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    what is the least possible number of cuts required to cut a cube into 80 identical pieces?

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    What is the least number of cuts required to cut a cube into 84 identical pieces,

    What is the least number of cuts required to cut a cube into 84 identical pieces, assuming all cuts are made ... faces of the cube? 11 12 13 10

    What is the least number of cuts required to cut a cube into 84 identical pieces,

    asked in Quantitative Aptitude Aug 22, 2015

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    स्रोत : aptitude.gateoverflow.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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    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.

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    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

    We want five identical vertical layers (front to back), which requires four cuts

    We want five identical vertical layers (left to right), which requires four cuts

    one cut + four cuts + four cuts = nine cuts

    Again, I don’t

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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.

    Aaron Jantzen

    Loophole developer and exploiterAuthor has 3.4K answers and 14M answer views4y

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    What is the maximum number of identical pieces a cube can be cut into by 4 cuts?

    Any Japanese chef can give you a far better answer to this than your typical mathematician will. A mathematician might tell you 12 or 16 depending upon whether they rearrange the pieces before cutting. However a Japanese chef will tell you “thousands” and that is meant literally. Not only can it be done, it is indeed regularly done by these chefs. The secret is called the Katsuramuki technique. Quite simply, the first cut is a spiral, which is normally done by hand with a sharp knife on a log-shaped vegetable such as Daikon, resulting in a long sheet, but could work with a cube by gradually ro

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    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

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    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..

    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. Bu...

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    Mohammed 2 month ago
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