# number Puzzle #71

How will u divide 1000 one Rs. Coins in ten bag so that u can give any amount between 1-1000 by just giving the bags without changing the no of coins in each bag.

Think of the binary numbers of 8 digits

11 1111 1111 => represents 1023, every number upto 1023 can be represented by a combination of 0’s and 1’s.

for example 5 => 00 0000 0101 => 4 + 1

39 => 00 0010 0111 => 32 + 4 + 2 + 1

so if we put 1,2,4,8,16,32,64,128,256,512 coins in 10 bags, we can always make any number by just taking the bag where it is 1 in the binary representation of that number.

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Sunil Jaz says

Each bags 100 as coins …. So problem solved….

Lokesh Nayak says

1, 2, 3, 5, 10, 20, 40, 80, 160, 320, 359

This can be also a one of the solution.

LarchOye says

1,2,4,8,16,32,64,128,256,512… My first guess, after manually counting our the numbers from 1 to 32, and presuming that the exact same trend/pattern would repeat with each power of 2.

LarchOye says

AH HAH! I of course have almost certainly forgotten to check and see that my total added up to 1000, and not say 1023!!

SO, the correct answer *should* be 1,2,4,8,16,32,64,128,256,489

looks like everything works to me, (starting with 1000, subtract 256, which should give you 744. Then, subtract each of the remaining values from 128,64,32,16,8,4,2,1: which should leave 489. BUT, we’re already able to make 489 out of the other bags (256+128+64+32+8+1=489) right? What about making all of the values from 489 to 511- better check those and make sure (511=256+128+…+2+1), which means we should be able to make 510,509,508,507,506… and since we already know we can make 11,10,9,8,7,6,5,4,3,2, and 1- then it appears as though we’re able to make any number between 1 and 511, or any number between 512 (489+16+4+2+1) and 744 (512+128+64+32+8), but how about the last set- everything from 745 to 1000? Since we’ve already used up all of the other bags except for 256, we’re going to have to change things around and do 745=(489+256). Then, we should be able to create any of the values between 745 and 872 by adding combinations of the 1,2,4,8,16,32,and 64 bags. In order to create 873, we’ll again have to switch to using the 489+256+128 bags. We should then be able to create 874 through 1000 by adding combinations of the 1,2,4,8,16,32,and 64 bags – with the number 1000 being the sum of all ten bags combined. (489 + 256 + 128 + 64 + 32 + 16 + 8 + 4 + 2 + 1) = 1,000.

This is of course, the absolute most exhaustive proof imaginable (without actually building a table of the values and how to create them), but it *should* be simple enough for most people to understand- and hopefully something that the average high school student can verify will work.

IT MOST CERTAINLY IS *NOT* PROOF THAT THIS IS THE ONLY SOLUTION, OR THERE ARE NO OTHER POSSIBLE ANSWERS/SOLUTIONS!!

soman kumar says

thanks to u tooo

Shoaib says

bag1 = 1 coin

bag2 = 2 coins

bag3 = 4 coins

bag4 = 8 coins

bag5 = 16 coins

bag6 = 32 coins

bag7 = 64 coins

bag8 = 128 coins

bag9 = 256 coins

bag10 = Remainig coins

soman kumar says

thank u so much

piyush says

Thannk u so much