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For a given NxN square there are N diagonals, starting with N in the top right corner. Each number in the sequence is below and one to the left of the previous number. So it is N - 1 more than the previous number. For a given N this forms an arithmetic progression with d = N -1. So S = (n/2) (2a + (n-1)d) where S = 5335, n = N, a = N, d = N - 1 Substitute in and arrange to get: N3 + N = 10670 Solve for N and find N2 for the bottom right entry. This equation is a depressed cubic (no N2 term) and can be solved by the Cardano method ( http://www.sosmath.com/algebra/factor/fac11/fac11.html ) to give N = 22. Alternatively, note that, for N much greater than 10, N is less than 1% of N3 and can be ignored. This gives N3 ~ 10670 N ~ 22 So the bottom right entry is 222 = 484
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Since you mentioned France, pyramid makes me think of the Louvre. Not a clue what to do with those times though. It's well past May 11th. Do you have the answer?
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Drummer : Mix Juggler : Mix Jester : Truth Bear : Mix Piper : Liar
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Namdevvikas2@gmail.com BIU
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I think I can improve your solution to 10 toys instead of 9, You are welcome to improve better! But your solution as well as mine both are subject to a big IF - the acceptance of my note below. 1: 100 cash 100 pay 5 with 100 get 50 20 20 5 2: 50 20 20 5 cash 95 pay 3 with 5 get 1 1 3: 50 20 20 1 1 cash 92 pay 5 with 20 get 10 5 4: 50 20 10 5 1 1 cash 87 pay 3 with 5 get 1 1 5: 50 20 10 1 1 1 1 cash 84 pay 8 with 10 get 1 1 6: 50 20 1 1 1 1 1 1 cash 76 pay 9 with 20 get 10 1 7: 50 10 1 1 1 1 1 1 1 cash 67 pay 8 with 10 get 1 1 8: 50 1 1 1 1 1 1 1 1 1 cash 59 pay 10 with 50 get 20 20 9: 20 20 1 1 1 1 1 1 1 1 1 cash 49 pay 18 with 20 get 1 1 10: 20 1 1 1 1 1 1 1 1 1 1 1 cash 31 pay 18 with 20 get 1 1 Balance all small change ......... cash 13 END With the best utilization of 87%!! Notes: Problem-setter's solution (of 8 toys) seems to interpret the rule 4 as "...............return the change with as less banknotes as possible,............ at the end it turns out that each seller always return at least two banknotes to George". (BOTH OF THESE HIGHLIGHTED CONDITIONS MUST BE COMPLIED IN TOGETHER AND HENCE COMPLIED ACCORDINGLY). Whereas in your as well as mine (of 9 toys & 10 toys respectively), the interpretation of rule 4 is as " .......... always return at least two bank notes. Therefore, if at any stage return of 1 banknote alone becomes possible, ignore such option and simply stick to return 2 banknotes instead....." I have interpreted so (with more liberty than the problem-setter), especially because the word "always" has NOT been used by the problem-setter in the first part of the highlighted conditions . Word "always" has been used by him in second part of the highlighted condition only.
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This is a puzzle from an Usborne puzzle books called Logic Puzzles, and though it gives the correct answer, the method for how it reaches that answer eludes me. I’m posting it here to see if anyone can explain the logical steps to reach the answer (I’ll refrain from posting the answer for now if people with to try and solve it. The puzzle goes as follows: There is a troupe of five mummers: a drummer, a dancing bear, a piper, a juggler, and a jester. One always tells the truth, one always tells lies, and the remaining three tell a mixture of truth and lies. To find out who tells what, the following clues are presented: Drummer: I always tell a mixture of truth and lies. Juggler: That is not true. Jester: If the bear is always truthful, the juggler tells nothing but lies. Bear: That is false. Piper: The drummer always tells the truth. Jester: The piper tells nothing but lies. Can you figure out who tells what based on this information, and how?
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And what about this:
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Ramakant joined the community
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I think I have a much better answer. 01: 100 pay 3 with 100 02: 50 20 20 10 5 2 pay 3 with 5 03: 50 20 20 10 2 1 1 @harey Rule 3 in the puzzle stipulated "..........always gives the nearest bank note (of which he had at the moment)....". You have broken this rule at shop 02, by tendering a note of denomination 50. You MUST have tendered note of denomination 20 intead!!
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They both leave on day 2. The first day each one sees the other doesn’t leave, which proves he himself has a tattoo. Then the next day they both leave.
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For interest's sake, if no true fractions were involved at all, there is an alternative solution ...
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Correct answer? Correct answer?
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Thinking it over and over again, I always finish with a problem I cannot solve. The holes are numbered 1, 2, 3.... Three hunters check: 1 2 3 3 4 5 5 6 7 .... On day n, they will have checked up to the hole 2 * n +1 The groundhog starts in the hole k and moves to the right. On day n, he will be in the hole k+n Question 1: Will the hunters catch the groundhog (and if so, when)? 2 * n +1 grows faster than k + 1, I already have the answer. Nevertheless: 2 * n + 1 = k + n n ⁼ k + 1 No matter how high the number of the starting hole the groundhog choses, it will be caught. Question 2: Is there a starting hole so that the groundhog is not discovered on day n? k + n > 2 * n + 1 k > n + 1 No matter how long the hunters hunt, it always is possible that the groundhog started in a hole leading him outside the checked area.
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I assume a is 1/4 of x (?). It means that x^2 = 4*95=380. x= square root of 380
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Emoji based math puzzle - level easy!
Archit replied to Solvemoji's question in New Logic/Math Puzzles
I think that the ans. is 39 because the blink eye emoji is of 10, the heart eye emoji is of 7, and the purple emoji is of 5 think about it 114 is the correct ans sorry because i plus in last instead of multipling the digits -
? =19
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Hello i‘m RiddleRocket and i‘m currently setting up a youtube account. I‘m thanksfull for any suggestions forimpovement! And very Happy about every subscriber! https://youtu.be/eJ2J88X9bFk
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Groundhog in a Hole - Yet Again
EventHorizon replied to EventHorizon's question in New Logic/Math Puzzles
While the hole that the groundhog starts in is unknown, it is one specific hole and its movements are as described. No matter what hole it starts in, given its described movement, the strategy given will eventually find it. You make it seem like the groundhog can teleport just because it's starting hole was unknown. Given your "about 17 holes outside" example, it would take just one or two (depending on parity) rounds of area clearing/expansion to get it. It does not matter that there are more holes outside the area, it will eventually be trapped and caught. -
Never mind how big the area you chose, I always can say "Bad luck, the groundhog is somewhere about 17 holes outside the defined area." I agree that the size of the cleared area will -> inf. Just the size is expressed as a number. No matter how big it is, there always is a bigger number.
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Ohhh... I get what you mean now. So it was a language barrier. Thanks a lot for a clarification!
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The kitchen scale: I’m trying to understand the 2g precision. Does that mean that if Ng is placed on it, it will display any number from N-1 to N+1 inclusive? Or N-2 through N+2? Or something else?
