Question 136664: Jason is filling grab bags for the school festival. Two hundred
bags are lined up on a long table. He has already placed crackers and
other food items in each bag and now has a limited amount of prizes to
add to some of the bags. If he places prize A in every 8th bag, prize B in
every 12th bag, and prize C in every 15th bag, which bag will have all
three prizes?
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Jason is filling grab bags for the school festival. Two hundred
bags are lined up on a long table. He has already placed crackers and
other food items in each bag and now has a limited amount of prizes to
add to some of the bags. If he places prize A in every 8th bag, prize B in
every 12th bag, and prize C in every 15th bag, which bag will have all
three prizes?
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You need the bag number that is divisible by 8,12,and 15.
That is the "least common multiple".
Write
8 = 2^3
12 = 2^2*3
15 = 3*5
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The lcm must have each of the prime power numbers raised to its highest power.
lcm = 2^3*3^1*5^1
lcm = 8*15 = 120
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The 120th bag will have all three prizes because it is an 8th bag,
a 12th bag and a 15th bag.
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cheers,
Stan H.
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