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DCCXXI = 721 D=500 C =100 X =10 I = 1 you can use these to work it out
No.The sum of the first n consecutive natural numbers is:sum = n(n+1)/2So if the sum is 500, then:n(n+1)/2 = 500⇒ n(n+1) = 1000Which means that 1000 must have a pair of factors with a difference of 1 if it is possible.The pairs factors of 1000 are:1 x 10002 x 5004 x 2505 x 2008 x 12510 x 10020 x 5025 x 40No two are separated by 1, thus the sum is impossible.
Sum of first n numbers = n/2(n +1) = 500 x 1001 = 500500
2 x 2 x 5 x 5 x 5 = 500
502