31.how do you solve?
9000 integers.
128!
1x2x3=6 2x3x4=24 3x4x5=60 4x5x6=120 5x6x7=210 6x7x8=336 7x8x9=504 8x9x10=720 9x10x11=990
If you mean "between" 100 and 1000 to mean excluding 100 and 1000 then 449. If you want to include 100 and 1000 then 451.
No.Let the four numbers be (n-1), n, (n+1), (n+2).Their sum is 4n + 2 = 2(2n + 1)If 2000 is the sum of four consecutive integers, then:2(2n+1) = 2000⇒ 2n + 1 = 1000⇒ 2n = 999but 999 is odd, not even and so n cannot be an integer; therefore 2000 is not the sum of four consecutive integers.
There are 30 such integers.
1024
You need 5.
1049 x 1051 = 1102499
do not know i thought i asked that question your fake
There are 2828 integers between 1000 and 9999.
9000 integers.
333 integers.
899
There are 128 integers between 100 and 1000 that are divisible by seven.
128!
90 palindromes.