## Special Six-Digit

There exists a special six-digit number such that when this number is multiplied by four, its digits are reversed.  Determine this special six-digit number.  Note:  digits can be used more than once.
Source: mathcontest.olemiss.edu 1/25/2010

SOLUTION
Let $A$ be the special six-digit number and $B$ be the reversed digit number. We make the following observations about $A$.

1. The most significant digit of A can only be {1, 2}, because if it is greater than or equal to 3, the multiplication by 4 will make A a seven-digit number.

2. The most significant digit of A cannot be 1 because there is no way to produce the digit 1 by a multiplication by 4. Thus, the most significant digit of A must be 2.

3. If $A=2abcde$, then the most significant digit of $B$ is 8 because $4\times 2=8$. Thus, $e=8 \textup{ and}A=2abcd8$ .

4. Let $a=0$. Then, $A=20bcd8$. Since $4\times 8=32$, we have a “carry” of 3. In order to get a zero digit out of a multiplication by 4 we have three possibilities:
$7+3=10$
$17+3=20$
$27+3=30$
But, it is imposssible to get 7, 17, or 27 out of a multiplication by 4.

5. Let $a=1$. Then, $A=21bcd8$. We have two choices for digit $d$:
$4\times d+3=4\times 2+3=11$
$4\times d+3=4\times 7+3=31$
We have two cases to consider: $d=2 \textup{ and }d=7$.

6.1 First case: $d=2,A=21bc28$.
6.1.0 Let $b=0$. Then, $A=210c28$. This time, the carry is equal to 1. In order to get a zero digit out of a multiplication by 4 we have three possibilities:
$9+1=10$
$19+1=20$
$29+1=30$
But, it is impossible to get 9, 19, or 29 out of a multiplication by 4.

6.1.1 Let $b=1$. Then $A=211c28$. There is only one choice for digit $c$:
$4\times c+1=4\times 5+1=21$
Check
$4\times 211528=846112$. Not an answer.

6.1.2 Let $b=2$. Then $A=212c28$. In order to get a 2 digit out of a multiplication by 4 we have three possibilities:
$11+1=12$
$21+1=22$
$31+1=32$
But, it is impossible to obtain 11, 21, or 31 out of a multiplication by 4.

6.1.3 Let $b=3$. Then, $A=213c28$. There are two choices for digit $c$:
$4\times c+1=4\times 3+1=13$
$4\times c+1=4\times 8+1=33$
Check
$4\times 213328=853312$. Not an answer.
$4\times 213828=855312$. Not an answer.

6.1.4 Let $b=4$. Then, $A=214c28$. In order to get a 4 digit out of a multiplication by 4 we have three possibilities:
$13+1=14$
$23+1=24$
$33+1=34$
But, it is impossible to get 13, 23, or 33 out of a multiplication by 4.

6.1.5 Let $b=5$. Then, $A=215c28$. There are two choices for digit $c$:
$4\times c+1=4\times 1+1=5$
$4\times c+1=4\times 6+1=25$
Check
$4\times 215128=860512$. Not an answer.
$4\times 215628=862512$. Not an answer.

6.1.6 Let $b=6$. Then, $A=216c28$. In order to get a 6 digit out of a multiplication by 4 we have four possibilities:
$5+1=6$
$15+1=16$
$25+1=26$
$35+1=36$
But, it is impossible to get 5, 15, 25, or 35 out of a multiplication by 4.

6.1.7 Let $b=7$. Then, $A=217c28$. There are two choices for digit $c$:
$4\times c+1=4\times 4+1=17$
$4\times c+1=4\times 9+1=37$
Check
$4\times 217428=869712$. Not an answer.
$4\times 217928=871712$. Not an answer.

6.1.8 Let $b=8$. Then, $A=218c28$. In order to get a digit 8 out of a multiplication by 4 we have three possibilities:
$7+1=8$
$17+1=18$
$27+1=28$
But, it is impossible to get 7, 17, or 27 out of a multiplication by 4.

6.1.9 Let $b=9$. Then, $A=219c28$. There are two choices for digit $c$:
$4\times c+1=4\times 2+1=9$
$4\times c+1=4\times 7+1=29$
Check
$4\times 219228=876912$. Not an answer.
$4\times 219728=878912$. Not an answer.

6.2 Second case: $d=7,A=21bc78$

6.2.0 Let $b=0$. Then, $A=210c78$. This time, the carry is equal to 3. In order to get a zero digit out of a multiplication by 4 we have three possibilities:
$7+3=10$
$17+3=20$
$27+3=30$
But, it is impossible to get 7, 17, or 27 out of a multiplication by 4.

6.2.1 Let $b=1$. Then, $A=211c78$. There are two choices for digit $c$:
$4\times c+3=4\times 2+3=11$
$4\times c+3=4\times 7+3=31$
Check
$4\times 211278=845112$. Not an answer.
$4\times 211778=847112$. Not an answer.

6.2.2 Let $b=2$. Then, $A=212c78$. In order to get a 2 digit out of a multiplication by 4 we have three possibilities:
$9+3=12$
$19+3=22$
$29+3=32$
But, it is impossible to get 9, 19, or 29 out of a multiplication by 4.

6.2.3 Let $b=3$. Then, $A=213c78$. There is only one choice for digit $c$:
$4\times c+3=4\times 5+3=23$
Check
$4\times 213578=854312$. Not an answer.

6.2.4 Let $b=4$. Then, $A=214c78$. In order to get a 4 digit out of a multiplication by 4 we have four possibilities:
$11+3=14$
$21+3=24$
$31+3=34$
But, it is impossible to get 11, 21, or 31 out of a multiplication by 4.

6.2.5 Let $b=5$. Then, $A=215c78$. There are two choices for digit $c$:
$4\times c+3=4\times 3+3=15$
$4\times c+3=4\times 8+3=35$
Check
$4\times 215378=861512$. Not an answer.
$4\times 215878=863512$. Not an answer.

6.2.6 Let $b=6$. Then, $A=216c78$. In order to get a 6 digit out of a multiplication by 4 we have three possibilities:
$13+3=16$
$23+3=26$
$33+3=36$
But, it is impossible to get 13, 23, or 33 out of a multiplication by 4.

6.2.7 Let $b=7$. Then, $A=217c78$. There are two choices for digit $c$:
$4\times c+3=4\times 1+3=7$
$4\times c+3=4\times 6+3=27$
Check
$4\times 217178=868712$. Not an answer.
$4\times 217678=870712$. Not an answer.

6.2.8 Let $b=8$. Then, $A=218c78$. In order to get an 8 digit out of a multiplication by 4 we have four possibilities:
$5+3=8$
$15+3=18$
$25+3=28$
$35+3=38$
But, it is impossible to get 5, 15, 25, or 35 out of a multiplication by 4.

6.2.9 Let $b=9$. Then, $A=219c78$. There are two choices for digit $c$:
$4\times c+3=4\times 4+3=19$
$4\times c+3=4\times 9+3=39$
Check
$4\times 219478=877912$. Not an answer.
$4\times 219978=879912$. YES! Finally!