Consecutive Integers

We know that to find the prime factorization of a number, we need to consider primes only upto the square root of that number (for example : consider primes only upto $\sqrt{71}$ for 71.)

If we had been given a square number, then we would be done after taking the square root.

But since the number is not a square root, we can consider whole numbers a little smaller than the square root and a little greater than the square root.

We also have that they are consecutive whole numbers, hence they will be a little less and a little greater than the square root. (why?)

Moreover, the unit digit is 2 . Hence the unit digits of the numbers we need must give a product ending in 2 (forget about carrying over for now).

Now $\sqrt{567762} = 753.499$ Therefore, we need to consider the whole numbers a little less than and a little greater than 753.499 whose unit digits give a product 2.

Hence we get the numbers 754 and 753, whose sum is $\boxed{1507}$ . And we are done !!

We know that 567762 is the product of two positive consecutive integers mean x(x+1)=567762

So, the sum of the two numbers is x+(x+1) = 2x+1

Assume 2x+1 = y, we'll find that x = ½(y-1)

So, x(x+1) = 567762 equals with :

½(y-1)[½ (y-1)+1] = 567762

½(y-1)½(y+1) = 567762

y^2 - 1 = 4 x 567762

y = ±√2271049

Because y is the sum of two positive numbers so y = 1507

X(X+1) = 567762

X^2 + X = 567762

X^2 + X - 567762 = 0

X = 753 & X = -754

so

X = 753 & (X+1) = 754

754+753 = 1507

suppose the two numbers are x and x+1. so we can write x*(x+1)=567762 or,x^2+x-567762=0 or, x^2+754x-753x-567762=0 or, x(x-753)+754(x-753)=0 or,x=753 and x=754 [As the numbers are consecutive integers ,x=-754 cannot possible] so the two numbers are 753,754.

I did this in my head in about 2 minutes without a pen/paper and without doing any difficult math. It did take me 2 attempts though. Basically, 7x8=56 so 700x800=560,000 so the 2 numbers are between 700 and 800, and seem to be around 750. Last 2 digits produce a 2 in the units so I first tried 1x2 (761 and 762 - which were wrong) and then 3x4 (which was right). Because this produced a +1 in the 10s digit, the 10s digit needed must be 5 to produce 5+1. So the 2 numbers are 753 and 754 and the sum is 1507.

Let the numbers be (x-.5) and (x+.5)

Therefore x = √(567762+.25) = 703.5

And the required sum is 2x =1507

754*753=567762 754+753=1507

x*(x+1)=567762 so x^2+x-566762=0 so x(x-753)+754(x-753) the numbers are thus 753 and 754 .sum =1507