tag:blogger.com,1999:blog-4115025577315673827.post3201188810434872102..comments2019-11-21T20:33:48.870+05:30Comments on CSE Blog - quant, math, computer science puzzles: Gold Silver Numbers PuzzleUnknownnoreply@blogger.comBlogger7125tag:blogger.com,1999:blog-4115025577315673827.post-14498146732902457182017-03-08T22:49:37.644+05:302017-03-08T22:49:37.644+05:30I think the soultion is much more simple: Either 1...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.<br />Based on first case:<br />1) gold number * gold number -> gold<br />2) gold number + gold number -> gold<br />3) silver number * silver number -> silver<br />4) silver number + silver number - gold<br />Based on second case the result is of course always either silver or goldAlkis Piskashttps://www.blogger.com/profile/01258797863535815094noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-30614498987076152452015-04-02T15:13:31.275+05:302015-04-02T15:13:31.275+05:30gold are all those numbers which are multiples of ...gold are all those numbers which are multiples of a given number(this number may or may not equal 2)prateekhttps://www.blogger.com/profile/06600580372273192646noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-56697977322114452822015-01-23T14:28:52.435+05:302015-01-23T14:28:52.435+05:30ANS:
(1) gold number * gold number = gold number
(...ANS:<br />(1) gold number * gold number = gold number<br />(2) gold number + gold number = gold number<br />(3) silver number * silver number = gold/silver number<br />(4) silver number + silver number = gold/silver number<br /><br />"1" is silver (as BOTH colours are used and if "1" was gold => all are gold).<br />Given:<br />(i) s + g = s<br />(ii) s * g = g<br /><br />1:<br /> Consider:<br /> (s*g) + (g*g) = (s*g)<br /> g + (g*g) = g<br /> if (g*g) -> silver<br /> => g + s = g (which does not satisfy (i)).<br /> => (g*g) = gold number<br /><br />let "a" be the least gold number (>1)<br />=> 1..(a-1) -> silver<br />=> (a+1) .. (2a-1) -> silver (from (i))<br />From (ii) & (1) we have n*a = gold (For all n>0)<br />From (i) we have (na+1) .. ((n+1)a-1) -> silver<br />=> only n*a = gold (for all n>0)<br /><br />2:<br /> m*a + n*a = (m+n)*a = gold (For all m,n>0)<br /><br />3:<br /> Eg a = 6<br /> 2(silver) * 3(silver) = 6(gold)<br /> 2(silver) * 4(silver) = 8(silver)<br /><br />4:<br /> Eg a = 6<br /> 2(silver) + 3(silver) = 5(silver)<br /> 2(silver) + 4(silver) = 6(gold)Aditya Tirumalahttps://www.blogger.com/profile/05892898666205305657noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-20272759474976610162015-01-21T06:44:40.794+05:302015-01-21T06:44:40.794+05:30It must be true 1 is silver: if 1 were gold, it wo...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.<br /><br />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. Mike Earnesthttps://www.blogger.com/profile/10706561663990058978noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-62622489072381940802015-01-20T19:22:46.906+05:302015-01-20T19:22:46.906+05:30S*G=G -- 1
S+G=S -- 2
2*G
S*G + G*G = S*G
G + G*G ...S*G=G -- 1<br />S+G=S -- 2<br />2*G<br />S*G + G*G = S*G<br />G + G*G = G<br />if G*G = S<br />then G+S=G which stands antithetical to --1<br />hence,<br />G*G = G<br />AND G+G=G<br /><br />Now if S*S = G then<br />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.<br /><br />hence, S*S = S<br /><br />NOW,<br />IF S+S=S<br />THEN,<br />S+S = S and<br />S+G = S -- 2 <br />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<br />hence S+S=Snitin siwachhttps://www.blogger.com/profile/04752930302372361487noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-33733188649375150942015-01-20T16:12:46.189+05:302015-01-20T16:12:46.189+05:30all odd numbers with silver and even numbers are p...all odd numbers with silver and even numbers are painted with goldGreeshma Balabhadrahttps://www.blogger.com/profile/03561048946430437650noreply@blogger.comtag:blogger.com,1999:blog-4115025577315673827.post-72010948993916623942015-01-20T11:36:30.258+05:302015-01-20T11:36:30.258+05:30First we show 1 is silver. If 1 is gold, let x be ...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.<br /><br />Let x be the smallest gold number. x>=2. <br />First note that using (silver + gold) = (silver) repeatedly, we can show (silver + multiple of gold) = silver (mathematical induction).<br />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.<br /><br />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. <br /><br />Thus the set of gold numbers are multiples of x, and the set of silver numbers are the remaining numbers.<br /><br />We can now give the answers. <br />1. gold*gold = gold.<br />2. gold+gold = gold.<br />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).<br />4. silver+silver can be either silver or gold. For example, if gold is multiples of 6, 2+4=6 and 2+3=5.GoKuhttps://www.blogger.com/profile/05674743004147477362noreply@blogger.com