# Math Help - number puzzle

1. ## number puzzle

If you multiply a 4 digit #, ABCD by 4, the digits get reversed. that is
ABCD X 4=DCBA find ABCD (#'s cant repeat) I found out that 2469 is the largest and 1023 is the lowest.

If you multiply a 4 digit #, ABCD by 4, the digits get reversed. that is
ABCD X 4=DCBA find ABCD (#'s cant repeat) I found out that 2469 is the largest and 1023 is the lowest.
what's the question?

3. The question: what is ABCD?

4. Originally Posted by MathGuru
what's the question?
the question was stated, the two numbers he gave form the possible range for the answer.

5. any number multiplied by 4 does not ever end in 1 so all of the numbers 1000-1999 are out of the question because in reverse they end in 1.

6. since it is in the 2000's range we know it has to end in 8 so that when multiplied by 4 there is a 2 in the ones coumn. (8*4=32)

7. and BA is divisible by 4, and as stated A is 2.

Therefore the possible combinations for BA are:

12
32
52
72
92

That should narrow down your search

8. There are only about 50 numbers between 2008 and 2468 that end in 8.

After checking them in an excel column you can find:

2178*4=8712

9. thx a lot!

10. I'm going to redo the entire thing for you and show you how to do it without using excel.

1. The range of the number is 1023<ABCD<2469

2. Make note that DCBA is divisible by 4

3. No number divisible by 4 ends in 1, therefore A cannot be 1, therefore A is 2.

4. D can be either 3 or 8 since both numbers multiply by 4 to a number that ends in 2, however in the number DCBA, we know that D has to be greater than or equal to 8 because 2 times 4 (4A) equals 8. Thus D is 8.

5. Because D has to be 8, it shows that 4B can't be greater than 10 because otherwise it would carryover and mess up the whole number. The only times that 4B is less than 10 is when B=1 or 2. As said in my previous post, combination BA must be divisible by 4 and end in 2. Therefore B must be 1.

6. Now use guess and check. But all you have to check is if C works when it's 3, 4, 5, 6, 7, or 9 (there is an algebraic way to find this but you have to ask yourself "is it really worth it")

The reasoning is simpler than you think . . .

If you multiply a 4-digit number ABCD by 4, the digits get reversed.
That is: ABCD X 4 = DCBA. .Find ABCD.

Number the columns for reference.
I will use ? for a possible "carry".
Also, → will mean "ends with" . . . as in: 4 x 7 → 8
Code:
      1 2 3 4

A B C D
x     4
- - - -
D C B A

In column-1, we have: 4 x A + ? = D
. . Hence, A = 1 or 2.

But if A = 1, column-4 says: 4 x D → 1 . . . which is impossible.
. . Therefore: A = 2

So we have:
Code:
      1 2 3 4

2 B C D
x     4
- - - -
D C B 2

In column-1, we have: 4 x 2 + ? = D
. . Hence: D = 8 or 9.

In column-4, we have: 4 x D → 2
. . Therefore: D = 8 (and there is no "carry" from column-2)

So we have:
Code:
      1 2 3 4

2 B C 8
x     4
- - - -
8 C B 2

If there is no "carry" from column-2, then B = 0 or 1.

If B = 0, then column-3 has: 4 x C + 3 → 0 ... which is impossible.
. . Therefore: B = 1

So we have:
Code:
      1 2 3 4

2 1 C 8
x     4
- - - -
8 C 1 2

In column-3, we have: 4 x C + 3 → 1
. . Hence: C = 2 or 7
Since 2 is already used, C = 7
Code:
    Solution

2 1 7 8
x     4
- - - -
8 7 1 2