Real Numbers Test - 4

Largest such number will be the HCF of [(70 - 5) , (125 - 8)] = HCF (65,117)

65 = 13 x  5

117 =13 x 9

So, HCF =13

We know that 2 is a factor of all the even numbers.

Consider an even number, say 2q.

This number will have its next consecutive even number as 2q+2

Further, this number will have its consecutive even number are 2q + 2 + 2= 2q + 4

2q + 2 can be written as 2(q+1), while 2q + 4 can be written as 2(q+2).

So, the only common factor is 2.

The HCF of two consecutive even numbers is 2.

n² –n= n (n-1)

Now, if n is odd , then (n – 1) is even .

If n is even , then ( n – 1 ) is odd .

In either case n (n -1 ) is the product of an odd and even integer , which we know is even.

18 x 25 x 25 + 25 = 25(450 + 1)

                           = 25 x 451

which is composite since it has 2 factors other than one.

p and q are said to be co-primes if they have no common factor other than 1.

The HCF of two numbers that have no common factor other than 1 is 1.

7 x 11 x 13+13

= 13(7 x 11 + 1)

=  13(78)

Now this number has more than two factors.

Therefore it is a composite number.

A real number is an irrational number when it has a non-terminating non-repeating decimal representation.

Thus, 0.101100101010……. is an irrational number.

We see that 273 >105

So,

273 = (105×2) + 63

105 = (63×1) + 42

63 = (42×1) + 21

42 =(21×2) + 0

The divisor in the step where the remainder is 0, is 21.

Hence, the HCF is 21.

If the number that divides 460 and 50 leaving remainder 5 in the first case and 8 in the second case,

then it divides 460 – 5= 455 and 50 – 8 = 42 , exactly.

We have to find the HCF of 455 and 42.

Using Euclid's Division Algorithm,

455 = 42 x 10 + 35

42 = 35 x 1 + 7

35 = 7 x 5 + 0

So, HCF (455, 42) = 7 ....(Since the divisor at the last stage is 7)

Therefore, the largest number that divides 460 and 50 leaving remainder 5 in the first case and 8 in the second case is 7.