Question

the square of a number has four digits one of which is 0 and the other three digits are even if the square of the number is less than 7000 find the number​

Answer 1

Answer:

the less than 7000

square number =80 square

Answer 2

If the square of the number is less than 7000 then the number​ is 78

Solution:

Given, the square of a number has four digits one of which is 0  and the other three digits are even and  the square of the number is less than 7000  we have to find the number

You don’t need to check many numbers; just the 19 even numbers between 46 and 82 inclusive. As square of number has 4 digits and is less than 7000

we can also skip checking 50, 60, 70 and 80 because we know the square will end in two zeros.  

You can also skip 56 to 62 and 72 to 76 because their squares start with an odd number.  

This leaves only nine reasonable candidates to check: 46, 48, 52, 54, 64, 66, 68, 78 and 82.

So now we get

$\begin{array}{l}{46^{2}=2116} \\ {48^{2}=2304} \\ {52^{2}=2704} \\ {54^{2}=2916} \\ {64^{2}=4096} \\ {66^{2}=4356} \\ {68^{2}=4624} \\ {78^{2}=6084} \\ {82^{2}=6724}\end{array}$

As we can see $78^2$ = 6084  is the only one with all even digits and one zero

Hence, the required number is 78.