The Apple Vendor


An apple vendor has 1000 apples and 10 empty boxes. He asks his son to place all the 1000 apples in all the 10 boxes in such a manner that if he asks for any number of apples from 1 to 1000, his son should be able to pick them in terms of boxes. How did the son place all the apples among the 10 boxes, given that any number of apples can be put in one box.

For Solution SCROLL DOWN...

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1, 2, 4, 8, 16, 32, 64, 128, 256, 489

Let's start from scratch.
• The apple vendor can ask for only 1 apple, so one box must contain 1 apple.
• He can ask for 2 apples, so one box must contain 2 apples.
He can ask for 3 apples, in that case box one and box two will add up to 3.
• He can ask for 4 apples, so one box i.e. third box must contain 4 apples.
• Now using box number one, two and three containing 1, 2 and 4 apples respectively, his son can give upto 7 apples. Hence, forth box must contain 8 apples.
• Similarly, using first four boxes containing 1, 2, 4 and 8 apples, his son can give upto 15 apples. Hence fifth box must contain 16 apples.
You must have noticed one thing till now that each box till now contains power of 2 apples. Hence the answer is 1, 2, 4, 8, 16, 32, 64, 128, 256, 489. This is true for any number of apples, here in our case only upto 1000.


pg... said...

1,2,4,8,16,32,64,128,256, 489

Ankur Gupta said...

same as pg

Jayant said...

Box(n) = 2^(n-1) , n : 1-9
Box(10) = 1000 - Sum(Box(n)), n:1-9

As in, binary representation (basis vectors)...

saahith said...

1+2+4+8+......+2^n = 2^(n+1) - 1

and the effect is inductive.................

Suren said...

With 1,2,4,8,16,32,64,256,512 we could have covered all values from 1 to 1023...

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David Jackman said...

Another good math puzzle. I thought this one would be hard until I reasoned it out a little bit.

I knew that there would have to be a box with just one apple in it, since the father could ask for just one apple.

That's a good logical judgment, but it doesn't quite get to the heart of the puzzle. I needed a way to break down the number 1000 into increasingly smaller numbers.

I don't know how exactly I came up with it, but it came to me to try 500 apples in one box, 250 in another, 125 in another, etc... so that I was dividing by two each time. When a number like 125 was divided in two (62.5) I would just round down and continue on.

That solution was so close, but wrong. I found I couldn't make the number 13.

The right way is to start with the same idea (increasingly smaller/bigger numbers of apples per box), but go UP instead of down. Box 1 has 1 apple, the next has 2, then 4, 8, 16, 32, 64, 128, 256, and since 512 is too many apples (it puts the total over 1000) I just calculated the difference that would make up for the remaining apples: 1000 - (1+2+4+8+16+32+64+128+256) = 489

So, in the 10 boxes there are the following numbers of apples:

1, 2, 4, 8, 16, 32, 64, 128, 256, 489

ANY number from 1 to 1000 can be expressed as the sum of the above numbers.

47 = 32 + 8 + 4 + 2 + 1
376 = 256 + 64 + 32 + 16 + 8

There's also a lesson in how binary works in the above numbers, but I've written quite a bit already. Until tomorrow!