SOLUTION: there are coconuts in a basket. when they are removed seven at a time six remains at the end,when they removed six at a time, five remains, five at a time four remains and so on ti

Question 1026255: there are coconuts in a basket. when they are removed seven at a time six remains at the end,when they removed six at a time, five remains, five at a time four remains and so on till two at a time and one remains. what is the smallest number of coconuts in the basket originally?
Answer by ikleyn(52776) (Show Source): You can put this solution on YOUR website!
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there are coconuts in a basket. when they are removed seven at a time six remains at the end, when they removed six at a time, five remains, five at a time four remains and so on till two at a time and one remains. what is the smallest number of coconuts in the basket originally?
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Let me re-formulate the problem in this, equivalent way:

   There are coconuts in a basket. 
   When they are arranged by groups of 7, six coconuts remain in the last group. 
   When they are arranged by groups of 6, five coconuts remain in the last group.   
   When they are arranged by groups of 5, four coconuts remain in the last group.   
   And so on till two coconuts in groups and 1 as a remainder. 

   What is the smallest number of coconuts in the basket originally?


Solution


Let n be the smallest number of coconuts in the basket originally.
Add one coconut. You will have (n+1) coconuts.

Then (n+1) is divisible by 7 without a remainder,
           is divisible by 6 without a remainder,
           is divisible by 5 without a remainder,
           is divisible by 4 without a remainder,
           is divisible by 3 without a remainder,
           is divisible by 2 without a remainder.

Thus (n+1) is divisible by  = 420 without a remainder.

The smallest such a number is n+1 = 420.

Hence, n = 419.

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