# what is the least possible number of cuts required to cut a cube into 343 identical pieces?

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## A cube is divided into 343 identical cubelets. Each cut is made parallel to some surface of the cube. But before donig that cube is coloured with green colour on one set of adjacent faces, red on the second, and blue on the third set.How many minimum cuts you have made?

A cube is divided into 343 identical cubelets. Each cut is made parallel to some surface of the cube. But before donig that cube is coloured with green colour on one set of adjacent faces, red on the second, and blue on the third set.How many minimum cuts you have made?

Byju's Answer Standard IV Mathematics

Different Views of Looking at a solid

A cube is div... Question

A cube is divided into 343 identical cubelets. Each cut is made parallel to some surface of the cube. But before donig that cube is coloured with green colour on one set of adjacent faces, red on the second, and blue on the third set.

How many minimum cuts you have made?

A 15 B 18 C 21 D 9 Open in App Solution

The correct option is **B** 18

n 3 = 343 = 7 3 ⇒ n = 7

Minimum number of cuts = 3(n-1)

=3(7-1) =3 × 3 × 6=18

As we can see in above figure. 3 faces are visible in 3-diff colours, out of hidden faces, bottom is red, one is green and another is blue.

Suggest Corrections 0 SIMILAR QUESTIONS

**Q.**57. A cube is cut in two equal parts along a plane parallel to one of its faces. One piece is then coloured red on the two larger faces and green on the remaining, while the other is coloured green on two smaller adjacent faces and red on the remaining. Each is then cut into 32 cubes of same size and mixed up. How many cubes have only one coloured face each?

**Q.**A cube is divided into 343 identical cubelets. Each cut is made parallel to some surface of the cube. But before donig that cube is coloured with green colour on one set of adjacent faces, red on the second, and blue on the third set.

How many minimum cuts you have made?

**Q.**A cube is divided into 343 identical cubelets. Each cut is made parallel to some surface of the cube. But before donig that cube is colored with green colour on one set of adjacent faces, red on the second and blue on the third set

How many cubelets are colored with exactly two colors?

**Q.**A cube is divided into 343 identical cubelets. Each cut is made parallel to some surface of the cube. But before donig that cube is colored with green colour on one set of adjacent faces, red on the second and blue on the third set.

How many cubelets are thin which are colored with exactly three colors?

**Q.**A cube is divided into 343 identical cubelets. Each cut is made parallel to some surface of the cube. But before donig that cube is colored with green colour on one set of adjacent faces, red on the second and blue on the third set.

How many cubelets are painted with multiple colors?

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## 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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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

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 417.1K 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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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

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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

Guys, does anyone know the answer?