Find the four digit numbers #abcd# that satisfy, #2(abcd)+1000=dcba#?

2 Answers
Oct 4, 2016

Answer:

See below.

Explanation:

Number #abcd# can be represented as

#n_1 = a x^3+b x^2+ c x + d# also

#n_2 = d x^3 + c x^2 + b x + a#

and

#n_0 = 1000 = 1 x^3+0 x^2+0 x+0# so

#2 n_1+n_0 =n_2# requires

#{(2 d-a=0), (2 c-b=0), (2 b - c=0), (1 + 2 a - d=0):}#

and this cannot be accomplished with #a,b,c,d# integers.

Oct 4, 2016

Answer:

#abcd=2996#

Explanation:

write:
#[[a, b, c,d],[a, b, c, d], [1,0,0,0],["-","-", "-", "-"],[d,c,b,a]] #
this implies that: #2a+1=d#
We also know that:
#2a+1 <= d<= 9# why? d us a digit number...
also notice from the last expression #2d=a# that is #a# is even.
This limits a to be #{2,4}#
So now let's try if #a# can be 2 or 4:
Pick #a=4# then
#d<= 2a+1=9# this means the last digit of #2d=8# a contradiction.
If we pick #a=2# then #d<=2a+1=5#
The last digit of #2d=>2# thus #d=6#
#d+d=6+6# which will write 2 carry #1#. Now with #a=2 and b=6# the carry over the remaining equation is:
#[[" ",b,c],[" ",b,c],[" "," ",1],["-","-","-"],[1,c,b]]#

We have 2 scenarios:

Scenario :
#2c + 1 = b and 2b = 10 + c#, unfortunately no integer solution

Scenario:
#2c + 1 = 10 + b and 2b + 1 = 10 + c#, this yields
#b = c = 9#.

Therefore the answer is: #abcd=2996#