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A tale of The Rich Arab man.

There was a rich Arab man who had four wives. One day he decided to give away all his diamonds to all of his wives.

He carried all his diamonds in a huge bag, walked into the house of his most favorite wife and told her about his intention.

This lady, a fair and understanding woman, counted the diamonds and took a quarter for herself and returned the remaining diamonds to her husband to be shared among the other wives.

He walked into the house of his next most favorite wife, told her his intention but he didn't tell her his most favourite wife had already taken her share. She counted exactly one quarter of the sum and returned the balance to the man to be shared among the other three wives.

The same thing happened with the other two wives too.

Walking out from the last house, he found the remaining quantity of diamonds was exactly divisible by four. He split the sum into four equal parts and gave it to the wives separately.

How many diamonds (the smallest number) was the Arab man carrying initially?

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  • $\begingroup$ This question is actually the simpler version of more tougher one. I will post it next. $\endgroup$ – soosai steven Aug 30 '16 at 15:58
  • $\begingroup$ This is a 4 wives with diamonds variation on 3 kids with chocolates puzzle. $\endgroup$ – axavio Aug 30 '16 at 16:26
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – GentlePurpleRain Aug 31 '16 at 14:34
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10 to begin with.

10/4 = 2 so wife 1 gets 2 diamonds with 8 left over.
8/4 = 2 so wife 2 also gets 2 diamonds with 6 left over.
6/4 = 1 so wife 3 gets just 1 diamond - 5 now remain.
5/4 = 1 so wife 4 also gets 1 diamond.
4 now remain - give 1 to each of the four wives.

The man had

10 diamonds.

However -

all were fakes.

He now has

10 pieces of glass and 4 angry wives.

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The answer is

1024

Reasoning:

Let $x$ be the number of diamonds. After wife $n$, the remaining number of diamonds is $(\frac{3}{4})^n x$ - after the fourth wife, it is $\frac{81}{256}x$. For that number to be divisible by four, $x$ needs to be at least 1024, so that is the answer to the question.

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  • $\begingroup$ Each wife left a multiple of 3. $\endgroup$ – donjuedo Aug 30 '16 at 20:30
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The number of diamonds carried by the Arab man must be divisible by 45.

So the minimum is 45 = 1024, or 0 if we don't assume that the man actually had any diamonds to share.

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  • 1
    $\begingroup$ Most elegant answer ! So elegant I signed up just to upvote this. It's missing spoiler tags though. $\endgroup$ – Skippy le Grand Gourou Aug 31 '16 at 6:14
  • $\begingroup$ @SkippyleGrandGourou This answer does not explain how it arrives at its first sentence. This makes it short and not elegant but incomplete. $\endgroup$ – miracle173 Aug 31 '16 at 6:37
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    $\begingroup$ @miracle173 I agree the first sentence is a bit condensed, though if you think about it it's self-explanatory (the number of diamonds must be divisible by 4, 5 times). Still, noticing that the last step is not different from the first four makes it the most elegant solution. $\endgroup$ – Skippy le Grand Gourou Aug 31 '16 at 8:30
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If he had $x$ diamonds to begin with, after the first wife he would have:

$\dfrac{3x}{4}$

After the second, third and fourth:

$\dfrac{9x}{16},\dfrac{27x}{64},\dfrac{81x}{256}$

As the numerator needs to be divisible by $4$:

$x=1024k, k\in\mathbb{Z_{>=0}}$

Assuming $k>0$, $k=1$, in which each wife received:

256, 192, 144 and 108, and then an additional 81.

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I believe the answer is

1024

Explanation:

When we reach each wife the number of diamonds needs to be divisible by 4. Let $x_i$ denote the number of diamonds when he reaches each wife ($x_5$ being the amount left after all wives are done taking). Every $x_i$ has to be divisible by 4 and can be found by the following recursive formula $$x_i - x_i/4 = x_{i+1}$$. So, we choose initially $x_5$ to be the smallest multiple of 4; 4! This doesn't work as each time we recursively find the previous $x_i$ we have to do the following: $$A: x_{i-1} = 4x_i/3$$ so, we need to multiply $x_5$ by $3^4$ to give us $x_5 = 4*3^4 = 324$. We can then recursively work backwards using A to get $$x_4 =432,x_3=576,x_2=768,x_1=1024$$

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With each wife, we multiply the total by 3/4, and end with a number divisible by 4. Therefore, we must have at least

3*3*3*3*4 at the end = 324.

Backtracking, we multiply 4/3 four times, yielding

1024 initially

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    $\begingroup$ Someone with this much rep should know better about spoilers. $\endgroup$ – Nij Aug 30 '16 at 18:04
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That man is so smart!

He got his favorite wife the highest number of diamonds without the others knowing! I believe the answer is 1024, he got his first wife 256 diamonds, his next wife 192 diamonds, the third one 144 diamonds, and the last one 108 diamonds And he got 324 diamonds left, which is divisible by 4 I should also note that any number that is 1024*(4^n) where n is bigger than 0, this should also work, but this puzzle needs the smallest amount required :)

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