Question

Positive integers x and y satisfy the equation
sqrt x - sqrt 11 = sqrt y
What is the maximum possible value of x/y?

A 2 B 4 C 8 D 11 E 44

i spent a couple of hours on this and i think the answer is 4 but i'm far from sure.

Answer 1

the integer part is throwing me off, but if you have only 4 choices should be easy enough to check which works right?

Answer 2

oh never mind that is the output, not input sorry

Answer 3

yes - i did it by guessing and working out - not in a very formal way
and figured that x = 44 and y = 11 giving x/y = 4

Answer 4

on the other hand both x and y are integers right? that means you have very few choices for x.
\[y=x-2\sqrt{11x}+11\] so your only choices for x are
\[11,11^3,...\]

Answer 5

its deceptively difficult - for me anyway!

Answer 6

oh right, what happens if you ignore the integer part?

Answer 7

sorry - dont know what you mean

Answer 8

Nothing in the problem says x and y are integers.

Answer 9

Stumped!!

Answer 10

looks like 4 to me

Answer 11

x and y are positive integers

Answer 12

yea - i think it is - other options gave irrational values for x and y

Answer 13

if my arithmetic is correct - its worn me out!

Answer 14

4 can't be right. Take x=16. y=0.467. Ratio is 34.26.

Answer 15

yes i think 4 is it. if you try
\[x=99\] you get
\[y=44\] so
\[\frac{x}{y}=\frac{9}{4}\]

Answer 16

right

Answer 17

@GT
\[x=n^2\times 11, n \in \mathbb Z^+\] or else y is not an integer

Answer 18

Sorry - i missed the first line of the problem. duh! :)

Answer 19

ahhh - right satellite

Answer 20

Thats ok GT

Answer 21

ok that was a mistake but i think it is the right approach

Answer 22

how did you get - oh ok

Answer 23

n^2 and (n-1)^2 are the values for coefficient of 11 for x and y.
So, the value of x/y has to be n/(n-1). Maximum value is 2?

Answer 24

\[x=n^2\times 11\]
\[\sqrt{n^2\times 11}-\sqrt{11}=\sqrt{y}\]

\[n\sqrt{11}-\sqrt{11}=\sqrt{y}\]
\[(n-1)\sqrt{11}=\sqrt{y}\]
\[11(n-1)^2=y\]
\[\frac{x}{y}=\frac{n^2\times 11}{11(n-1)^2}\]
\[\frac{x}{y}=\frac{n^2}{(n-1)^2}\]

Answer 25

now your choices for n are 1 which gets you zero for y so that is undefined.
n = 2 which gives you 4, and this function is decreasing for all n > 1, so that is your best bet. namely n = 2 and your max is 4

Answer 26

right - i see - but - excuse me if this is a silly question how did you get the first line?
x = n^2 x 11?

Answer 27

oh you gave it to me!

Answer 28

you said x could be 11,44 , etc. so we know it has to be n^2*11 else the square root is not an integer

Answer 29

right - thanks a lot

Answer 30

yw

Answer 31

this was a question on a UK maths challenge for ages 16-17. some question