MHB Find Integer $n$: Reversed Digits = 999 - Prime < 6000

[lfdahl]

Determine the integer $n$ with the properties:

a). $n$ is a prime less than $6000$,
b). the number formed by the last two digits of $n$ is $< 10$, and
c). if the decimal digits of $n$ are reversed to obtain $N$, then $N − n = 999$.

 

[Greg]

Spoiler

2003

 

[lfdahl]

Spoiler

2003

Thankyou very much, greg1313, for a correct answer!(Yes)

Spoiler

Would you perhaps like to show, how you came up with $2003$? (Talking)

 

[I like Serena]

Determine the integer $n$ with the properties:

a). $n$ is a prime less than $6000$,
b). the number formed by the last two digits of $n$ is $< 10$, and
c). if the decimal digits of $n$ are reversed to obtain $N$, then $N − n = 999$.

Spoiler

Let $n=\overline{abcd}$.
Then due to (a) we have a=0-5 and d=1,3,7,9.
And due to (b) we have c=0.
From (c) we get that $\overline{ab0d} + 999 = \overline{d0ba}$, so that:
- $d+9\to a$, implying that $d=a+1$, and thus $a=0,2$.
- $0+9+1\to b$, implying $b=0$.

0001 is not a prime and doesn't actually satisfy (c), so that only leaves 2003, which is the answer.

 

[Greg]

Thankyou very much, greg1313, for a correct answer!(Yes)

Spoiler

Would you perhaps like to show, how you came up with $2003$? (Talking)

[sp]From c), n is four digits, From a), the last digit of n is 3. The solution follows.[/sp]

 

[lfdahl]

Spoiler

Let $n=\overline{abcd}$.
Then due to (a) we have a=0-5 and d=1,3,7,9.
And due to (b) we have c=0.
From (c) we get that $\overline{ab0d} + 999 = \overline{d0ba}$, so that:
- $d+9\to a$, implying that $d=a+1$, and thus $a=0,2$.
- $0+9+1\to b$, implying $b=0$.

0001 is not a prime and doesn't actually satisfy (c), so that only leaves 2003, which is the answer.

Thankyou, I like Serena! - for your participation and nice solution!

P.S.:

Spoiler

- isn´t condition (c) satisfied by $0001$? $1000 - 0001 = 999$?

 

[I like Serena]

Thankyou, I like Serena! - for your participation and nice solution!

P.S.:

Spoiler

- isn´t condition (c) satisfied by $0001$? $1000 - 0001 = 999$?

Spoiler

- isn´t condition (c) satisfied by $0001$? $1000 - 0001 = 999$?

Isn't the number actually $1$ instead of $0001$?
Then, if we reverse its digits and subtract, don't we get $1-1=0\ne 999$?

After all, if we'd reverse and subtract $02003$, we'd get $30020 - 02003 = 28017 \ne 999$.
So it seems to me that it can't be the intention to have leading zeroes, can it?

 

[lfdahl]

Spoiler

Isn't the number actually $1$ instead of $0001$?
Then, if we reverse its digits and subtract, don't we get $1-1=0\ne 999$?

After all, if we'd reverse and subtract $02003$, we'd get $30020 - 02003 = 28017 \ne 999$.
So it seems to me that it can't be the intention to have leading zeroes, can it?

Thanks, I like Serena, I see your point.(Yes)