Generalized infinite square roots
Determine the sum of all possible integer values of x < 1 0 0 such that y is an integer.
x + x + x + x . . . = y
The answer is 330.
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5 solutions
*squares. :v
where did get the value of Y? which is equal to 1+square root of 1+4(x)/2?
Squaring both the sides and subtracting x from both sides, y 2 − x = x + x + x + ⋅ ⋅ ⋅ = y
doesn't the square root have a (-) and a (+) root. If yes, then its sum yields (+)1 for every eligible 'x'. There are only 9 eligible x's. So, shouldn't answer be 9
I had nearly the same solution as you but halfway through finding the x values, I noticed that the difference between the n th and the ( n − 1 ) th value of x is always equivalent to the n th even number and so I was able to deduce the remaining values of x fairly quickly. I was wondering if we can prove this pattern. Perhaps through induction?
We do the normal method and change the equation to
x + y = y
Next, we remove the radical and put everything on one side of the equation.
y 2 − y − x = 0
For this to be factorable and because y is dependent upon x,x has to be able to be written as a ( a + 1 ) . Implying that it can be factored into
( y + a ) ( y − ( a + 1 ) ) ⇒ y 2 − y ( a + 1 ) + y a − a ( a + 1 ) ⇒ y 2 − y − a ( a + 1 )
Thus we need to plug in a = [ 1 , 2 , 3 . . . 9 ] since a=10 implies x > 1 0 0
After plugging and chugging, we find our answer to be 330.
Normal method? Isnt it logic? :P O r is it that too many problems of this type have made it a method? Nice problem though.
nice problem ........ i didnt see that it was sum of solutions and typed the number of solutions :p
i found the sum of the possible values y instead!!.... :P
square both sides, substitute for y, rearrange in terms of y. we see that x is a product of two consecutive integers and therefore must always be even, 10th partial sum of sequence y(y-1) is your answer=330.
As y^2=x+y y*(y-1)=x Thus put values of yfrom 2-10 and get respective values of x as 2+6+12+20+30+42+56+72+90=330
all satisfying numbers below 100 are 2, 6, 12, 20, 30, 42, 56, 72, 90 thir sum is 330 hence answered
Instead of simply adding these numbers we can observe that 2= 1^2 +1, 6=2^2 +2, 12=3^2+3, 90=9^2+9
This is same as x + y = y . S o l v i n g f o r y : − y = 2 1 + 1 + 4 ∗ x . y i s t o b e a n i n t e g e r . ⟹ 1 + 4 ∗ x m u s t b e a n o d d i n t e g e r .
⟹ 4 ∗ x + 1 m u s t b e s q u a r e o f a n o d d i n t r g e r . F o r x < 1 0 0 , t h e s e s u q a r e s a r e : − 1 , 9 , 2 5 , 4 9 , 8 1 , 1 2 1 , 1 6 9 , 2 2 5 , 2 8 9 , 3 6 1 .
T h e y m u s t b e e q u a l t o > > > 4 ∗ x + 1 . ⟹ x = 0 , 2 , 6 , 1 2 , 2 0 , 3 0 , 4 2 , , 5 6 , 7 2 , 9 0 . . . a l l < 1 0 0 . T h e i r s u m = 3 3 0 .