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FIND THE LEAST NUMBER TO BE MULTIPLIED TO GET A PERFECT SQUARE

A perfect square is a number that can be expressed as the product of an integer by itself or as the second exponent of an integer.

For example consider 72,

Decomposing 72 into prime factors, we get

72 = 2 x 2 x 2 x 3 x 3

72 is not a perfect square, to make this as prefect square we need one more 2.

72 x 2 ==> 144

144 = 122

Problem 1 :

The least number by which 294 must be multiplied to make it a perfect square, is

a)   2        b)   3        c)   6    d)   24

Solution :

From the questions, it is clear 294 is not a perfect square. To make 294 as perfect square, we will multiply some numerical value or values.

Prime factorization of 294 is,

294 = 7 × 7 × 2 × 3

Here we need one more 2 and one more 3.

To make it a perfect square, it must be multiplied by 2 × 3 = 6

Required number = 6

So, option (c) is correct.

Problem 2 :

Find the smallest number by which 5808 should be multiplied so that the product becomes a perfect square.

a)   2      b)   3       c)   7        d)   11

Solution :

Prime factorization of 5808 is,

5808 = 2 × 2 × 2 × 2 × 3 × 11 × 11

Therefore, when 5808 is multiplied by 3, then it will be perfect square number.

So, option (b) is correct.

Problem 3 :

By which least number should 22050 be multiplied such that the result is a perfect square.

Solution :

Decomposing 22050,

22050 = 5 x 5 x 2 x 21 x 21

To make it as perfect square, we need one more 2. So, the required number is 2.

Problem 4 :

Find the smallest common multiple of 48, 72 and 32 that is a perfect cube.

Solution :

Decomposing 48, 72 and 32, we get

48 = 2 x 2 x 2 x 2 x 3 ==> 24 x 3

72 = 2 x 2 x 2 x 3 x 3 ==> 23 x 32

32 = 2 x 2 x 2 x 2 x 2 ==> 25

LCM = 25 x 32

= 288

Decomposing 288, we get

25 x 32

that is, 2 x 2 x 2 x 2 x 2 x 3 x 3

To make it as perfect cube, we need one more 2 and one more 3.

So, the required number is 2 x 3 ==> 6.

Problem 5 :

What is the smallest common multiple of 72 and 108 that is a perfect square ?

Solution :

Least common multiple of 72 and 108.

72 = 2 x 2 x 2 x 3 x 3 ==> 23 x 32

108 = 2 x 2 x 3 x 3 x 3 ==> 22 x 32

= 23 x 32

= 72

Now decomposing 72, we get

= 23 x 32

To make it as perfect square

= 23 x 32 x 2 x 3

So, the required number is 6.

Problem 6 :

By which least number should 4375 be multiplied such that the result is a perfect square ?

Solution :

Decomposing 4375, we get

4375 = 5 x 5 x 5 x 5 x 7

To make it as perfect square, we need one more 7.

4375 = 5 x 5 x 5 x 5 x 7 x 7

So, the required number is 7.

Problem 7 :

By which least number should 72000 be multiplied such that the result is a perfect square ?

Solution :

Decomposing 72000, we get

72000 = 25 x 52 x 32

To make it as perfect square, we need one more 2.

So, the required number is 2.

Problem 8 :

Find the smallest number by which 375 must be multiplied to obtain a perfect cube.

Solution :

Decomposing 375, we get

375 = 5 x 5 x 5 x 3

= 53 x 3

To get cube of 3, two more 3's should be included. So, 9 should be multiplied by 375 to make it as perfect cube.

Problem 9 :

You have 275 square inches of wrapping paper. Do you have enough wrapping paper to wrap the gift box shown? Explain.

perfect-square-or-not-q2

Solution :

Volume of box = 343 cubic inches

Side length is x.

x3 = 343

x =  3√343

x =  3√(7 x 7 x 7)

= 7 inches

Total surface area of cube = 6x2

= 6(7)2

= 6(49)

= 294 square inches

To wrap the gift box, we need 294 square inches of paper. Since it is provided 275 square inches of paper, it is enough.

Problem 10 :

A cube-shaped end table has a volume of 216,000 cubic centimeters. Does the end table fit in the corner shown? Justify your answer

perfect-square-or-not-q3

Solution :

Volume of cube shaped table = 216,000 cubic centimeters

Let x be the side length of table.

x= 216000

x = 3√216000

x = 3√(60 x 60 x 60)

x = 60 cm

Area of square = 6400 cm2

x= 6400

x = √6400

x = √(80 x 80)

= 80 cm

Since the side length of square shape is 80 cm and side length of cubic shape table is 60 cm, the table will fit there.

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