Question

find a four digit square whose first two and last two are equal​

Answer 1

Step-by-step explanation:

Let N=abcd , m=ab , and n=cd be a perfect squares. Since N has four digits, we will assume a>0 . Therefore m∈{42,…,92} .

If n=00 , then m can take any value from the set {42,…,92} , resulting in N taking corresponding values from the set {402,…,902} .

Now assume n≠00 .

If m≥52 , then N=(10m+ℓ)2=100m2+20mℓ+ℓ2 , with m≥5 and ℓ≥1 . But now we have the contradiction

m=⌊N100⌋>m .

We conclude that m=42 and N=(40+ℓ)2=1600+80ℓ+ℓ2 . Since m=⌊N100⌋ , the first two digits of N are 16 . Hence ℓ=1 , and N=412=1681 . ■

Answer 2

Answer:

7744

Step-by-step explanation:

hope it helps,

Mark me as brainliest if it did.d