Barbara has boxes in three
sizes: large, standard, and small. She puts 11 large boxes on a table. She
leaves some of these boxes empty, and in all the other boxes she puts 8 standard
boxes. She leaves some of these standard boxes empty, and in all the other
standard boxes she puts 8 (empty) small boxes. Now, 102 of all the boxes on the
table are empty. How many boxes has Barbara used in total?
By putting 8 boxes in a box, the total number of empty boxes increases by 8 - 1
= 7. If we call x the number of times that 8 boxes have been put in a box, we
know that 11 + 7x = 102. It follows that x=13. In total, 11 + 13 × 8 = 115
boxes have been used.
Here is a sequence of
numbers: 1 11 21 1211 111221 It seems to be a strange sequence, but yet there is
a system behind it... What is the next term in this sequence?
Again, the system behind the sequence is that each number (except the first one
of the sequence) "describes" the previous number. Now, however, the number of
occurrences of each cipher is counted. So 1231 means one "2" and three times a
"1", and 131221 means one "3", one "2", and two times a "1". The number
following on 131221 is therefore 132231 (one "3", two times a "2", and three
times a "1"). The complete sequence is as follows: 1 11 21 1211 1231 131221
132231 232221 134211 14131231 14231241 24132231 14233221 14233221 etcetera .
A light bulb is hanging in a
room. Outside of the room there are three switches, of which only one is
connected to the lamp. In the starting situation, all switches are 'off' and the
bulb is not lit. If it is allowed to check in the room only once to see if the
bulb is lit or not (this is not visible from the outside), how can you determine
with which of the three switches the light bulb can be switched on?
To find the correct switch (1, 2, or 3), turn switch 1 to 'on' and leave it like
that for a few minutes. After that you turn switch 1 back to 'off', and turn
switch 2 to 'on'. Now enter the room. If the light bulb is lit, then you know
that switch 2 is connected to it. If the bulb is not lit, then it has to be
switch 1 or 3. Now touching for short the light bulb, will give you the answer:
if the bulb is still hot, then switch 1 was the correct one; if the bulb is
cold, then it has to be switch 3.
Using the ciphers 1 up to 9,
three numbers (of three ciphers each) can be formed, such that the second number
is twice the first number, and the third number is three times the first number.
Which are these three numbers?
There are two solutions:
192, 384, and 576.
327, 654, and 981.
A man has a wolf, a goat,
and a cabbage. He must cross a river with the two animals and the cabbage. There
is a small rowing-boat, in which he can take only one thing with him at a time.
If, however, the wolf and the goat are left alone, the wolf will eat the goat.
If the goat and the cabbage are left alone, the goat will eat the cabbage. How
can the man get across the river with the two animals and the cabbage?
There are two solutions: First, the man takes the goat across, leaving the wolf
with the cabbage. Then he goes back. Next, he takes the wolf across. Then the
man goes back, taking the goat with him. After this, he takes the cabbage
across. Then he goes back again, leaving the wolf with the cabbage. Finally, he
takes the goat across. First, the man takes the goat across, leaving the wolf
with the cabbage. Then he goes back. Next, he takes the cabbage across. Then the
man goes back, taking the goat with him. After this, he takes the wolf across.
Then he goes back again, leaving the wolf with the cabbage. Finally, he takes
the goat across.
Of all the numbers whose
literal representations in capital letters consists only of straight line
segments (for example, FIVE), only one number has a value equal to the number of
segments used to write it. Which number has this property?
This is the only solution that satisfies the requirement that the capital
letters shall consist only of straight line segments.
Greengrocer C. Carrot wants
to expose his oranges neatly for sale. Doing this he discovers that one orange
is left over when he places them in groups of three. The same happens if he
tries to place them in groups of 5, 7, or 9 oranges. Only when he makes groups
of 11 oranges, it fits exactly. How many oranges does the greengrocer have at
least?
Assume the number of oranges is A. Then A-1 is divisible by 3, 5, 7 and 9. So,
A-1 is a multiple of 5×7×9 = 315 (note: 9 is also a multiple of 3, so
3 must not be included!). We are looking for a value of N for which holds that
315×N + 1 is divisible by 11. After some trying it turns out that N = 3.
This means that the greengrocer has 946 oranges.
A number is called a
palindrome when it is equal to the number you get when all its digits are
reversed. For example, 2772 is a palindrome. We discovered a curious thing. We
took the number 461, reversed the digits, giving the number 164, and calculated
the sum of these two numbers: 461 164 + ------- 625 We repeated the process of
reversing the digits and calculating the sum two more times: 625 526 + -------
1151 1511 + ------- 2662 To our surprise, the result 2662 was a palindrome. We
decided to see if this was a pure coincidence or not. So we took another 3-digit
number, reversed it, which gave a larger number, and added the two. The result
was not a palindrome. We repeated the process, which resulted in another 3-digit
number which was still not a palindrome. We had to repeat the process twice more
to finally arrive at a 4-digit number which was a palindrome. What was the
3-digit number we started with the second time?
Because the reverse of the starting number is greater than the starting number
itself, the first digit of the starting number must be less than the last digit.
Therefore, the starting number must be at least 102. Secondly, we know that
after two summations, the result has still only 3 digits.
abc
cba +
-------
def
fed +
-------
ghi
We know that def is not a palindrome. Therefore, d differs from f. This is only
possible if d=f+1 (d can only be one greater than f, because b is at most 9).
Since abc is at least 102, def is at least 403, so d+f will be at least 7. Since
ghi is still a 3-digit number but not a palindrome, i can be at most 8, so d+f
can be at most 8. Since d=f+1, d+f can only be 7, from which we conclude that
a=1 and c=2. Now we have:
1b2
2b1 +
-------
4e3
To make the first digit of 4e3 a 4, b must be 5, 6, 7, 8, or 9. Now calculate
the sum of 4e3 and 3e4:
4e3
3e4 +
-------
8h7
Because the first digit of the sum must be 8, e must be at least 5. Therefore,
the only remaining candidates for b are 8 (8+8=16) and 9 (9+9=18). Now it can
easily be found that b must be 9 and the starting number we are looking for is
192:
192
291 + (291 is greater than 192)
-------
483
384 +
-------
867 (still a 3-digit number)
768 +
-------
1635
5361 +
-------
6996 (the 4-digit palindrome).
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