (iii), (vii) : B is three fruits eater. So not possible.
SET-II: (i), (iv), (vii) : B is three fruits eater. So not possible.
SET-III: (i), (v), (vii) : B is three fruits eater. So not possible.
SET-IV: (i), (vi), (vii): A is three fruits eater. So not possible.
SET-V: (ii), (iii), (vii): B is three fruits eater. So not possible.
SET-VI: (ii), (iv), (vii): B is three fruits eater. So not possible.
SET-VII: (ii), (v), (vii): B is three fruits eater. So not possible.
SET-VIII: (ii), (vi), (vii): A is three fruits eater. So not possible.
SET-IX: (i), (iii), (viii): C is three fruits eater. So not possible.
SET-X: (i), (iv), (viii): C is three fruits eater. So not possible.
SET-XI: (i), (v), (viii): C is three fruits eater. So not possible.
SET-XII: (i), (vi), (viii): D is three fruits eater. So not possible.
SET-XIII: (ii), (iii), (viii): C is three fruits eater. So not possible.
SET-XIV: (ii), (iv), (viii): C is three fruits eater. So not possible.
SET-XV: (ii), (v), (viii): C is three fruits eater. So not possible.
SET-XVI: (ii), (vi), (viii): C is three fruits eater. So not possible.
LASTLY only set possible is:
SET-XVIII: (ii), (vi), (viii): All but B, eat two fruits as under
A & E eat Apple + Orange
C eats Apple + Banana
D eats Orange + Banana
B eats Apple.
Therefore B eats SWEETS.
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Thank youGood job to you too bhrammaraj. Also, this evidence can connect us to everyone's favorite suspect...
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"Well, sir," she said timidly, "Mr. Bennett never really let me into his office, and I've only been here a week, so I don't know how much a help I can be." This was the Lucy,s statement when lawyer hoped to get some help from lucy during investigation.
But Later during investigation, Smith trapped Lucy: To Lucy, you ask: "Can I get Samuelson's file? I want to know all about the little 'disagreement'."
And Lucy is trapped ....!
Lucy jumps up and opens a filing cabinet. A few seconds later she pulls out the file and hands it to you.
How could she so easily got the file...? while according to her earlier statement Mr Bennett never allowed her into his office!
She had the opportunity to KIll the lawyer as she admitted that she did leave the front desk once or twice. Also the old couple stated that she ran out of room screaming. But except Lucy nobody is the witness of the time she entered lawyer's office.
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Five friends Anita, Bina, Chitra, Dimpy, & Ekta all have a habit of eating fruits or sweets before and after their meals.
[1] Four friends eat Apple before or after their meals.
[2] Three friends eat orange before or after their meals.
[3] Two friends eat banana before and after their meals.
[4] One friend eats sweet before or after her meals.
[5] Either Anita & Bina, both or Chitr & Dimpy both eat banana before or after their meals.
[6] Of Bina & Chitra either both eat or neither eat orange before or after their meals.
[7] Dimpy and Ekta did not both eat apple before or after their meals (i.e. only one of them eats apple before or after her meals).
Now, with the help of above clues, you are required to tell, Who eats SWEETS.
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Samuelson had appointment at 7:45. He stated "Well, Detective, I arrived early, knowing Mr. Bennett's crowds. I waited for my name to be called, but it didn't. Being suspicious that he was avoiding me, I, well, when Miss Lucy wasn't looking I ducked behind the desk and went into the workroom. The room has an air vent that looks into Mr. Bennett's office. I noticed it when he tried to print my credit report. So, I stood on a copy machine," "anyway, I look in and see nothing but the front of the room."
So at that time lawyer was not hanged. After he left, only Lucy was there to murder, and came out crying at 8:45.
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If [1] is true, then G and D could be: (a) son-daughter, (b) son-mother, © father-daughter.
If [2] is true, then H and G could be: father-son
If [3] is true, then V and H could be (a) daughter-father, (b) mother-father, © daughter-son.
If [4] is true, then V and D could be: mother-daughter.
If [2] & [4] were the exactly two true statements, then [1] also becomes true, which violates the condition given in OP.
[1] & [3] have multiple choices, and if they were the exactly two true statements, then: CASE-I: [1a] & [3b], or CASE-II: [1b] & [3a], could be true.
In both the above cases, [2] also becomes true, which is violation of the condition given in OP.
It is therefore clear that only one of the statement out of [1} & [3] and [2] & [4] must be True.
If [1] Is True, then [3] has to be false.
But only if [1c] was true, only then [3] could be false; and only if [3c] was true, only then [1] could be false.
So, let us assume [1c] as true, i.e. G – father, D – daughter; then [3] & [2] become false and [4] could be true; which satisfies the condition in OP. Therefore, George is father, Dorothy is daughter, Virginia is mother, and Howard is son.
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Got it. Real good puzzle. Thank you.I undersatnd you people had got the answer... you are real genuises... but I would like to put here again explanation for others...
If the ant crawls 1 inch down, then 30 inches across the bottom, then 11 inches up, it will travel 42 inches. But this is not the shortest distance. The solution is found by unfolding the box and then finding the shortest path between the ant and the honey. There are actually 4 ways to "flatten" the box (shown here). But only one method corresponds to the the shortest distance as follows:
The 4th figure corresponds to correct answer: sqrt(322 + 242) = 4 sqrt(82 + 62) = 4 x 10 = 40.
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But it is nowhere mentioned in OP that the box will be broken for fascilitating the ant.That's exactly what I was going to say.....(but my picture wasn't as nice as yours and it wouldn't download)
I hope my logic given in Post # 13, may be the correct answer to this tricky puzzle.
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As per clue [2], M could neither be civil Engineer nor Dentist, and as per clue [1], M & N could not be the Botanist. Therefore M was Artist.
Now we have to find profession of K, L, & N. As per clue [3], K & L could not be civil Engineer, therefore N was civil Engineer.
Now the debate is between professions of K & L _ which one of them was dentist and the other was botanist..?
It is customary in all marriages that the guests used to stand near the bride and groom, for the photo session. It is therefore not logical to assume that the bride or groom will shoot photo.
Considering this aspect, we could safely deduce from clue [3] that K was neither bride nor groom. Therefore, as per clue [2], K could not be dentist. Hence K was botanist.
All other eliminated, therefore remaining L was dentist.
Also from clue [2] & [3], party was for engagement of M and L.
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I could not think of any other route shorter than 42 inch. However if
we assume that the aunt will fall on to the floor, from its present location, it will have to crawle 1 inch lesser, i.e. 41 inch.
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My first answer was totally wrong. I blame my flu meds.
i got 40 inches. That making the poor ant walk on the floor, side wall and ceiling.
sqrt(322 + 242)
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Wow, nice one MikeD, care to explain ?
Could it be so simple....? I think there should be some complexity.....! Still trying to work out......!42 ? (1+30+11) The diagonal route would bring about an inch more.
A Bingo Card Puzzle (A very lengthy math puzzle)
in New Logic/Math Puzzles
Posted