Playing with Numbers Test - 4

A number is divisible by both 6 and 13.
Factors of 6 = 1, 2, 3, 6

Factors of 13 = 1, 13

As the common factors of these numbers is 1, the given two numbers are co-prime numbers and they will also be divisible by their products 13 x 6 = 78

Sum of odd places = 7 + 0 + 6 = 13
Sum of even numbers = * + 5 + 4 = 9 + *

If the number is divisible by 11, then either 13 - (9 + *) = 0
Or 13 - (9 + *) = 11 m (where m is any integer)

Now, for the least value of *, m should be zero
13 - 9 - * = 0
* = 4.

Factors of 17 = 1 and 17.
Factors of 23 = 1 and 23.
Therefore, 17 and 23 are prime numbers.
H.C.F of 17 and 23 = 1
Therefore, they are co-prime numbers also.

Let two consecutive numbers be 2n and 2n + 2.
So, both have HCF 2.
So, HCF of two consecutive even numbers is 2.

If a number has exactly two factors, 1 and itself, then the number is a prime number.

According to the given pattern the sum of 1st 'n' odd natural numbers is n².

Number of terms in 1 + 3 + 5 +.......+ 21 is 11.

So, the sum of odd numbers till 21 is equal to 11² = 121.

A fraction is in its lowest terms or in lowest form, if the HCF of its numerator and denominator, is 1.

The two numbers nearest to 10,000, which are exactly divisible by each of 3, 5, 7, 8 and 9, are the multiples of LCM of the given numbers.
LCM = 3 x 5 x 7 x 8 x 9 = 2,520

Now, dividing 10,000 by 2,520, we get remainder = 2,440

Number just less than 10,000 and exactly divisible by the given numbers = (10,000 - 2,440) = 7,560

Number just greater than 10,000 and exactly divisible by the given numbers = 10,000 + (2,520 - 2,440) = 10,080