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A man had to pack a sack of apples into packets but as each packet had to have exactly the same number of apples he was having difficulty.
If he packed 10 apples per packet, one packet only had 9.
If he packed 9 apples per packet, one packet only had 8.
If he packed 8 apples per packet, one packet only had 7.
If he packed 7 apples per packet, one packet only had 6.
And so on down to 2 apples.
How many apples did he start with?
[Ref: ZZQR]
Hint: The answer involves times tables.
Answer: 2519 apples.
The amount of apples divided by 10 leaves a remainder of 9, the amount of apples divided by 9 leaves a remainder of 8, etc. So we're after a number of apples that divided by 10, 9, etc leaves a remainder of one less. This can be found by using the lowest/least common multiple of 10, 9, 8, 7, 6, 5, 4, 3 and 2, and then subtracting 1.
Puzzle 602
Using the BrainBashers Ecommerce solution, James ordered a fishing rod, priced at $3.56.
Unfortunately, James is an fisherman who lives in a very remote part of Greenland and the import rules there forbid any package longer than 4 feet to be imported.
The fishing rod was 4 feet and 1 inch, just a little too long, so how can the fishing rod be mailed to James without breaking the rules? Ideally James would like the fishing rod to arrive in one piece!
[Ref: ZFEL]
Hint: The fishing rod does end up in a real box.
Answer:
One solution is to use a box which measures 4 feet on all sides, the fishing rod will fit within the diagonal of the box with room to spare.
Puzzle 603
During the recent BrainBashers cipher convention, a Morse code contest took place. The contest consisted of a Morse code transmission where the spaces between the letters and words were missing. Can you find the ten boy's names?
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
Luckily, BrainBashers has provided you with a listing of the Morse code characters:
Answer:
1. James
2. Arthur
3. Victor
4. Englebert
5. Tarzan
6. Jonathan
7. Christopher
8. Alexander
9. Gregory
10. Patrick.
Puzzle 604
In farmer Brown's hay loft there are a number of animals, in particular crows, mice and cockroaches.
Being bored one day, I decided to count the animals and found there were exactly 150 feet and 50 heads in total, and there were twice as many cockroaches as mice.
10M = 50
M = 5
So we have 5 mice (and 10 cockroaches). We can use M = 5 in (1) to give:
B + 3 x 5 = 50
B + 15 = 50
B = 35
So, C = 10, M = 5 and B = 35. Checking that 10 + 5 + 35 = 50, and 10 x 6 + 5 x 4 + 35 x 2 = 150. QED.
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