Question

find the least number which must be subtracted from 6409 to make it a perfect square
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Answer 1

Answer:-

The least number which must be subtracted from 6409 to make it a perfect square is 9.

Step-by-step explanation:-

Let's find the square root of 6409.

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,2)(1,1)(1,6)\qbezier(1,6)(1,6)(4,6)\put(0,3.8){\large\sf 160}\put(0.5,5.5){\large\sf 8}\put(1.3,5.5){\large\sf 64 \quad09}\put(1.8,6.4){\large\sf 80}\put(1.3,4.8){\large\sf 64}\put(1.3,3.8){\large\sf 00 \quad09}\put(1.3,3.1){\large\sf 64 \quad09}\qbezier( - 0.1,4.5)(1,4.5)(4,4.5)\qbezier(1,2.8)(1,2.8)(4,2.8)\put(1.8,2.2){\large\sf 9}\end{picture}$

We got remainder = 9

Since, 6409 remainder is not 0,

So, 6409 is not a perfect square,

Clearly, ( 80 )² is less than 6409 by 9

Now, we need to find the least number that must be substracted from 6409 to make it a perfect square:-

Perfect square = 6409 - 9

= 6400

And also, if we do long division with 6400,

$\setlength{\unitlength}{1 cm}\begin{picture}(20,15)\thicklines\qbezier(1,2)(1,1)(1,6)\qbezier(1,6)(1,6)(4,6)\put(0,3.8){\large\sf 160}\put(0.5,5.5){\large\sf 8}\put(1.3,5.5){\large\sf 64 \quad00}\put(1.8,6.4){\large\sf 80}\put(1.3,4.8){\large\sf 64}\put(1.3,3.8){\large\sf 00 \quad00}\put(1.3,3.1){\large\sf 64 \quad00}\qbezier( - 0.1,4.5)(1,4.5)(4,4.5)\qbezier(1,2.8)(1,2.8)(4,2.8)\put(1.8,2.2){\large\sf 0}\end{picture}$

We got 80 as square root.

Therefore, The least number which must be subtracted from 6409 to make it a perfect square is 9.

Answer 2

Answer:

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