Question

N is a four digit perfect square whose decimal digits are each less than 7. If each digit in N is increased by 3, then the new number is also a perfect square. What is the square root of N?

Answer 1

@ivanapen8 we c u

Answer 2

$$N=A\times10^3+B\times10^2+C\times10+D$$is a perfect square. We're given that \(A,B,C,D<7\). We're also given that:$$N'=(A+3)\times10^3+(B+3)\times10^3+(C+3)\times10+(D+3)=N+3333$$ is also a perfect square. Observe their difference:$$N'-N=3333$$is a difference of squares, so it follows that our right-hand side be factorable like so:$$(\sqrt{N'}+\sqrt{N})(\sqrt{N'}-\sqrt{N})=3333$$It boils down then to a problem of factoring \(3333\); it is obvious we can factor it like so:$$3333=3\times11\times101$$

Answer 3

Observe that for 4-digit \(N\) with digits less than \(7\), we have \(1000\le N\le6666\) hence we expect \(\sqrt{1000}\le\sqrt{N}\le\sqrt{6666}\). This tells us that \(32\le\sqrt{N}\le81\) hence we have a good starting point to look for potential factors.

Answer 4

We therefore expect \(4333\le N'\le9999\) hence \(66\le\sqrt{N'}\le99\) hence \(32\le\sqrt{N'}-\sqrt{N}\le34\).

\(3333\) gives the following possible factorizations into 2 integers:$$3333=1\times3333\\3333=3\times1111\\3333=11\times303\\3333=33\times101$$observe that we want the difference between these be our smaller factor hence we pick \(33\):$$\sqrt{N'}-\sqrt{N}=33$$Our greater factor is then:$$\sqrt{N'}+\sqrt{N}=101\\2\sqrt{N'}=134\\\sqrt{N'}=67\\N'=4489$$

Answer 5

To determine our original number, then, by deducting \(3333\):$$N=4489-3333=1156$$

Answer 6

Hi Tks for your answer, but the correct answer shows in myanswer paper as 34. but i dont know how is that?

Answer 7

@GPEE11 once we find \(N=1156\), it is trivial to determine its square root is just \(\sqrt{N}=34\)

Answer 8

Tks ...