There is an equation to find the sum of a list of numbers . It is called the Arithmetic Progression.
The equation is
S(n) = (n/2)[a + (n-1)d] 4
Where
n = number of terms (1000)
a = first term (1000)
d = difference between terms (1)
Hence
S(1000) = (1000/2)[1000 + ( 1000 - 1)(1)]
S(1000) = 500[ 1000 + 999]
S(1000) = 500[1999]
S(1000) = 999500 The answer!!!!!
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The answer is 500,500. (I did it on my calculator. It took forever. YOUR WELCOME!)
there are many symmetric numbers. 111, 121, 454, 252, 999, 545 etc
999. Think about it 001 002 003.....etc. It just goes in order up to the 999 plus 000 makes 1,000 To calculate an ordered sequence (a permutation), you take the number of options for the first choice, times the number of options for the second choice, times the number of options for the third choice (etc). In this case, you have 10 options for each digit (0 thru 9), so: 10 x 10 x10 = 1,000
500+501, 1000+1, etc
To calculate how many times 28 can go into 1000, you would divide 1000 by 28. The result is approximately 35.714. Since you cannot have a fraction of a division, you would have to round down to the nearest whole number. Therefore, 28 can go into 1000 approximately 35 times.
An infinite number of ways. 99+9/9 (99+9/9)*9/9 (99+9/9)*99/99 (99+9/9)*999/999 (99+9/9)*9/9*9/9 etc