is there a number which is half (or double) of itself when written backward?

Question

backwards like 21 and 12, or 2004 and 4002.

Answer 1

There may or may not be. Let's see.

First, we need to decide on the the number of digits of our number to solve this equation.

Let's start with 2 digits, then we can do it for 3, 4, 5, ... digits.

(Apparently, there is no such number with a one digit number, because no number is the half of itself.)

So we start our search with 2 digits. Let's assume our number is XY. What we want is:

10X + Y = 2(10Y + X)

where X and Y are integers between 1 and 9. (With 2 digits we cannot have 0 included, but with more digits we will be able to add 0 selectively to some digits.)

Solving for X and Y:

8X = 19Y

or

X = 2.375Y

Since X is an integer between 1 and 9, for any value you pick for X, Y will be a non integer. But Y also needs to be an integer. Therefore, we do not have such a number with 2 digits.

OK, so we do not have such a number with 2 digits. How about 3 digits? Let's look at 3 digits now. Let's assume our number is XYZ.

100X + 10Y + Z = 2(100Z + 10Y + X)

where X and Z are integers between 1 and 9, and
where Y is an integer between 0 and 9

Solving for the equation, we get

98X = 10Y + 199Z

Let's start with Y and try each one of the 10 alternatives (0 to 9) one by one.

Now, let's start with 1 for Y (I will handle 0 later).

If Y was a 1, in that case, when XYZ was multiplied by 2, we would either have a 2 or a 3 for Y in the ending number - depending the value of Z. If Z was between 0 to 4, the ending Y would be a 2, contradicting our initial assumption that Y was a 1. If Z was between 5 to 9, Y would a 3, again contradicting our initial assumption. Therefore Y cannot be a 1.

The argument above also holds for Y being 2 to 8. Y would be contradicting its initial assumed value. I put a small table below for shorthand:

Initial Y = 1 ==> Ending Y = 2 or 3 (depending on Z)
Initial Y = 2 ==> Ending Y = 4 or 5 (depending on Z)
Initial Y = 3 ==> Ending Y = 6 or 7 (depending on Z)
Initial Y = 4 ==> Ending Y = 8 or 9 (depending on Z)
Initial Y = 5 ==> Ending Y = 0 or 1 (depending on Z)
Initial Y = 6 ==> Ending Y = 2 or 3 (depending on Z)
Initial Y = 7 ==> Ending Y = 4 or 5 (depending on Z)
Initial Y = 8 ==> Ending Y = 6 or 7 (depending on Z)

But with initial Y = 9, we have an interesting situation, because there is a possibility for not contradicting our initial assumption:

Initial Y = 9 ==> Ending Y = 9 or 0 (depending on Z)

If Z = 1 to 4, both initial and ending Y's are 9.

Please remember Z cannot be a 0.

So far we have these candidates that may satisfy your requirements:

X91
X92
X93
X94

and the requirements are:

X91 = 2(19X)
X92 = 2(29X)
X93 = 2(39X)
X94 = 2(49X)

Solving for:

100X + 90+ 1 = 200 + 180 + 2X (and the other three similarly constructable formulas) we see that all values for X are non-integers, meaning there is no X if Y is a 9.

Now, our last hope for 3 digit numbers, initial Y = 0. So far we have exhausted all digits for Y between 1 and 9. If we find that Y cannot be a 0 either, then we have to conclude that there is no available integer for Y. This will also mean that we cannot have a number that is the twice value of itself when written backwards.

Initial Y = 0 ==> Ending Y = 0 or 1 (depending on Z)

Similar to the argument we made above, if Z = 5 to 9, initial Y will be contradicting ending Y. So Z has to be between 1 to 4, and Z cannot be a 0.

So our options are:

X01
X02
X03
X04

and the requirements are:

X01 = 2(10X)
X02 = 2(20X)
X03 = 2(30X)
X04 = 2(40X)

Solving for:

100X + 0 + 1 = 200 + 0 + 2X (and the other three similarly constructable formulas) we see that all values for X are non-integers, meaning there is no X if Y is a 0.

Sooooo ... We do not have any possible value for Y for 3 digit numbers either. Meaning, we cannot have such a number that has 3 digits.

What do we do now? Do we do the same analysis with 4 digits, 5 digits, 6 digits, etc.?

Luckily, our analysis with 3 digits has revealed us something: the digit in the middle stays the same. So, we can easily convert the approach above with 3 digits to 5, 7, 9, 11, ... digits.

Meaning, our number cannot have an odd number of digits.

Now ...

The only thing that is left is to analyze whether our number can have an even number of digits.

But I have already written too long, and I have to leave now.

Hope to continue later.

Answer 2

Let the number be ABC.....Z. The addition looks like this:

ABC....Z

ABC....Z
_________

ZYX.....A

Here A must be even. On the left we see A<5 or else A+A

would give another digit to the left of Z. So, A = 2 or 4.

Then, Z = 4 or 8 or else 5 or 9 if there's a carry (at most 1)

on the left. But on the right Z + Z has to end in A = 2 or 4

but it ends in 0, 6, or 8 because Z = 4 or 8 or 5 or 9.

Therefore, there is no such number.

Answer 3

let us assume that there exists a two digit number with this property,let the digits be x,y,then 10x+y=1/2(10y+x)

on solving this relation we get 19x=8y
the only possible solutions for this equation are x=8,y=19
but y=19 is not possible because y is a digit.therefore there exists no two digit number with this property.
now let us check for three digit number.
then,100x+10y+z=1/2(100z+10y+x)
on solving this relation we get
199x=98z-10y
199x=2(49z-5y)
there is a factor of 2 in the right hand side but 199 is not a factor of2,which implies x=2
the next equation takes shape as
199=49z-5y
if y is odd z should be even, coz even-odd=odd
and if y is even z should be odd " " ' " "
therefore knowing that such a number can exist only in two hundred (as x=2) on checking all the values from 200 to 299 we don't get such number, with the same logic we can prove that we won't get any more numbers of that kind in four digited and more than four digited numbers.hence the only possible solution is 0 and not 1 coz on reversing i we get 10 which is not a possible solution

Answer 4

i don't think so because at the number reverses, the larger number (is in 4002) will reverse to 2004, making the last number larger than the in the original number - 4 in 2004 is larger than 2 in 2004, so to have one large number have an inverse that is half of it is impossible because to make one number larger would be to make the smaller number's last number digit than the larger number's last digit.

Answer 5

1 (one) and 0 (zero)

Answer 6

1 double 1 = 1
Same backwards

Edit-n Why did i get 2 bad thumbs down, my answer is right :(