A 24-hour digital clock occasionally shows times that are palindromes such as 0:34:30, 21:33:12, and 1:20:21, if we ignore the colons.
What is the smallest interval between two such times?
Clarification: If the hour and/or minute and/or second have leading zeroes, only the leading zero for the hour is dropped. For example, one second after midnight would be displayed as 0:00:01.
Attribution: momath.org
The smallest possible interval is
just over 1 second.
The other answers correctly note that the nearest palindromic times are
9:59:59 and 10:00:01
But they incorrectly conclude that
2 seconds must elapse between them. This is incorrect. If we look at the clock in the last possible moment that it reads 9:59:59, we just have to wait 1 full second while the clock reads 10:00:00 before it reads 10:00:01. It is possible to see two palindromic times on the clock while looking at it for just over 1 second (although you couldn't actually measure this sub-second interval with the clock you're looking at).
The smallest interval I can find is
2 seconds long:
9:59:59 - 10:00:01
My reasoning:
Whenever we change the seconds we also have to change the hours. That would result in a big interval.
So I thought it must have to do with some kind of overflow. I first thought about the overflow around midnight. 0:00:00 looked promising but the closest palindromic time before midnight would be 23:55:32.
Next thing I thought about was an overflow from ?:??:59 to ?:??:00. The only possible time ending in 59 is 9:59:59. Adding 1 second gives 10:00:00 which is no palindrome. Adding one more second gives the palindrome 10:00:01. Which is the smallest possible interval.
It's:
$2$ secs.
Why?
If a time can be written as $a:bc:ba$ ($c<9$) and $c$ increases by $1$, it'll still be a palindrome.
Can the interval be shorter than a minute?
For $a:bc:ba + 0:xy$, if the addition of $xy$ seconds changes the hour to $A=a+1$, then we have $a:59:5a$, which turns into $A:00:0A$, taking $11$ seconds. If it takes even fewer seconds, the last digit of the second will change, also altering the very first digit of the time. We can do so by switching to a different number of digits, like $9:59:59$ to $10:00:01$ ($2$ seconds).
$1$ second would also alter the first and last digits, requiring _$:59:59$, but then only $9$ could be the hour without a leading zero, and $10:00:00$ wouldn't be palindromic.
Another way:
If it's $a:bc:ba$, it corresponds to $3601a+610b+60c$ seconds, with $b<6$, meaning we can add or subtract at most $5$ to/from $b$ when we take another $b$. If we're working out the difference between two such numbers, even if you try to compensate for a change in $a$ by $1$ ($+-3601$) by pushing the other digits to the limit - changing $b$ by $5$ ($+-3050$) and $c$ by $9$ ($+-540$), the difference will be $3601-3590=11$ seconds. Only changing $b$ and $c$ will result in $610-540=70$ seconds, and only changing $c$ by $1$ in $60$.
If it's $xy:zz:yx$, it corresponds to $36001x+3610y+660z$ seconds, with $x<3$, $y,z<6$ meaning we can add or subtract at most $5$ to/from $y$ or $z$ when we take another, and $2$ to/from $x$. If we're working out the difference between two such numbers, the best we can get is $3610-3300=310$ seconds.
What if we switch between $3601a+610b+60c$ and $36001x+3610y+660z$?
Assuming the difference is less than $11$, if we subtract $10b+a$ from the former and $10y+x$ from the latter, the difference between them can be $5\times10+9+10=69$ at most. However, both would then be divisible by $60$, so their division by that ($60a+10b+c$ vs. $600x+60y+11z$) will give at most a difference of $1$.
If equal:
$c=z$ ($mod$ $10$)
$10b+c=11z$ ($mod$ $60$) => $10b+c=11c$ => $b=c=z$
$60a+11b = 600x+60y+11b$ => $a=10x+y$ => $x=0$ (contradiction).
If unequal:
$c-z=+-1$ ($mod$ $10$)
$10b+c-11z=+-1$ ($mod$ $60$)
$x=1$, $a=9$, $b=5$, $c=9$, $y=z=0$ ($9:59:59$ to $10:00:01$)
Pedantic answer:
Zero
because you didn't say
two different times
