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Q: What are the multiples of 7 from 1000 to 9999?

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There are 2828 integers between 1000 and 9999.

1001, 1014, 10027 and just keep adding 13 till you get to 9997.

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

They are members of the set of numbers of the form 7*k where k is an integer which takes 1000 different values..

128!

10,000

from 1000 to 9999, ie 9999 - 999 = 9000

There are 143 such numbers, too many to list.

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

0.999

That would be 10,000.

9,999, 19,998, 29,997, 39,996, 49,995

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.

1,000 multiplied by 9,999 is 9,999,000.

The numbers from 1000 to 9999 are all 4 digits, as are -1000 to -9999. Therefore there are 18,000 four-digit numbers in all.

The numbers from 1000 to 9999 are all 4 digits, as are -1000 to -9999. Therefore there are 18,000 four-digit numbers in all.

All the factors of 1000 have an infinite number of multiples.

There are 12 multiples of 77 in that range.

That's too much to list

90 palindromes.

The multiples of 1,000 are an infinite number of integers in the set that begins 1000, 2000, 3000, 4000, and so on.

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