### Gold Silver Numbers Puzzle

Source: Mailed to me by JDGM ("regular commenter JDGM")

Problem:

The integers greater than zero are painted such that:

• every number is either gold or silver.
• both paints are used.
• silver number + gold number = silver number
• silver number * gold number = gold number

Given only this information, for each of the following decide whether it is a gold number, a silver number, or could be either:

1.) gold number * gold number
2.) gold number + gold number
3.) silver number * silver number
4.) silver number + silver number

1. First we show 1 is silver. If 1 is gold, let x be silver. Then x*1=x, but x*1 must be gold and x is silver, a contradiction. Therefore, 1 is silver.

Let x be the smallest gold number. x>=2.
First note that using (silver + gold) = (silver) repeatedly, we can show (silver + multiple of gold) = silver (mathematical induction).
Since {1,2,3,....x-1} are silver, any non multiple of x can be written as nx+r (multiple of gold + silver) where 1<=r <= x-1. Thus any non-multiple of x is silver.

Now we prove that multiples of x are gold. If k is a non-multiple of x, then kx is gold (because silver*gold=gold). We only have to worry about the case when k is a multiple of x. In that case, (k+1) is a non-multiple of x and is therefore silver. Hence, (k+1)x is gold. Suppose kx is silver. Then we have (k+1)x = kx+x = silver+gold=silver, contradicting the fact that (k+1)x is gold. This shows that kx must be gold for any k.

Thus the set of gold numbers are multiples of x, and the set of silver numbers are the remaining numbers.

We can now give the answers.
1. gold*gold = gold.
2. gold+gold = gold.
3. silver*silver can be either silver or gold. For example, if gold is multiples of 6, we can have 2*3=6 (silver*silver=gold), but 2*2=4 (silver*silver=silver).
4. silver+silver can be either silver or gold. For example, if gold is multiples of 6, 2+4=6 and 2+3=5.

2. all odd numbers with silver and even numbers are painted with gold

1. gold are all those numbers which are multiples of a given number(this number may or may not equal 2)

3. S*G=G -- 1
S+G=S -- 2
2*G
S*G + G*G = S*G
G + G*G = G
if G*G = S
then G+S=G which stands antithetical to --1
hence,
G*G = G
AND G+G=G

Now if S*S = G then
S*S=G, S*G=G, G*G=G which is not possible since all the numbers are either silver or gold and both the colors are used.

hence, S*S = S

NOW,
IF S+S=S
THEN,
S+S = S and
S+G = S -- 2
which is again not possible as every number after first silver would be silver and which is a contradiction as (last GOLD)*(first SILVER) = GOLD
hence S+S=S

4. It must be true 1 is silver: if 1 were gold, it would contradict 1 * silver = silver. Let G be the smallest gold number, so that 1,2, . . . (G-1) are all silver. Then adding G to each of these, we see that (G+1), (G+2), . . . , 2G-1 are all silver. Adding G to these shows 2G+1, 2G+2, . . . 3G - 1 are all silver, and proceeding in this fashion, you can show any number of the form a*G + b, with 0 1. When c is not a multiple of G, so c is silver, then silver * gold = gold implies that c*G is gold. When c is a multiple of G, so c = k*G, I claim it is still true that c*G is gold. Assume c*G was silver: then since G is gold, this would imply G + c*G = G + k*G^2 = (kG + 1)G is silver, which is a contradiction, since kG + 1 is silver (by first paragraph), G is gold, and silver * gold = gold.

Thus, the gold numbers are just the multiples of G, for some G>1, and it's easy to see that picking any G will work. This implies that 1) and 2) are both always gold. When G = 4, we see that 3) can be either, since 1+1 is silver while 2+2 is gold, and 4) can be either, since 1*1 is silver while 2*2 is gold.

5. ANS:
(1) gold number * gold number = gold number
(2) gold number + gold number = gold number
(3) silver number * silver number = gold/silver number
(4) silver number + silver number = gold/silver number

"1" is silver (as BOTH colours are used and if "1" was gold => all are gold).
Given:
(i) s + g = s
(ii) s * g = g

1:
Consider:
(s*g) + (g*g) = (s*g)
g + (g*g) = g
if (g*g) -> silver
=> g + s = g (which does not satisfy (i)).
=> (g*g) = gold number

let "a" be the least gold number (>1)
=> 1..(a-1) -> silver
=> (a+1) .. (2a-1) -> silver (from (i))
From (ii) & (1) we have n*a = gold (For all n>0)
From (i) we have (na+1) .. ((n+1)a-1) -> silver
=> only n*a = gold (for all n>0)

2:
m*a + n*a = (m+n)*a = gold (For all m,n>0)

3:
Eg a = 6
2(silver) * 3(silver) = 6(gold)
2(silver) * 4(silver) = 8(silver)

4:
Eg a = 6
2(silver) + 3(silver) = 5(silver)
2(silver) + 4(silver) = 6(gold)

6. I think the soultion is much more simple: Either 1) silver numbers are odd and gold numbers are even or 2) both silver and gold numbers are even.
Based on first case:
1) gold number * gold number -> gold
2) gold number + gold number -> gold
3) silver number * silver number -> silver
4) silver number + silver number - gold
Based on second case the result is of course always either silver or gold