One year, on Sue's birthday, Sue started a collection of thimbles. The following birthday she added to her collection, which went from strength to strength. In all subsequent years when she counted the thimbles on her birthday the total had increased from the previous year’s total by a number equal to the total she had on her birthday the year before that. (So, for example, her 1983 total equalled her 1982 total added to her 1981 total).
Now, by coincidence, her daughter was born on her birthday. And, with her collection growing following the described pattern, on their birthday in 1983 the number of thimbles Sue owned had reached exactly four times her daughter’s age on that day. On her birthday this year the total of thimbles was four times her age. On only one other occasion has the total been divisible by four, and that was in the year Sue's son was born.
How many thimbles were there in Sue's collection on her birthday this year? How many (if any) did she have the day her daughter was born?
AD Year MageDageSage Thim 1/4*Thim
1980 0 42 0 - 0 0
1981 1 43 1 - 6 1.5
1982 2 44 2 - 6 1.5
1983 3 45 3 - 12 3 12=4*3
1984 4 46 4 - 18 4.5
1985 5 47 5 - 30 7.5
1986 6 48 6 0 48 12 Son born
1987 7 49 7 1 78 19.5
1988 8 50 8 2 126 31.5
1989 9 51 9 3 204 51 204=4*51
There is a logical issue with the problem, for it states, 'The following birthday she added to her collection.'
If Sue had 0 thimbles and bought t thimbles in some year, then the 'following year' Sue must buy 0 thimbles, as t+0=t
I therefore assume that Sue started buying thimbles the year after she 'started the collection" e.g. by buying an ornate container.
Then Sue can add 6 thimbles in the 'following year' 1, and none in year 2.
If so, then 'This year' is 1989, the number of thimbles is 204, and Sue had 0 thimbles when her daughter was born.
Edited on July 16, 2024, 1:36 am
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Posted by broll
on 2024-07-16 01:33:01 |
Possible Solution
