Question

Another fun question from my 5th grade

Answer 1

Solutions 

The given question is a doable program.

Def pal():
    for a in range $(0,10):$
        for b in range$(0,10):$
            for c in range$(0,10):$
                for d in range$(0,10):$

   $c^1 =int(d*d)$ %10)

                   $y= (c*d + c^1)$ %10

                   $c^2=int((c*d+c^1)/10)$

                 $z =(b*d +c^2)$ %10

                    $c^3=int((b*d+c^2)/10)$

                    $w=(a*d+c^3)$ %10 

if (a!=b and a!=c and a!=d and b!=c and b!=d and c!=d and x==a and y==b and z==c and w==d ):
                       print("a = ", a)
                       print("b = ", b)
                       print("c = ", c)
                       print("d = ", d)

$a= 1 b= 0 c=8 d=9$ 

Plug in the numbers to solve the problem.

The 'int' function is the same thing as floor.
The % symbol is the modulo operator


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In base 10 expanded form $ABCD=A*1000+B*100+C*10+D$
and $DCBA=D*1000+C*100+B*10+A$ , where A,B,C,D are counting numbers. 

We are given $ABCD*D=DCBA$
Converting to decimal expanded forms we get 

$(A*1000+B*100+C*10+D)*D =D*1000+C*100+B*10+A$ 

Distributing the above equation

$AD*1000+BD*100+CD*10+D^2 =D*1000+C*100+B*10+A$

brute force solution works, but is best done by a computer not by hand. 

We could also try to use modular math. 

$d^2 mod 10 = a$ 

$cd+[d^2/10]mod10=b$ 

$bd+[(cd+[d^2/10])/10]mod10=c$ 

$ad+[(bd+[(cd+[d^2/10])]/10)/10]mod10=a))$ 

$1089 * 9 = 9801$

Answer 2

A can only be 1, because D*A=D, and only *1 can get to that.
So B is 0, if not then the thousands place might not be D
In DCBA, the one's place is 1
D squared is 1, so D is 9 (because if D=1, then ABCD*D=ABCD)
So C is 8 because it's the only one that's valid.
The final answer is A=1 B=0 C=8 D=9, and 1089*9=9801, so the statement hold true.