Math, asked by heeraaleeza, 1 month ago

the length of a room is 960cm and its breadth is 1120cm. What is the measurement of the largest square tile that can be fixed on the floor so that no tile needs to be cut of floor.

Pls show the method too

Answers

Answered by mehraniayashika
1

Answer:

Required Basics

\red{\bigstar\ \text{The Sum of the First }n\text{ Integers}}★ The Sum of the First n Integers

The sum of the first nn positive integers, \purple{S_{n}}S

n

is,

\implies\purple{\dfrac{n(n+1)}{2}}⟹

2

n(n+1)

\red{\bigstar\ \text{The Number of Integers Between }x\text{ and }y}★ The Number of Integers Between x and y

The number of integers between xx and yy , with y\geq xy≥x is,

\implies\purple{y-x+1}⟹y−x+1

\large\text{\underline{Solution}}

Solution

The sum of the first nn integers;

\implies S_{n}=\dfrac{n(n+1)}{2}⟹S

n

=

2

n(n+1)

The sum of the first n+11n+11 positive integers;

\implies S_{n+11}=\dfrac{(n+11)(n+12)}{2}⟹S

n+11

=

2

(n+11)(n+12)

The way to obtain the sum of consecutive 11 natural numbers is using the subtraction between the two sums;

\implies S_{n+11}-S_{n}=\dfrac{n^{2}+23n+132}{2}-\dfrac{n^{2}+n}{2}⟹S

n+11

−S

n

=

2

n

2

+23n+132

2

n

2

+n

\implies S_{n+11}-S_{n}=\dfrac{22n+132}{2}⟹S

n+11

−S

n

=

2

22n+132

\implies S_{n+11}-S_{n}=11n+66⟹S

n+11

−S

n

=11n+66

Hence, this should be less than 100;

\implies11n+66 < 100⟹11n+66<100

On solving,

\implies 11n < 100-66⟹11n<100−66

\implies 11n < 34⟹11n<34

\implies n < \dfrac{34}{11}⟹n<

11

34

We are adding numbers starting from n+1n+1 ;

\implies n+1\geq1⟹n+1≥1

\implies n\geq0⟹n≥0

The common interval of two inequality is,

\implies 0\leq n < \dfrac{34}{11}⟹0≤n<

11

34

So, the solution is,

\implies\boxed{n=0,1,2,3}⟹

n=0,1,2,3

\large\text{\underline{Conclusion}}

Conclusion

There are four numbers in the form of 11n+6611n+66 , which are 6666 , 7777 , 8888 , or 9999 .

\large\text{\underline{Verification}}

Verification

We are adding numbers from n+1n+1 to n+11n+11 ;

n=0n=0

\implies66=1+2+3+4+5+6+7+8+9+10+11⟹66=1+2+3+4+5+6+7+8+9+10+11

n=1n=1

\implies77=2+3+4+5+6+7+8+9+10+11+12⟹77=2+3+4+5+6+7+8+9+10+11+12

n=2n=2

\implies88=3+4+5+6+7+8+9+10+11+12+13⟹88=3+4+5+6+7+8+9+10+11+12+13

n=3n=3

\implies99=4+5+6+7+8+9+10+11+12+13+14⟹99=4+5+6+7+8+9+10+11+12+13+14

Hence verified.

Answered by Ryder420
1

\large\green{fuc  \: u}

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