Question

Another fun question from my 5th grade

A,B,C,D are DIFFERENT counting numbers (1,2,3,4,5,6,7...)

ABCD doens't mean A times B times C times D, it means like 1,234 if A=1,B=2,C=3,D=4

Find A, B,C and D such that
ABCD times D=DCBA

represented like this:
ABCD
        D X
DCBA

remember, A,B,C,D are all different

Find A,B,C,D       
show all work and logic

Answer 1

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.

Answer 2

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`