Question 168625: Find a six-digit number in which the first digit is two less than the second.
The second digit is two less than the third.
The third digit is two less than the fourth.
And the fourth digit is two less than the fifth.
The first and last digits are the same.
Please show the work, i would appreciate it!
Answer by Mathtut(3670)   (Show Source):
You can put this solution on YOUR website! lets call the six digits consecutively A,B,C,D,E,F....these must all be single digits
A=B-2
B=C-2
C=D-2
D=E-2
E=unknown
F=A
now plug in B,C,D values(one at a time), into 1st equation and you arrive at
A=E-8
now in order for A to be a single digit number E-8>0 or E>8 since 9 is the only single digit remaining above 8 then E=9
since E=9 it follows that D=7,C=5,B=3, and A=1, and since A=F F=1 also
so we have 1,3,5,7,9,1 as the six digit number.
:
I got to thinking that 0 is also a digit so the above E-8>0 should really be 
so E can be 8 or 9 ...we already have the answer if E is 9..If E is 8
it would be 0,2,4,6,8,0
so two possibilities in my opinion because I see nothing that narrows it down to one or the other.
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