Question 210928: Tany saved a dollor in the first month. She saved b dollors in the second month. The money she saved the third month was the sum of money that she saved in the first and second month.the money she saved in the fourth month was the sum of the money she saved in the second and third months and so on. If she had saved $280 in the ninth month, rhat were the values of a and b given that they were whole numbers?
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! a = 7
b = 9
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13a + 21b = 13*7 + 21*9 = 91 + 189 = 280
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i did it by brute force rather than any finesse.
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progression is as follows:
1: a
2: b
3: a+b
4: a+2b
5: 2a+3b
6: 3a+5b
7: 5a+8b
8: 8a+13b
9: 13a+21b = $280.00
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formula to solve was 13a + 21b = $280.00
the answer had to be an integer.
i would solve it if i wasn't looking for an integer but i was not able to solve it when i was looking for an integer.
i then went to the brute force approach just to see even if there was an answer.
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i took 21 * (1 through 13)
i took 13 * (1 through 13)
i looked at the b column and took 280 minus that answer. i looked at the result and tried to find that result in the a column. i struck paydirt at 189.
280 - 189 equaled 91 and 91 was one of the answers in the a column.
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the answer in the b column was for b = 9
the answer in the a column was for a = 7
those were my answers.
i plugged them in the formula and they worked.
i checked for months 7 and 8 to see if this gave me the right answer.
in month 7, 5a + 8b = 5*7 + 8*9 = 35 + 72 = 107
in month 8, 8a + 13b = 8*7 + 13*9 = 56 + 117 = 173
173 + 107 = 280 which is what the sum should have been.
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the formula to solve was 13a + 21b = 280
i looked at values for b from 1 through 13 as follows:
possible integer values for b:
1: 21
2: 42
3: 63
4: 74
5: 105
6: 126
7: 147
8: 168
9: 189
10: 210
etc.
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i looked at values for a from 1 through 13 as follows:
possible integer values for a:
1: 13
2: 26
3: 39
4: 52
5: 65
6: 78
7: 91
8: 104
9: 117
10: 130
etc.
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i then looked at each b value to see if i could find a corresponding a value once i subtracted that value from 280.
example:
i looked at b = 8 which gave a b value of 168
i subtracted 168 from 280 to get 112
i then looked to see if there was an a value of 112.
there was none, so i looked at b = 9 which gave a b value of 189
i subtracted 189 from 280 to get 91
i looked at the a table for an a value of 91 and found it at a = 7
my potential answers were:
b = 9
a = 7
i then substituted in the equation of 13a + 21b = 280 using those values for a and b and had success.
i then checked values down the line to see that the rules were being followed (month 9 = month 8 + 7, month 8 = month 7 + 6, etc.
the results of that analysis are shown below:
1: a = 7
2: b = 9
3: a+b = 16 = 9+7
4: a+2b = 25 = 16+9
5: 2a+3b = 41 = 25+16
6: 3a+5b = 66 = 41+25
7: 5a+8b = 107 = 66+41
8: 8a+13b = 173 = 107+66
9: 13a+21b = $280.00 = 173+107
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everything checks out so the answer is good.
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if i ever figure out a formula for how to solve this, i'll update the answer.
no more time to play with it for now.
the answer is good, though.
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