Page Progression - Part 2

The sum of the first $n$ natural numbers is $\dfrac{n(n+1)}{2}$ , and by the problem we know that:

$\dfrac{n(n+1)}{2}+x=2000$

Where $n$ is the page number, and $x$ is the page conted twice. So, let's solve only for $n$ :

$\dfrac{n(n+1)}{2}=2000$

$n\approx62.74$

So, the actual page is $n=62$ . Now, let's replace that value to find $x$ :

$\dfrac{62(62+1)}{2}+x=2000$

$1953+x=2000$

$x=\boxed{47}$

So, the page conted twice is the 47th one.

Sum of n natural numbers = n(n+1)/2. Since a page is counted twice, let the twice counted page be x and n(n+1)/2 = 2000 - x This implies, n^2 +n -4000+2x=0 Thus, Discriminant should be a perfect square and n the number of pages and x should always be a natural number. Thus, D=16001-8x is a perfect square. Thus, by solving we get x = 47 ( whole number). Where D = 15625 = 125^2 We may also get D=15876 = 126^2 but 16001-15876 = 125 will not be divisible by 8 for x E W. Thus, x=47

Let, no page is counted twice &he; read x pages So,, x(x+1)/2=2000 or,x^2+x-4000=0 {√4000=63.24}[as the no. of pages never be a fraction, so we factorises the quadratic eqn x^2+63x-62x+3906=0] We can't write x as 64x-63x because then the summation of pages will be >2000(63×64/2=2016). It is't possible as Satvik count only 1page twice. So the actual sum of the pages is 62×63/2=1953 So he repeated the page (2000-1953)=47 twice. so easy...............

Let, no page is counted twice &he read x pages So,, x(x+1)/2=2000 or,x^2+x-4000=0 {√4000=63.24}[as the no. of pages never be a fraction, so we factorises the quadratic eqn x^2+63x-62x+3906=0] We can't write x as 64x-63x because then the summation of pages will be >2000(63×64/2=2016). It is't possible as Satvik count only 1page twice. So the actual sum of the pages is 62×63/2=1953 So he repeated the page (2000-1953)=47 twice.

Abide the clear solutions presented already, you can always counter proof by subtracting all the integers progressively from 2000. The left over is your result: 2000 - 1 - 2 - 3 - 4 - 5....- 62 = 47

Let number of pages be n, page counted twice be k. Then we have n(n+1)/2 + k = 2000. We can find the upper bound of n by letting k = 1 and solving the quadratic which gives the maximum possible value of of n as 63 ( rounded off). So now we try the lower values to see which works .. Trying n = 62 we find n(n+)/2 = 1953 , which leads to a valid value of k = 2000 -1953 which is 47. Also k satisfies the constraint , n => k.

I solved this sum by a formula- $\frac{n(n+1)}{2}$ ... so, lets assume the pages he read in total be 62(by calculation certainly) so by formula- = $\frac{62*(62+1)}{2}$ = $\frac{62*63}{2}$ = $\frac{3906}{2}$ =1953 then, subtracting 1953 from 2000, we get 2000-1953=47... that means he counted 47 twice.. therefore, 47 is the ans..

We all know the sum of hundred numbers is 5050. Therefore we know it's somewhere in between. For 50 numbers it's 1275 where we can check by adding more numbers which I found by brüte force 47

You can simply use the method $\frac{n(n+1)}{2}$ = 2000 to obtain approx 62.

Now replacing n = 62 you get 1953.

So the required page is 2000-1953 = $\boxed{47}$

Sum of 'n' terms of an A.P is "S=n/2 [2a + (n-1)d]

a = first term

d = common difference

Now,

=>2000=n/2[2(1) + (n-1)*1]

=>4000=n(n+1)

If you take n=62 you will get n(n+1)=62(63)=3906<4000(that is what , we need)

Then 4000-3906=94

=>Number of the page counted twice = 94/2=47!!

How do we solve this other than by trial and error method.

Can we arrive at a formula which could be applied to any (sum+repeated_number) ?