There is only one 7. Whether it's between 1-999, or between 1-999,999,999. Now, if the question was how many numbers that contains 7 in it (i.e., 17, 27, 70, 750) I am not sure. I believe it is 280.
There are 280 of them.
997
45 of them.
There are 107 numerical palindromes between the numbers 1 and 1000, starting from 2 to 999.
There are 179 multiples of 5 between 101 and 999. 101 ÷ 5 = 201/5 → first multiple of 5 ≥ 101 is 21 x 5 999 ÷ 5 = 1994/5 → last multiple of 5 ≤ 999 is 199 x 5 → number of multiples of 5 between 101 and 999 is 199 - 21 + 1 = 179.
If by "between" you mean "between but not including", then the answer is 9999 - 999 - 1 = 8999.If on the other hand, you mean to include both 9999 and 999, then the answer is 9999 - 999 + 1 = 9001.
997
999 999. 1 million and 2 million aren't counted. It is only the number inbetween. so from 1000001, 1000002,.........1999998, 1999999 which is 999 999 numbers
45 of them.
There are 179 multiples of 5 between 101 and 999. 101 ÷ 5 = 201/5 → first multiple of 5 ≥ 101 is 21 x 5 999 ÷ 5 = 1994/5 → last multiple of 5 ≤ 999 is 199 x 5 → number of multiples of 5 between 101 and 999 is 199 - 21 + 1 = 179.
999 MOD 1000 = 999
Easier way than what?I presume you mean long multiplication of 999 by 999, adding 998 and adding 1.One way:999 = 1000 - 1 ⇒ 999 x 999 + 998 + 1 = (1000 - 1) x (1000 - 1) + 998 + 1= 10002 - 2 x 1000 + 1 + 998 + 1= 1000000 - 2 x 1000 + 1000= 1000000 - 1000= 999000Another way:998 + 1 = 999 ⇒ 999 x 999 + 998 + 1 = 999 x 999 + 999= 999 x (999 + 1)= 999 x 1000= 999000
999
9 times 111 = 999, or 3 times 333 = 999
1.0 × 10^56
999 quadrillion, 999 trillion, and 1 billion.
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