A Mirror's Image When Multiplied By 4

Because A*4 <10 then A=1,2; but 4|DCBA so A = 2.

2 4(+ carried) = D -> D=8,9 but 4 D ends with 2 so D = 8, and there are no carry over.

So B*4 < 10, and A is already 2 so B = 1.

Then we have C*4 + 3 ends with 1, and A is already 2 so C = 7.

That we have ABCD = 2178

Well, we know that ABCD times 4 is still a 4 digit number, which means that ABCD times 4 is less than 10000 so ABCD must be less than 10000/4 = 2500. This means that A is either 1 or 2 (assuming no leading zeros are allowed) Also, since ABCD * 4 = DCBA, then 4 D has A as a units digit. This means that D can be 3 (4 3=12) or 8 (4 8 = 32) The point is, there's no number that, when multiplied by 4 has a units digit of 1 -----> A must be 2. This means that D must be 8 since 4 A = D (with no carry since result only has 4 digits) Since ABCD < 2500, then B can't be 5. B can be 0,1,2,3 or 4. Again 4 C + whatever is carried from 4 D has units digit = B Since we have 4 D =32, then the carry from 4 D is 3. So 4*C + 3 = XB (X is the tens digit, if any)) 4C=XB -3 If B were zero, then this means that 4C will have a units digit of 7. But no multiple of 4 has a units digit of 7 -----> B can't be 0 Similar arguments lead you to reject B = 2 and B=4 So either B=1 or B=3 which gives us the following possibilities:

D = 8 and B=1 -----> C= 7 ( C= 2 would've worked, but we already have A=2) D= 8 and B=3 ---> C= 0 or 5

So we have 3 possible numbers: 2178, 2308 or 2358 Multiplying each of them by 4, the only one that works is the first: 2178 * 4 = 8712

A = 2, B = 1, C = 7 and D = 8

1.Lets say the number is written as ABCD (1000a+100b+10c+d).
2.Multiplying it by 4, gives you DCBA (1000d+100c+10b+a)
3.By divisibility of 4, BA is a two-digit number divisible by 4.
4.Now, the largest possible value of ABCD is 2499, because anything more will result in a five digit number when multiplied by 4.
5.Out of the two possibilites 1 and 2 for A, [3] follows that A is even, so 2.
6.From [5], D is either 8 or 9. However, since D times 4 would end in 2, D is 8.
7.Using [2], 4000a+400b+40c+4d = 1000d+100c+10b+a, which would simplify to 2c-13b = 1 after substituting A = 2 and D = 8.
8.The only possible value of (b,c) where they would both be single digit integers is (1,7). Also to note from [3], B is odd. And from [4], B is either 1 or 3.

The answer you're looking for is 2178, multiplied by 4 gives 8712

ABCD

x 4

DCBA

this is trial and error method since this is multiplication, ABCD is greater than DCBA, then D>A and let T = tens digit 4xD = TA 4x1 = 4, but 1 is not > 4 4x2 = 8, but 2 is not > 8 4x3 = 12, possible since 3>2 4x4 = 16, not possible 4x5 = 20, not possible because A cannot be = 0, since it is the first digit 4x6 = 24, possible 4x7 = 28, not possible 4x8 = 32, possible 4x9 = 36, possible

lets add that 4A >= D so now we can say that D=8 and A=2

2BC8

x 4

8CB2

next is B&C, T = tens digit of the previous part and R = tens digit T=3 since 8x4=32 4xC+3=RB and K=ten digit 4xB+R=KC since 4A=D, then K=0 and 4B+R=C

4C+3=RB--(1) 4B+R=C---(2) we can say that 4B+R<=9, 0<=C<=9 and 0<=R<=9. 4B<10, B is might be 1 or 2 only.

lets try B=1: 4C+3=10R+1 C=(10R-2)/4--(3) (3)-->(2) 4+R=(10R-2)/4 16+4R=10R-2 R=3 R-->(3) C=7

2178

x 4

8712

you can try B=2, but R & C will be decimals and what we are looking for is whole numbers

so A=2,B=1,C=7,D=8

Similarly 1089 X 9 = 9801 Are there any other four digit numbers multiplied by a non-zero single digit has this characteristic? Can this be extended to 5 digits or more?