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Practice each question, then open the answer or worked solution.
There are 200 natural numbers between 1002 and 1012.
Solution:-
True.
We know that, natural numbers between a and b = b –a – 1
= 1012 – 1002 – 1
= (101 + 100) (101 – 100) – 1
= 201 – 1
= 200
The sum of first n odd natural numbers is n2.
Solution:-
True.
The sum of odd numbers = ∑ (2n – 1)
= (2 × n × (n + 1))/2 –n
= n2 + n – n
= n2
1000 is a perfect square.
Solution:-
False.
It is not a perfect square.
1000 is a perfect cube, i.e. 10 × 10 × 10 = 1000
A perfect square can have 8 as its unit’s digit.
Solution:-
False.
A perfect square can never have 8 as its unit’s digit.
For every natural number m, (2m –1, 2m2 –2m, 2m2 –2m + 1) is a Pythagorean triplet.
Solution:-
False.
For every natural number m > 1, 2m, m2–1 and m2 + 1 form a Pythagorean triplet.
All numbers of a Pythagorean triplet are odd.
Solution:-
False.
Three natural numbers a, b, c are said to form a Pythagorean triplet if a2 + b2 = c2.
Then,
Pythagorean triplet as, 72 = 62 + 52
6 is not an odd number.
For an integer a, a3 is always greater than a2.
Solution:-
False.
Let us take one negative integer, i.e. -2
Then, -23 = -2 × -2 × -2 = -8
-22 = -2 × -2 = 4
Here, 4 > -8
Hence, -22 is greater than -23.
If x and y are integers such that x2 > y2, then x3 > y3.
Solution:-
False.
Let us take negative integers -2 and -3
Then,
= (-22) > (-12)
= (4) > (1)
and = -23 < -13
= -8 < -1
Let x and y be natural numbers. If x divides y, then x3 divides y3.
Solution:-
True.
If x divides y = y/x is a natural number
and then x3 divides y3 = y3/x3 is a natural number
If a2 ends in 5, then a3 ends in 25.
Solution:-
False.
Let a2 be 152
So, 152 = 225
Then, 153 = 3375