Question

How many integers p satisfy the following properties: i) 1<=p<=5000
ii) [sqrt p]=[sqrt(p+125)] where "[]" is the greatest integer function

Answer 1

@pooja195

Answer 2

@imqwerty

Answer 3

Think about this-
[sqrt p]=[sqrt(p+125)] for this condition to hold (p+125) and p must lie between squares of two consecutive integers

Answer 4

correct

Answer 5

then

Answer 6

now (p+125)-p < (n+1)^2-n^2 from the figure

Answer 7

so n>62

Answer 8

Yeah thats correct

Answer 9

and now we need an upper bound

Answer 10

(p+125)-p < (n+1)^2-n^2

Will putting this "<=" instead of "<" work

Answer 11

yeah it will work...

Answer 12

but it will not give the exact solution

Answer 13

I gtgz dude :/ i'll check this when i come back

Answer 14

ok thanks.. :)...

Answer 15

@imqwerty

Answer 16

@Loser66 any idea?

Answer 17

\(p\leq 5000\\\sqrt p\leq \sqrt{5000}\\\lfloor\sqrt{p}\rfloor \leq \lfloor \sqrt{5000}\rfloor = 70\)

Answer 18

So min p =4900
then you have p+125= 5025
floor function of them are equal

Answer 19

but p+125 may not lie in the same range as p

Answer 20

hey, the condition is p <= 5000, it doesn't say p+125<=5000

Answer 21

yeah...

Answer 22

now, max p:

Answer 23

if p =5000, then p+125 = 71.589... then floor of p \(\neq \)floor of p=125
Hence p cannot excess 5000, ok?

Answer 24

yes right

Answer 25

I am sorry for deleting it. You need a firmer logic. Give me few minutes to arrange it

Answer 26

No problem. :). Take your time

Answer 27

Now, let A be max p, then \(\sqrt{A + 125} <71\)
Hence A <4916,
that gives us A =4915
hence there are 15 numbers from 4900 to 4915 which satisfy the conditions

Answer 28

@Loser66 the answer is 72

Answer 29

@Loser66 I did a computer program, and it is showing 72 solutions, 3970-4916

Answer 30

@mathmate any idea?

Answer 31

Well, the situation boils down to
k=63 to 70, such that
for each k that you have, there are (k+1)^2-(k^2+125) values of p that satisfy the given condition. If you sum the values, that should add up to 72.
This is the main idea, and have not check out the details, such as boundary conditions (like where it starts, and where it stops). Please do some work to verify.

Answer 32

Any explanation? The answer is correct though

Answer 33

@mathmate

Answer 34

:)

Answer 35

As I mentioned, the algo has to be checked out, which I did, and found that an adjustment is required.
It should be the sum of
\(((k+1)^2-125) - k^2\) for k equal to 62 to 70, which still adds up to 72.

I will let you work out the details, if not already done.
If you still prefer to see my reasoning, you just have to ask, but I will give you some time to think about it.

Hint: you have done a brute force calculation.
Compare the results with this one and the reasoning will be obvious.

Answer 36

@mathmate n^2<p<5000 and so n<70.71 i.e. n varies from 63 to 70

Answer 37

Exactly, that's the upper limit. Did you figure out how the lower limit was obtained?

Answer 38

yeah, the numbers p+125 and p must lie between the squares of two consecutive integers

Answer 39

Exactly.
So everything is clear?

Answer 40

why is the the total of each such numbers for is (k+1)^2-(k+125) ??

Answer 41

@mathmate

Answer 42

Take k=63.
p=k^2=3969, [sqrt(3969)]=63
p+125=4094, [sqrt(4094)]=[63.98]=63, so 3969 is a candidate
we can go on until the next perfect square, so
for p=3970, [sqrt(3970)]=[sqrt(4095)=63, so 3970 is a candidate.
We know that 4096 is not a candidate because it is the next perfect square.

Hence the formula:
((k+1)^2-125)-k^2 is the number of such candidates, ex.
((63+1)^2-125)-63^2)=4096-125 - 3969=3971-3969=2

Note: you quoted the outdated (and incorrect) formula I posted. Please read my recent post with the new formula.