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Mark is taking a trip and is taking along exactly $1020 in cash. His
father gives him cash for the trip in denominations of 20 dollar bills
and 50 dollar bills only. How many denominations of each
($20's and $50's) are possible?
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Let's decide the amount of $50 bills he takes first. Before the 20's get added, the amount of money Mark has will be either ??00 or ??50.
This means that in order to make the amount up to 1020, Mark needs either ??20 or ??70 respectively in $20 bills. The latter of these is clearly impossible, so the $50 bills must make up a multiple of 100.
From here, it is easy to see that there are 11 possible combinations - 2k 50's and (51-5k) 20's, where k is an integer between 0 and 10.
Why did the vector cross the road?
It wanted to be normal.
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