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    what is the highest possible number of cuts required to cut a cube into 150 identical pieces?

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    Monisha Sep 25, 2017 Related

    Minimum possible number of cuts required to cut a cube into 150 pieces

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    स्रोत : edurev.in

    Find the least number of cuts required which can cut a cube into 60 identical pieces?

    Click here👆to get an answer to your question ✍️ Find the least number of cuts required which can cut a cube into 60 identical pieces?

    Question

    Find the least number of cuts required which can cut a cube into 60 identical pieces?

    A

    9

    B

    12

    C

    15

    D

    5

    Medium Open in App

    Updated on : 2022-09-05

    Solution Verified by Toppr

    Correct option is A)

    We need to cut a ute to 60 identical pieces

    If we cut oru it become 2 identical pieces.

    If we cut twice on same face it becomes 3 identical pieces.

    If we cute once on one face and another on other face it becomes 4 identical pieces.

    ∴ general formula ⇒ number of identical pieces=(l+1)(m+1)(n+1)

    (n,m,l)= no. of cut on each face

    Given:- (l+1)(m+1)(n+1)=60

    (l+1)(m+1)(n+1)=3×4×5

    ⇒(l+1)=3,m+1=4,n+1=5

    ⇒l=2,m=3,n=4

    ∴ total no. of cuts =l+m+n=3+4+2

    =9

    ∴ minimum 9 cuts have to be made to get 60 identical pieces.

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    16 6

    स्रोत : www.toppr.com

    What is the minimum number of cuts required to cut a cube into 50 identical pieces?

    Answer (1 of 4): * The product of three numbers with 50 with minimum sum of the three numbers is the answer. * So a x b x c = 50 The value of a,b,c with minimum (a+b+c) value is 5 ,5 ,2. * Therefore if we have 5- parts along the length 5- parts along the breadth and 2-parts along the height wi...

    What is the minimum number of cuts required to cut a cube into 50 identical pieces?

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    Sort Taruna Agrawal

    Btech from The LNM Institute of Information Technology (LNMIIT) (Graduated 2021)2y

    Factorise 50 in form of a*b*c where a,b,c are as close as possible then subtract 1 from each multiplier and add them.

    a,b,c are the no of pieces when cut from one of dimensions of 3D respectively.

    a-1, b-1,c-1 are the cuts to get those pieces in each dimension

    Possible factors are-

    1*5*10 ( 0+4+9=13 ) 2*5*5 ( 1+4+4=9 )

    Later one is having least sum so 9 cuts required to get 50 pieces.

    Related questions

    How many cuts are needed to cut a cube to 60 pieces?

    How can I find out the number of cuts required to divide a cube into 120 smaller cubes?

    What is the least number of cuts required to cut a cube into 24 identical cuboids if you are not allowed to rearrange the cut pieces?

    What is the maximum number of smaller identical pieces a cube can be cut into by making 17 cuts?

    Can two identical cubes be cut into pieces and rearranged to form a single, larger cube?

    John K Williamsson

    Accredited (MS Educ) nerd who loves talking about mathAuthor has 6.4K answers and 16.7M answer views2y

    I don’t know if this is the best answer, but here’s how I got 9 as a potential answer:

    What is a factorization of 50 = a × b × c that has the smallest possible sum?

    * 1 × 5 × 10 → 1+5+10 = 16

    * 1 × 2 × 25 → 1+2+25 = 28

    * 2 × 5 × 5 → 2+5+5 = 12 ← we have a winner

    What does 50 = 2 × 5 × 5 and 1+4+4=9 tell us? (I subtracted 1 from each factor)

    * We want two identical layers, which requires one cut

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    Vasudevan A.N.S.

    Former Retired Professor of Chemistry. (1977–2014)Author has 7.8K answers and 9.1M answer views1y

    The product of three numbers with 50 with minimum sum of the three numbers is the answer.

    So a x b x c = 50 The value of a,b,c with minimum (a+b+c) value is 5 ,5 ,2.

    Therefore if we have 5- parts along the length 5- parts along the breadth and 2-parts along the height will give 50 - identical pieces.

    So actual number of required cuts along the length, breadth and height wise will be (5–1),(5–1) and (2–1) = 4,4 and 1 cuts.

    So a minimum of 9-cuts along the length, breadth and height wise will give the result.

    Tom Fyfield

    Studied Pure Elemental MathematicsAuthor has 1.2K answers and 417.2K answer viewsUpdated 1y

    What is the minimum number of cuts required to cut a cube into 50 identical pieces?

    13 is my answer.

    9 virtual cuts will give us 10 identical slices

    and 4 horizontal cuts gives us 5 the other way .

    5 times 10 equals 50

    And 9 plus 4 equals 13

    Edit… It appears I am mistaken here 4 x 4 x 1 gives 9 cuts.

    Nrusingha Charan Behera

    B.Tech(Agril. Engg ) in Mathematics & Physics, Orrisa University of Agriculture & Technology, Bhubaneswar (Graduated 1993)Author has 5.5K answers and 4M answer views3y

    Related

    What is the minimum number of cuts required to cut a cube into 420 pieces?

    As 420=10×7×6. So the number of cuts needed on the very sides is

    (10–1)+(7–1)+(6–1)=20.Ans..

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    Joe Schlessinger

    Works at RetirementAuthor has 92 answers and 35.7K answer views4y

    Related

    What is the least number of cuts required to cut a cube into 24 identical cuboids if you are not allowed to rearrange the cut pieces?

    All cuts must be parallel to one side and at right angles to two others; otherwise you don’t get cuboids. (This argument needs work).

    The easiest case is slicing down through the top to get 6x4 vertical cuboids. This gives you 24 identical elongated cuboids with (6–1) + (4–1) or 8 cuts.

    Systematically, how many ways can you factor 24 into 1, 2, or 3 factors? (3 for 3 dimensions)

    24 x 1: slice the top in one direction 24 times. 23 cuts

    12 x 2. Slice the top into 12 slices and the cut the middle once horizontally. 11 + 1 = 12 cuts

    6 x 4: slice the top in 2 directions into 6 and 4 slices. 5+3 = 8 cuts

    Doug Dillon

    I answered the question below.Author has 9.8K answers and 8M answer views4y

    Related

    What is the maximum number of smaller identical pieces a cube can be cut into by making 17 cuts?

    Let’s make cuts parallel to the face planes of the cube. If we make a cuts parallel to 2 opposite faces,

    a+1 a+1

    identical pieces might reveal themselves.

    Same with the other two pairs of opposite faces for a total of

    (a+1)(b+1)(c+1) (a+1)(b+1)(c+1) pieces. So we maximize (a+1)(b+1)(c+1) (a+1)(b+1)(c+1) subject to a+b+c=17 a+b+c=17

    . Unfortunately, I see no way to continue apart from a little brute force. But I hope the numbers

    a,b, a,b, and c c

    will be was close together as possible. Perhaps,

    स्रोत : www.quora.com

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    Mohammed 2 month ago
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