Puzzle Of The Day: 87
Collection of Coins
A large bottle contains a collection of coins. When you examine the contents you are amazed to discover that there is an equal number of each type of denomination: 1 p, 2 p, 5 p, 10 p, 20 p, 50 p, Re 1, and Rs 2. Even more remarkable is that if you remove one coin there will be an exact amount in rupees. What is the smallest amount, in rupees, that could be in the bottle after removing that coin?
0.01 + 0.02 + 0.05 + 0.10 + 0.20 + 0.50 + 1 + 2 = 3.88 ( value in rupees) 1+2+5+10+20+50+100+200 = 388 (value is paise) => y = 388x-100I => y = 4(97x-25I) hence y should be a multiple of 4 Clearly y can not be 1, 2 , 5 , 10 , 50 For y = 20, minimum x will be 15 So, total amount = 388 × 15 = 5820, after removing 20 paise remaining value 5800 paise = 58 rupees
So, Total amount is 388x = 100I+y, where I is an integer and y is one of the coin in paise
For y = 100 and 200, minimum x will be 25
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