## It is Time….

Starting today, we have a series of 4 **clock puzzles**. The first three are easy to medium in difficulty, and can be considered a warm up for the fourth one. This last is extremely difficult and should satisfy any of our math wizards.

Here is the 1st:

At 12 o’clock, the hour and minute hands of the clock are superimposed. At exactly what time will they next be superimposed?

*The 2nd will go up early Thursday am. Big Thanks go out to Bilbao for helping prepare this challenge!*

Also, don’t forget to check out the recently posted Click-A-Flick, as well as the Find the Fault.

Migrated| Profile February 23rd, 2009 - 1:22 amAt 1:05 AM/PM depending on whether 12 o’clock refers to midnight or midday.

cheddarmelt| Profile February 23rd, 2009 - 6:38 am1:05:27.27

Benson| Profile February 23rd, 2009 - 8:12 amim pretty sure its about 1:05 or 1:06

joe| Profile February 23rd, 2009 - 9:09 amThe hands will overlap 11 times as they travel round the full face of the clock (12 including the 12 oclock position again).

A full turn of the face of the closk is 360 degrees so they will superimpose every ( 360 / 11 =) 32.727272degrees as they travel round.

We can calculate how many minutes this equates to on our face in order toknow the position of the hands. So:

min 32.727272

—- = ———

60 360

Solving for min = 5.454545

So at every 5.4545 minute mark of the clock.

After 12 oclock this will be at 1.054545 minutes

So 1:05 and 27.27secs

BillyPilgrim| Profile February 23rd, 2009 - 9:34 am1:08

BillyPilgrim| Profile February 23rd, 2009 - 9:47 amSorry, disregard my last comment, my formula was wrong. I believe the answer is 1:05.45, with the 45 in the decimal repeating. Or if you want seconds too, 1:05:27.27 with the 27 repeating.

SharonH| Profile February 23rd, 2009 - 10:28 am1:05?

Shawn| PUZZLE GRANDMASTER | Profile February 23rd, 2009 - 10:49 am1:05:27 plus 27/100 of a second.

It is obvious that the next occurrence is sometime after 1:05, at which point the minuted hand will be at the “1” and the hour hand will be slightly past the “1.” To make this less confusing, I structured the math beginning at 1:05, so that all calculations are relative to the number “1” on the clock face.

The minute hand travels from one hash-mark on the watch to the next in 1 minute or 60 seconds. The hour hand travels the same distance 5 times per 60 minute interval, or once every 12 minutes.

At 1:05, the minute hand is at the “zero” point relative to the number “1”, and the hour hand is at 5/12 (5 minutes times 1/12) the distance from “zero” to the first hash-line between the numbers 1 and 2.

The formula is:

[starting point + (speed per minute)*(minutes passed)] minute hand

= equals =

[starting point + (speed per minute)*(minutes passed)] hour hand

minute hand starting point = 0

minute hand speed per minute = 1 hash-mark/minute

hour hand starting point = 5/12

hour hand speed per minute = 1/12 hash-mark/minute

number of minutes elapsed = X

0 + (1)*(X) = 5/12 + (1/12)*(X)

X = 5/11 = .454545 minutes = 27.27 seconds

michaelc| Profile February 23rd, 2009 - 12:47 pmMinute hand will be 60/11 past or 5 5/11 and the hour hand will be 1/11 past.

Time will be about 1:05::27.

Can’t wait for the other 3!

Oneiric| Profile February 23rd, 2009 - 1:06 pmAt five past one. (Or five past one, 25 seconds and some milliseconds, counting the movement of the hour hand according to the minutes and seconds passed.)

alderney girl| Profile February 23rd, 2009 - 4:23 pmthe next time the two clock arms will be superimposed is 5 minutes past one 1:05

doar823| Profile February 23rd, 2009 - 6:11 pmJust a little after 1:05…if the minute hand is rotating once every minute and the hour hand is rotating 1/60 every minute, I think it will be at 1:05:01

Sue| Profile February 23rd, 2009 - 7:03 pm1:05

zenitth| Profile February 23rd, 2009 - 7:11 pmat 1:05?

Obiwan| Profile February 23rd, 2009 - 8:51 pm1:05 1 hr=1/12 of the dial, 1/12 of 60 minutes=5

aaronlau| Profile February 23rd, 2009 - 11:01 pmGuess it is how precise should the answer be.

1 o’clock and 5 minutes.

And 45.45 seconds?

bach| Profile February 23rd, 2009 - 11:40 pmjust registered, my first time answering here… not sure if i’m right.

a “tick” is 1/60th of the circumference of the circle.

the minute hand moves 1 tick every minute.

the hour hand moves 1 tick every 12 minutes.

then, we can say the pace of the minute hand is x and the pace of the hour hand is 1/12x when x is the number of minutes elapsed.

using common sense we know the two hands will meet sometime after 1:00, so we can represent the time 1:00 in terms of ticks as 0 for the minute hand and 5 for the hour hand. since we want to find the time at which the ticks on both hands are equal, we factor in the rates: x + 0= 1/12x + 5. once solved, x= 60/11, where x is the time elapsed after 1:00. thus, the answer is 01:05:27:45, hh:mm:ss:ms.

bach| Profile February 24th, 2009 - 12:29 amoooops i meant 01:05:27:27!

deadskin| Profile February 24th, 2009 - 10:18 amAngle(Hour)=(H/12)*360 + (M/60)*30

Angle(Minute) = (M/60)*360

(H/12)*360 + (M/60)*30 = (M/60)*360

30*H + (1/2)*M = 6*M

60*H + M = 12*M

60*H = 11*M

(60/11)*H = M

For each Hour (1, 2, 3, … calculate the minutes that need to pass for the minute hand to catch up to the hour hand.

HH:MM::SS

12:00::00am

01:05::27+(3/11)am

02:10::54+(6/11)am

03:16::21+(9/11)am

04:21::49+(1/11)am

05:27::16+(4/11)am

06:32::43+(7/11)am

07:38::10+(10/11)am

08:43::38+(2/11)am

09:49::05+(5/11)am

10:54::32+(8/11)am

12:00::00pm

01:05::27+(3/11)pm

02:10::54+(6/11)pm

03:16::21+(9/11)pm

04:21::49+(1/11)pm

05:27::16+(4/11)pm

06:32::43+(7/11)pm

07:38::10+(10/11)pm

08:43::38+(2/11)pm

09:49::05+(5/11)pm

10:54::32+(8/11)pm

deejam| Profile February 24th, 2009 - 12:00 pmAt 1:05 the hands superimposed after 12 0’clock…

lograh| Profile February 24th, 2009 - 2:45 pmDepends somewhat on the mechanics of the hands, if they move in a smooth motion or in steps — I’ll assume they have a smooth movement. Also, I’ll assume we are talking about a standard 12-hour clock face (I’ve seen some 24-hour clock faces where the hour hand takes 24 hours to go around).

All that assumed, I’d guess they are superimposed next at somewhere around the 13:05-13:06 minute. Not precisely at 13:05, as the hour hand would have moved 1/12 of the distance between the “1” and the “2”, but that’s a really small movement so it wouldn’t be so late as 13:07 when they line up. Probably roughly 13:05:10 when it happens. Maybe?

Even more interesting, how much longer before this kind of puzzle becomes undo-able for the majority of people since they have been raised on digital clocks?

fuzzy| Profile February 24th, 2009 - 5:02 pmAt 1:05:25

hex| PUZZLE MASTER | Profile February 24th, 2009 - 5:24 pmm: minutes

h: hours

The angular position of the hours hand = 2 x Pi / 12 x (h + m / 60)

The angular position of the minutes hand = 2 x Pi x m / 60

The hands superimpose when both angular positions are equal (actually separated by 2 x k x Pi where k is an integer, but the conditions on the hours and minutes dictate that k=0):

m = 60 / 11 x h

h=1 -> m=60/11 = ~5.45 minutes

so ~1:05

Hendy| Profile February 25th, 2009 - 3:10 amThe actual answer is beyond the resolution of the standard dial-face clock. It would be sometime between 1:05 and 1:06. I am sure there is a formula that works, but I only look up formulas, I don’t generate them any more. I gave up that practice in my first attempt at college. However, I solved it via a form of interpolation and stopped at somewhere between 12567/138240 and

12568/138240 where the units are fractions of a rotation about the circle of the face (clockwise of course). To convert these to times it is somewhere between 1:05.45442708333333

and 1:05.45486111111111.

Here’s the premise: for each minute about the clock-face, the hour hand has to turn 1/60th of an hour beyond the current hour. In other words when the clock shows 1:01 to human recognition, the actual time is in separate hour hand rotation and minute hand rotation. The hour hand will have moved 1/60th of an hour past one o’clock and the minute hand has moved 1 minute (also 1/5th or 12/60th of an hour mark) past 12 o’clock on the dial. If you take the greatest common factor of 720 (12 hours * 60 minutes in a trip around the clock with the hour hand) and 60 (minutes in a trip around the clock with the minute hand) you get 720. Therefore, 1:01 is now 61/720th rotations on the hour hand and 12/720th (or 1/60th) rotations on the minute hand. Additionally, 1:02 is now 62/720th hours and 24/720th (or 2/60th) minutes. To further illustrate, 1:30 is 90/720th hours (half way between 1 and 2 on the clock-face) and 360/720th (30/60th or one-half) minutes.

Or more simply, the hour hand still moves while the minute hand moves. It does not stop exactly at 1 and wait until after 1:59 to click to 2 (unlike a digital clock).

If the question is changed to include, “within the resolution of a minute mark on the face of the clock”, then the answer would be only 12:00 (twice a day). This is the only spot where the two superimposed clock hands land on an exact minute mark.

Roel| Profile February 25th, 2009 - 4:29 amI went physical on this one, i.e. I got my physics book from the shelves.

I assumed the hands of the clock have a constant angular velocity, say v_h for the hour hand and v_m for the minute hand. The formulas for the angles of the hands at a given time t are thus given by:

h = h_0 + v_h * t

m = m_0 + v_m * t

Assuming that the starting position (t=0) is 12 o’clock, this means that

h_0 = m_0 = PI/2 (90 degrees, pointing up)

The velocities of the hands are

v_h = – 2*PI / (12*60*60) (it goes 360 degrees in 12 hours)

v_m = – 2*PI / (60*60) (it goes 360 degrees in 1 hour)

At the moment that the hands cross again for the first time, there is a difference of 360 degrees (=2*PI) between the angles of the hands. So equating h = m + 2*PI gives:

PI/2 – 2*PI*t / (12*60*60) = PI/2 – 2*PI*t / (60*60) + 2*PI

Solving this equation to t gives:

t = 3927.27272727… seconds

Which means that after 1h (3600s) 5 minutes (300s) 27.2727s the two hands are at the same position.

Roel| Profile February 25th, 2009 - 10:56 amBTW, Dr. Bilbao, I solved this one in Bilbao! )

Mashplum| PUZZLE MASTER | Profile February 25th, 2009 - 9:01 pm1:05:27 (and 3/11 seconds)

ludiavolo| Profile February 25th, 2009 - 9:28 pmOkay,

the correct time will occur shortly after 1 o’clock so I’ll start my problem when the hour hand at 1 o’clock and the minute hand at 12 o’clock.

i thought of this as a race problem where the two hands are chasing each other around a circle. the hour hand has a 30deg head start because it’s starting at 1 o’clock.

the angular frequency of the hour hand is 360deg/12hr and the angular frequency of the minute hand is 360deg/1hr.

i need to set the angular positions of the two hands equal:

30deg + 360deg*t/12hr = 360deg*t/1hr

solving for t

t = 1/11 hr (this is how long AFTER 1 o’clock the hands will coincide)

doing some conversions, the time when the minute hand and hour hand will coincide is approximately: 1:05:27

RK| Founder | Profile February 25th, 2009 - 11:36 pmVery good, quite a few survived the 1st round Some old names I recognize have returned for the challenge, and several excellent explanations are above.

#2 will be going up in just a couple minutes

RK| Founder | Profile February 25th, 2009 - 11:36 pmWherefore art thou Suineg!?!

APEX.JP| Profile February 27th, 2009 - 3:14 pmThe hands will next be superimposed at exactly 1:05:27.27