At a movie theatre, the manager announces that they will give a free ticket to the first person in line whose birthday is the same as someone who has already bought a ticket. You have the option of getting in line at any time. Assuming that you don't know anyone else's birthday and that birthdays are distributed randomly throughout the year, what position in line gives you the greatest chance of being the first duplicate birthday?

[Ref: ZBMH]

Hint: This is a very difficult puzzle.

Answer: 20th.

The solution requires quite a lot of knowledge of probability and is beyond the scope of a simple solution.

View the Wikipedia article on the Birthday Paradox for more information.

Puzzle 1487

What is the test to see if a number is divisible by nine?

[Ref: ZFJD]

Hint: What could we do with the digits?

Answer: If the sum of the digits is divisible by nine, so is the number.

Add up all of the digits in the number and see if the sum is divisible by 9. If you still can't tell, you can add those digits again to see if the new sum is divisible by 9. You can keep going until you the sum is obviously divisible by 9 or not.

For example, is 486451464 divisible by 9?

Do 4 + 8 + 6 + 4 + 5 + 1 + 4 + 6 + 4 = 42.

Is 42 divisible by 9? Not sure, you can then do:

4 + 2 = 6. Which clearly isn't divisible by 9. So our original number, 486451464, isn't either.

Puzzle 1488

Kevin is 5 years old, James is also 5 and Rebecca is 7. How old is Jacqueline?

Do the 3 hands on a clock ever divide the face of the clock into 3 equal segments, i.e. 120 degrees between each hand?

[Ref: ZJGA]

Hint:

Answer: No. The full solution is a little complicated...

First let us assume that our clock has 60 divisions. We will show that any time the hour hand and the minute hand are 20 divisions (120 degrees) apart, the second hand cannot be an integral number of divisions from the other hands, unless it is straight up (on the minute).

Let us use h for hours, m for minutes, and s for seconds. We will use =n to mean congruent mod n, therefore 12 =5 7.

We know that m =60 12h, that is, the minute hand moves 12 times as fast as the hour hand, and wraps around at 60. We also have s =60 60m. This simplifies to s/60 =1 m, which goes to s/60 = frac(m) (assuming s is in the range 0 <= s < 60), which goes to s =60 frac(m). So, if m is 5.5, s is 30.

Now let us assume the minute hand is 20 divisions ahead of the hour hand. So m =60 h + 20, so 12h =60 h + 20, 11h =60 20, and, finally, h =60/11 20/11 (read 'h is congruent mod 60/11 to 20/11'). So all values of m are k + n/11 for some integral k and integral n, 0 <= n < 11. s is therefore 60n/11. If s is to be an integral number of units from m and h, we must have 60n =11 n. But 60 and 11 are relatively prime, so this holds only for n = 0. But if n = 0, m is integral, so s is 0.

Now assume, instead, that the minute hand is 20 divisions behind the hour hand. So m =60 h - 20, 12h =60 h - 20, 11h =60 -20, h =60/11 -20/11. So m is still k + n/11. Therefore s must be 0.

But if s is 0, h must be 20 or 40. But this translates to 4 o'clock or 8 o'clock, at both of which the minute hand is at 0, along with the second hand.

Therefore the 3 hands can never be 120 degrees apart, QED

Puzzle 1490

How many times per day do the hour and minute hands of a clock form a right angle?

[Ref: ZRCV]

Hint: Think carefully at 3 and 9 o'clock.

Answer: 44.

Twice each hour equals 48, except 3am, 9am, 3pm, 9pm cause us to lose one each. i.e. we only have 3 from 2am to 4am, and not the normal 4.

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