There are 2828 integers between 1000 and 9999.
Smallest multiple of 7 greater than 100 is 15 Largest multiple of 7 less than 1000 is 142 So number of multiples of 7 = 142 - 15 + 1 = 128
68 of them.
The last "x7" under 100 is 14 x 7, the last under 1000 is 142 x 7, so there are 128 in the range.
428 of them.
There are 2828 integers between 1000 and 9999.
1001, 1014, 10027 and just keep adding 13 till you get to 9997.
To find the number of 4-digit numbers that are multiples of 9, we first determine the range of 4-digit numbers, which is from 1000 to 9999. Next, we calculate the first and last multiples of 9 within this range, which are 1008 and 9999 respectively. To find the total number of multiples of 9 in this range, we divide the difference between the last and first multiples by 9 and add 1, resulting in (9999 - 1008) / 9 + 1 = 999. Therefore, there are 999 4-digit numbers that are multiples of 9.
Prime No: between 1000 to 9999= 54_( 42x3)
From 1000 to 9999 (inclusive) there are 9999 - 1000 + 1 = 9000 integers. Half of which are even, half of which are odd. So the answer is 9000/2 = 4,500.
9*9*8*7 = 4536
128!
There are 10 palindromes divisible by 9 between 1000 and 9999.
They are members of the set of numbers of the form 7*k where k is an integer which takes 1000 different values..
Smallest multiple of 7 greater than 100 is 15 Largest multiple of 7 less than 1000 is 142 So number of multiples of 7 = 142 - 15 + 1 = 128
from 1000 to 9999, ie 9999 - 999 = 9000
There are 143 such numbers, too many to list.