Question 1130907
On day 1, there are 2^0 pennies. On day 2, there are 2^1 pennies. And so on. So basically, on day a, there are {{{2^(a-1)}}} pennies. So we need to find the smallest integer a such as {{{2^(a-1)>100}}}. 2^6 is 64 and 2^7 is 128. If a-1 is 7, than on day 8, there is more than $1.00 worth of pennies.