I am a four digit number. I am less than 4900 but more than 3600. I am a square number. I am a multiple of 5 and 25. What am I?

Question

To be answered by thursday 7th july 2016 (gb time)

Answer 1

4900 is 70^2 and 3600 is 60^2. So the square root of the number is between 60 and 70. The only integer between 60 and 70 ending in 5 (divisible by 5) is 65 and 65^2=4225. So the number must be 4225.

Answer 2

let be x the answer

x is a square than, we can write x = n^2 where n is an anteger

x is multiple of 25 than n is a multiple of 25 so we can write x = (25*p)^2 where p an integer

then x= 625*p^2

3600 <= x <= 4900 gives 5,76  <=  p^2 <= 7,84 then 2,4 <= p <= 2,8

no integer number verifies  2,4 <= p <= 2,8

so there are no integer that verifies the given conditions

Comments

If the number is a multiple of 25 then it will also be a multiple of 5, and 25 = 5^2.

The squares of all numbers ending in 5 always end with 25, and all numbers ending in 25 are divisible by 25. So all we need is a square between 3600 and 4900 ending in 25. Since 3600 and 4900 are the squares of 60 and 70, all we need is a number between 60 and 70 ending in 5, so there's only one: 65.

x=25p^2=n^2, so n=5p; 3600<25p^2<4900, 144<p^2<196, 12^2<p^2<14^2, so p=13 and n=65, x=4225.