Perhaps check it again
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Everything posted by Perhaps check it again
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Three 8's to make 24 (my modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
Regarding post # 3 -- DejMar, concatenation refers to only making multi-digit numbers here. It does not eliminate the double factorial sign. Your solution is one of the correct ones. I hadn't counted on that solution, so there are more than five other solutions. Regarding post # 4, all are valid. -
Three 8's to make 24 (my modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
Slight modification of the rules: An expression isn't valid unless if at least one operation, or set of parentheses, or both can be removed, and the expression is still equal to 24. -
Three 8's to make 24 (my modified version)
Perhaps check it again posted a question in New Logic/Math Puzzles
In each expression: Use exactly three 8's. Use at most three square root symbols. Use at most two factorial signs. You can use some combination of addition, subtraction/negation, multiplication, and/or division signs (+, -, *, /). You can use parentheses. One pair should be sufficient. Concatenation is not allowed. No other numbers, operations, or characters are allowed. To start off, I am giving you this expression for a solution: 8 + 8 + 8 > > > Try to come up with as many as five other expressions for solutions. < < < -
Quick Fun Geometric Puzzle
Perhaps check it again replied to BMAD's question in New Logic/Math Puzzles
BMAD, which of the seven divisions are you calling "outer divisions?" Can you label them in a diagram? -
Quick Fun Geometric Puzzle
Perhaps check it again replied to BMAD's question in New Logic/Math Puzzles
BMAD, please clarify what "outer divisions" mean. For instance, it look likes the division that has 2 in it and the three divisions that share parts of the perimeter of the division that has 2 in it are inner divisions. Then, it looks like the outer divisions are the three divisions that have 1 and 3 inside of them, respectively, and the division with the far right portion of the circle. -
Note: My upper bound was deliberately not meant to be thorough to the point of being the actual answer to the question in the first line of the original post of this thread.
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I have the prime numbers less than 100 memorized, but I could give an upper bound instead.
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Six sixes to make 1,000 (modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
gavinksong, you meant to type "Credit goes . . . and plainglazed)." -
Six sixes to make 1,000 (modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
plainglazed, you presented another correct solution. plainglazed, do you care to make another post where you post all five solutions in a spoiler, and then I can click "Mark Solved" next to your name? (If someone disagrees and claims there are more than my five solutions, then please post your solution(s) as soon as is convenient for you to do so, so we may analyze them for correctness.) -
Six sixes to make 1,000 (modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
I see exactly two more solutions that need to be given here beyond what gavinksong already gave. -
Six sixes to make 1,000 (modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
gavinksong, the solutions you presented so far are valid/correct. Everyone, there are still more solutions. The puzzle is still open. -
Six sixes to make 1,000 (modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
Purposely repetitively being argumentative and failing to follow instructions repeatedly, gets your past and any possible future posts/contributions in this thread ignored by me, DejMar. The puzzle is still open. -
Six sixes to make 1,000 (modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
No, the standard for these types of puzzles is the solver is permitted to only concatenate the original digits given in the puzzle at the outset, giving numbers to work with. The resulting numbers are then operated on. Any resulting numbers in the expression are not allowed to be concatenated with each other. Post # 4 was a clarification and emphasis. Please do not make a post like # 5 again, which goes against post #4, or the rest of your posts/solutions on here will be ignored, and credit will be given to other users who post any correct solutions that match the ones that you have already posted in this thread. Neither solutions in post #5 are correct due to their form. The solution given in post # 6 is one of the correct ones. -
Six sixes to make 1,000 (modified version)
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
DejMar, concatenation here only means that you can place two or more digits next to each other to form multi-digit numbers, such as ".66" or "66." ".6" is not properly represented by being preceded by a 0, because it is just as correct a form as "0.6" is. "0.6" is not permitted here for this puzzle. No "||" should be in the expression. Just the result of doing it should be in there. For your first expression, you should just type: It would be correct that way. For your second expression, I don't know what you intend for it to represent. -
Six sixes to make 1,000 (modified version)
Perhaps check it again posted a question in New Logic/Math Puzzles
Form as many horizontal-style arithmetical expressions as possible that equal 1,000. In order to limit the total number of possible solutions, the following rules of mine are in place: ---------------------------------------------------------------------------------------------------------------------------- Use exactly six sixes. Use no more than one pair of parentheses. Use no more than one "/" symbol. Use no more than two decimal points. Use no more than one minus (subtraction) sign. Concatenation is allowed. No other symbols/characters/operations are allowed. -
How many are there?
Perhaps check it again replied to Barcallica's question in New Logic/Math Puzzles
bonanova, you communicated wrong. You included 12 and 123. Those are strictly numbers and not parts of sequences of numbers. If you wanted numbers that began with those digits, then you need to put ellipses after those, as in: "Assuming strict increase and disallowing 1..., 11..., 12..., 112..., 122..., 123..., and numbers like that with repeats in them." Next time you can type out the word "sequence" or put in the ellipses so you don't pass on the wrong information. And no single digit numbers should be included, because they do not consist of numbers where the digits are increasing from left to right. -
How many are there?
Perhaps check it again replied to Barcallica's question in New Logic/Math Puzzles
Barcallica. the quoted pair of numbers in question are 12 and 123, not 12 and 24 as you stated. And here is bonanova's pertinent quote so you can see to which am referring: "Assuming strict increase and disallowing 1, 11, 12, 112, 122, 123, and numbers like that with repeats in them." -
How many are there?
Perhaps check it again replied to Barcallica's question in New Logic/Math Puzzles
bonanova, I see no sense in you disallowing 12 or 123, because each of those numbers have strict increases in their digits. -
Total number of equilateral triangles
Perhaps check it again replied to Perhaps check it again's question in New Logic/Math Puzzles
bonanova and other users, there are additional equilateral triangles with side lengths different from the correct (so far) three types you gave, and they have different orientations from the ones you listed. This problem is still open. -
Total number of equilateral triangles
Perhaps check it again posted a question in New Logic/Math Puzzles
* * * * * * * * * * * * * * * * The 16 points above lie in a plane on an equilateral triangular lattice. Certain sets of three points of the figure correspond to the vertices of equilateral triangles. Suppose you were to form all of the equilateral triangles possible, such that, for any given equilateral triangle, it must have its three vertices coincide with three of the 16 points. How many total equilateral triangles can be formed this way in the figure? -
Actually, I'm going to have to amend my answer to the revised answer of: because I left out the obvious dual combinations. ****** Actually, bonanova, your keyboard, I'm sure, does not look quite like that. For instance, the "Q" key is not perfectly adjacent diagonally to the "S" key. And, for instance, the "W" key is not perfectly adjacent diagonally to the "D" key. To go from Q to S in the shortest path, you must pass through either A or W. And to go from W to D in the shortest path, you must pass through either S or E. You just look more closely and you'll see that the top row is not in line with the middle row, I'm sure.
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Ten digits => remarkable number
Perhaps check it again replied to bonanova's question in New Logic/Math Puzzles
aristo, your candidate for an answer fails, for instance, when seeing that 943 is not divisible by 3. -
Ten digits => remarkable number
Perhaps check it again replied to bonanova's question in New Logic/Math Puzzles
but these are 9 digits and not 8 , so you should devide the number by 9. You're right of course. I meant to point out that 96325814 / 8 = 12040726.75 It is sufficient to check that 814 divided by 8 does not equal an integer to know that 96325814 divided by 8 will not give an integer. Suppose that an integer is at least four digits long. That integer is divisible by 8 if the number formed by the three rightmost digits is divisible by 8. If the number formed by the three rightmost digits is not divisible by 8, then the original number is not divisible by 8.