Question

The difference of squares of two consecutive natural numbers is 2011. Find the numbers.
152 a perfect square? Explain.
2.
3.
Find the Pythagorean triplet whose smallest number is 20.​

Answer 1

(x+1)²-x²=2011

x²+2x+1-x²=2011

2x+1=2011

2x=2010

x=1005

and the other no.is 1006

the pythagoras triplets 20²+21²+22²=400+441+484=1325

Answer 2

Solution :-

(1)

Let us assume that, smaller number is x .

So,

→ Greater number = (x + 1)

then,

→ (x + 1)² - x² = 2011

→ (x² + 1 + 2x) - x² = 2011

→ x² - x² + 1 + 2x = 2011

→ 2x = 2011 - 1

→ 2x = 2010

→ x = 1005

therefore,

→ Smaller number = 1005

→ Greater number = 1005 + 1 = 1006 .

Hence, required two consecutive natural numbers are 1005 and 1006 .

(2)

We know that,

• A perfect square number unit digit is 0, 1, 4, 5, 6 and 9 only .

So,

→ 152 = unit digit 2 .

Therefore, 152 is not a perfect square number .

(3)

We know that

• For all natural number n > 2 , (2n , n² - 1 and n² + 1) is a pythagorean triplet .

So,

→ Smallest number = 2n

→ 2n = 20

→ n = 10

then,

→ n² - 1 = 10² - 1 = 100

→ n² + 1 = 10² + 1 = 101

therefore, the pythagorean triplet whose smallest number is 20 :- 20, 99 and 101 .

Learn more :-

let a and b positive integers such that 90 less than a+b less than 99 and 0.9 less than a/b less than 0.91. Find ab/46

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