LCD's and LCM's are pretty much all you need yeah. I remember learning them in 4th, 5th grade. But gee that seems advanced. It is the one part of math where you pretty much are left to the mercy of writing them out until you find a match.
http://www.mathplayground.com/wordproblems.html Go to the one that says hoops, if you get five right you get to play a basketball game.
The last statement in the problem, "what is the smallest number of dimes she could have, if she has more than 10", pretty much says it needs to be solved via some combination of equation(s) and inequality constraint(s). You have: [1] 4n + 2 = 5m + 1 [2] 4n + 2 > 10 [3] 5m + 1 > 10 From [2], n > 2; from [3], m > 1 From [1], we can see (5m + 1) is an even number, which means m has to be an odd number, i.e., 3, 5, 7, ... This observation alone would guarantee to cut the computation (or you may call it trial and error) time by half. First try m = 3, n = 3.5 which is not an integer. Next try m = 5, then n = 6. Bingo. Basically my approach is of the same principle as moe's, only with one additional shortcut based on one more restraint on the nature of number m. Also, even if there is no odd-even restraint on either number, you want to pick up the number that is preceded by a bigger factor to start your trial and error, since that would also cut your computation time. Say, in 3n + 1 = 5m + 3, you start with m.
Who uses algebra in school after highschool anyways. TI-89. GG. Where you can type notes into your calculator. Muahahhahahah!!!
If t is total number of dimes, this statement gives: (t-2)/ 4 = [1, 2, 3, 4, 5, 6, 7, ...] (set of natural numbers) t-2 = [4, 8, 12, 16, 20, 24, 28, ...] t = [6, 10, 14, 18, 22, 26, 30, ...] (t-1)/5 = [1, 2, 3, 4, 5, 6, 7...] t-1 = [5, 10, 15, 20, 25, 30, 35, ...] t = [6, 11, 16, 21, 26, 31, 36, ...] This requires t to be 26.
I like this answer, but it still involves brute forcing or intuition to figure out the number. There isn't a simpler method?
It is a weird question. I don't see any other way than brute force. Questions like this usually have a third part that compares x and y. Then you can use substitution to solve the equation. For instance you could say the dimes in groups of 4 had one more group than the dimes in groups of five. Then you would have: 4x+2 = z 5y+1 = z x = y+1 So, 4x+2 = 5y+1 4(y+1)+2 = 5y+1 4y+6 = 5y+1 -1y = -5 y = 5 Then, 5(5)+1 = z z = 26
Werd man. My TI-89 has become the biggest crutch ever. I hated the TI-89 back in HS because I thought it was too complicated, but man, that thing is powerful.
yeah you can't do it without just trial and error. there is no equation. and some of y'all need to learn the modulus function (not that it would've really helped, just people were writing the problem wrong at the beginning).
