Question 24674: A community food pantry had cans of vegetables to distribute. When they put 10 cans in each bag, one bag was left with 9. When they put 9 cans per bag, one was left with 8. When they put 8 cans per bag, one was left with 7. This continued until 2 cans per bag left one bag with one can. What is the least number of cans the pantry had to begin with?
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! A food pantry had cans of vegetables to distribute. When they put 10 cans per bag, one bag would get 9. When they put 9 cans per bag, one bag would get 8. 8 cans per bag would leave a remainder of 7, and so on, until 2 per bag left one can in a bag. What is the least number of cans the pantry had to distribute? Please explain how you got the answer. Thank you!
THIS MEANS IF N IS THE NUMBER OF CANS ,IT LEAVES A REMAINDER OF 1,2,3,4,5,6,7,8 AND 9 WHEN DIVIDED BY 2,3,4,5,6,7,8,9 AND 10.THAT IS N LEAES A REMAINDER THAT IS JUST ONE LESS THAN THE NUMBER WE ARE DIVIDING WITH EACH TIME.OR PUT IN OTHER WORDS HAD THE NUMBER BEEN 1 MORE THAT IS N+1,IT WOULD HAVE BEEN PERFECTLY DIVISIBLE BY ALL THESE NUMBERS FROM 2 TO 10.SO WE JUST NEED TO FIND THE GCD OR GREATEST COMMON DIVISOR OF 2,3,4,5,6,7,8,9 AND 10 AND SUBTRACT 1 FROM IT TO GET OUR ANSWER OF MINIMUM NUMBER OF CANS SATISFYING THE ABOE CRITERIA.
GCD OF THE ABOVE NUMBERS IS
......2|2,3,4,5,6,7,8,9,10
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......2|1,3,2,5,3,7,4,9,5
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......3|1,3,1,5,3,7,2,9,5
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......5|1,1,1,5,1,7,2,3,5
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........1,1,1,1,1,7,2,3,1
GCD = 2*2*3*5*7*2*3=2520
HENCE N+1=2520
N=2519
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