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!

 Answer: 219978.

 

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About mvtrinh

Retired high school math teacher.
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