There's gold in there!

Let's think simple, look at the facts.

If x is the number of gold pieces in the treasure chest, then x+1 must be divisible by 3,4,5.

Because we require smallest possible number of gold pieces, x+1 should be equal to the least common multiplier (LCM) of 3,4,5 which is 60. So, the smallest possible number of gold pieces in the chest would be x=59

Let the number required be $n$ , then by Chinese Remainder Theorem , we have:

$\begin{aligned} n & \equiv 2 \pmod{3} \\ \Rightarrow n & = 2 + 3\color{#3D99F6}{t} \quad \quad \quad \quad \color{#3D99F6}{\text{where }t \text{ is an positive integer}} \\ 2 + 3t & \equiv 3 \pmod{4} \\ 3t & \equiv 1 \pmod{4} \\ t & \equiv 3 \pmod{4} \\ \Rightarrow t & = 3 + 4x \\ \Rightarrow n & = 2+3t = 2 + 3(3+4x) = 11+12x \\ 11 + 12x & \equiv 4 \pmod{5} \\ 2x & \equiv 3 \pmod{5} \\ x & \equiv 4 \pmod{5} \\ \Rightarrow x & = 4 + 5y \\ \Rightarrow n & = 11+12x = 11 + 12(4+5y) = 59 + 60y \end{aligned}$

$\Rightarrow$ the smallest $n = 59+60(0) = \boxed{59}$

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#include<iostream>
int main()
{ for (int i=1; i<200; ++i)
{
    if (i%3==2)
        {
            if (i%4==3)
                {
                  if (i%5==4)
                    { std::cout<<i<<"\n";
                    }
                 }
          }
}
return 0;
}

59

119

179

This is an AP with common difference $60$ .

You need a number that's not the product of 2 of the given number

The cycle of 5 fluctuates between 0 and 5; so the number would be X4 or X9

But the cycle of 4 fluctuates between 0,2,4,6,8, so we will eleminate X4 and it'll be X9

So for 3 we need X7,for 4 X6,for 5 X5

The cycle of 3 repeates every 30, and X7 appears at cycle end

So first we try 29

3 .. valid as 27 less by 2

But 4 not valid as 28 is less by 1

Ok now try 59

3 .. valid as 57=3×19 less by 2

4 . .valid as 56=4×14 less by 3

Let x be the smallest possible number of gold pieces in the chest

The question states the following (in simplified terms):

• x mod 3 = 2
• x mod 4 = 3
• x mod 5 = 4

Since x mod 5 = 4, the units digit of x must be a 4 or 9 , as a multiple of 5 + 4 will always have its units digit 4 or 9

x also must be odd, as x mod 4 = 3. This means that x is equal to a multiple of 4 + 3 , which is equal to Even + Odd , which in turn is Odd . Therefore, the units digit of x must be 9

Then possible values of x (has units digit of 9) are all checked to ensure that it satisfies the equation x mod 3 = 2, and we see that the lowest possible value for x is 59 (other lower values are a multiple of 3, or has a remainder of 1 when divided by 3)

Therefore x = $\boxed{59}$

The information about 2s is already true given the information about 4s, so we know the number of stones must be in the form $4n+3.$ (3 more than a multiple of 4.)

When we group this into 3's, we can make $n+1$ complete groups, with $n$ left over -- so $n$ must be of the form $3m+2$ (2 more than a multiple of 3), so the total number of stones is $4(3m+2)+3 = 12m + 11$ .

We could continue this logic for 5s, but it is easy enough to check $m=1,2,\ldots$ and see that $m=4$ gives 59 stones, which leaves 4 left over when grouped into 5s! So 59 is the fewest number of stones needed.

If you'd like to learn a more "methodical" way to approach these problems, check out the Chinese Remainder Theorem ! In fact, using this method, we can quickly get the answer, since CRT says that the number must leave a remainder of $(-1)^3=-1$ when divided by $3\cdot 4\cdot 5=60.$

x=-1 (mod 3) x=-1(mod 4) x=-1(mod 5) x=-1 (mod 3 4 5) (because they are relatively prime) 59

LCM of 3,4&5 minus 1, That is 60 - 1 = 59