Just Found Out About This Urgent Question

[coatsy]

This is due very soon and I didn't know about it until 10 minutes ago.


2) Find all 4 digit numbers that are complete squares and have the first 2 digits equal and the last 2 as well. eg. 1122.


Please, please help. It's not that I'm bad at maths, I'm just very short of time. Working and reasons would be appreciated too.

Thanx.

 

[Deleted member 4993]

Please show us your work - so that we can help you properly.

How many ways can you fill the thousndth digit?

After tht how many ways can you select the hundreth digit?

and continue....

If you expect that we will do your omework - without you lifiting your arm - I think you'll be disappointed?

 

[Denis]

x^2 = 1000a + 100a + 10b + b
x^2 = 1100a + 11b
x^2 = 11(100a + b)
x^2 / 11 = 100a + b

What does that tell you?
You'll get only 1 solution!

 

[Deleted member 4993]

Since the urgency has subsided - I'll present the full solution to this problem

Denis has shown that the number must satisfy the following equation:

x^2 = 11*(100*a + b)

So (100a +b) is a three digit number whose middle digit is '0' (like 201, 307, 704, etc.)

It is also divisble by 11 and the quotient after dividing by 11 must be a "two digit square number"

two digit square nubers are

16, 25, 36, 49, 64, 81

multiply those by 11 and you get

176, 257, 396, 539, 704, 891

of these only 704 fits the bill for (100a+b)

so the number is = 11*704 = 7744