dont kno so just helpp friendds!
The first whole number divisible by 3 is 102 and the last one 399. Let n be the number of whole numbers between 102 and 399 102 + (n - 1)x3 = 399 (this is an arithmetic progression) Solving n, n-1 = (399 - 102)/3 = 99 n = 100 Since even whole numbers among these 100 will be divisible by 6, the number not divisible is half of 100, i.e. 50. regards, lpokbeng
There are many numbers between 500 and 1000 divisible by 3 and 9. Any number divisible by 9 is divisible by 3. How about 900?
33 numbers between 1 and 100 are divisible by 3.
251
7 is only divisible by itself and one because it is a prime number.
An infinite number.
The first whole number divisible by 3 is 102 and the last one 399. Let n be the number of whole numbers between 102 and 399 102 + (n - 1)x3 = 399 (this is an arithmetic progression) Solving n, n-1 = (399 - 102)/3 = 99 n = 100 Since even whole numbers among these 100 will be divisible by 6, the number not divisible is half of 100, i.e. 50. regards, lpokbeng
There are many numbers between 500 and 1000 divisible by 3 and 9. Any number divisible by 9 is divisible by 3. How about 900?
251
18 numbers are there
2
7 is only divisible by itself and one because it is a prime number.
Quite a few. You can generate as many as you want. Here is a generator: [ 48 times 'x' ]. Make 'x' any whole number - positive or negative - do the multiplication, and Shazam! You have a number divisible by 48.
1004 nope that ain't right
The first number in this range divisible by 9 is 9 itself...9 = 1 x 9 The last number in the range is 999 = 111 x 9 So there are 111 numbers between 1 and 1000 that are divisible by 9.
Only one, and that is 456.
How many numbers are divisible by 9 between 5 and 1000²