A 24-hour digital clock shows the time using four digits: two for the hour and two for the minutes.
For example, 09:45
or 17:32
.
Some times read the same forwards and backwards, like a palindrome.
For example, 12:21
reads the same forwards and backwards.
How many palindromic times are there in a 24-hour day?
Can you use the digits on the left of this clock along with any mathematical operations to equal the digits on the right (also with any mathematical operations)?
Which times is this possible and which times impossible?
Use the digits of the current time to make as many different calculations as possible all with different answers.
All four digits must be used and each must appear once only in each of your calculations.
Next choose a different time that you think will produce more calculations than the present time.
Topics: Starter | Arithmetic | Mixed | Number | Problem Solving | Puzzles
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The time is written in the form HH:MM
. For it to be palindromic, the digits must satisfy:
H1H2:M1M2 = M2M1:H2H1
That gives the following valid palindromic times:
Total: 16 palindromic times
09:01 | 0x9=0 | 0x1=0 |
09:02 | 0x9=0 | 0x2=0 |
09:03 | 0x9=0 | 0x3=0 |
09:04 | 0x9=0 | 0x4=0 |
09:05 | 0x9=0 | 0x5=0 |
09:06 | 0x9=0 | 0x6=0 |
09:07 | 0x9=0 | 0x7=0 |
09:08 | 0x9=0 | 0x8=0 |
09:09 | 90 | 90 |
09:10 | 0x9=0 | 1x0=0 |
09:11 | 90=1 | 11=1 |
09:12 | 90=1 | 12=1 |
09:13 | 90=1 | 13=1 |
09:14 | 90=1 | 14=1 |
09:15 | 90=1 | 15=1 |
09:16 | 90=1 | 16=1 |
09:17 | 90=1 | 17=1 |
09:18 | 90=1 | 18=1 |
09:19 | 90=1 | 19=1 |
09:20 | 0x9=0 | 0x2=0 |
09:21 | √9+0=3 | 2+1=3 |
09:22 | 90=1 | 2÷2=1 |
09:23 | 0+9=9 | 32=9 |
09:24 | √9+0=3, 3!=6 | 2+4=6 |
09:25 | √9+0=3 | 5-2=3 |
09:26 | √9+0=3 | 6÷2=3 |
09:27 | 0+9=9 | 2+7=9 |
09:28 | 90=1 | 82=64, 6+4=10, 1+0=1 |
09:29 | 0+9=9 | 92=81, 8+1=9 |
09:30 | 0x9=0 | 0x3=0 |
For more ideas see Alan Sturgess' Maths Puzzle Investigation video.
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The digital clock used on this page is adapted from code provided by Radoslav Dimov copyright © 2009.