Math teacher told john to find sum of n natural numbers. John missed a number and got the sum as 3932. Find the number he missed.
Thanks in advance
Thanks in advance
I believe it is sum of first n natural numbers otherwise it is not possible to solve.Ramalakshmi wrote:Math teacher told john to find sum of n natural numbers. John missed a number and got the sum as 3932. Find the number he missed.
Sum of first n natural numbers is given by n(n + 1)/2.
Say, the number he missed is r.
Hence, [n(n + 1)/2 - r] = 3932
=> n(n + 1) - 2r = 7864
=> n(n + 1) = (7864 + 2r)
=> 2*(Ideal Sum) = (7864 + 2r)
Thus if we can determine the ideal sum, then we can easily answer the question. Note that twice the ideal sum can be expressed as the product of two consecutive integers, n and (n + 1). Hence we need to find the minimum possible number which is greater than 7864 and can be expressed as the product of two consecutive integers.
Now it is obvious that the consecutive integers will be some integers near the square root of 7864. √7864 = 88.(Something)
Now, 88*89 = 7832 < 7864 NO
........ 89*90 = 8010 > 7864 YES
Hence the ideal sum is (8010/2) = 4005
Hence, the number he missed, r = (4005 - 3932) = 73
