Was Fifth Grade math always this hard?
January 28, 2014 7:36 AM Subscribe
Can you show us how to solve a math word problem?
My granddaughter was given this question as a "Stretch Your Thinking" section of a fifth grade math assignment. She's turning the homework in without an answer, so we're not "cheating".
However, neither she nor her parents nor I could come up with a solution process. I told them MetaFilter ALWAYS has an answer. Please......!
Here's the question:
"I am a number less than 3,000. When you divide me by 32, my remainder is 30. When you divide me by 58, my remainder is 44. What number am I?"
Please show your work.
My granddaughter was given this question as a "Stretch Your Thinking" section of a fifth grade math assignment. She's turning the homework in without an answer, so we're not "cheating".
However, neither she nor her parents nor I could come up with a solution process. I told them MetaFilter ALWAYS has an answer. Please......!
Here's the question:
"I am a number less than 3,000. When you divide me by 32, my remainder is 30. When you divide me by 58, my remainder is 44. What number am I?"
Please show your work.
Brent Parker's solution method is the only one that I would expect to be within the grasp (even the "stretch your thinking") of a fifth grade class.
I defined a function y=(58x-14)/32 and used Wolfram Alpha to calculate its integer (whole number) solution, which can be found here. As you can see it's a general solution and there are a lot of valid answers less than 3000. I believe 1058 is the lowest.
Somehow I think that wouldn't get full credit.
posted by telegraph at 7:48 AM on January 28, 2014
I defined a function y=(58x-14)/32 and used Wolfram Alpha to calculate its integer (whole number) solution, which can be found here. As you can see it's a general solution and there are a lot of valid answers less than 3000. I believe 1058 is the lowest.
Somehow I think that wouldn't get full credit.
posted by telegraph at 7:48 AM on January 28, 2014
Brent Parker's solution method makes sense, and I find three numbers that way: 798, 1726, and 2654
posted by JumpW at 7:53 AM on January 28, 2014 [1 favorite]
posted by JumpW at 7:53 AM on January 28, 2014 [1 favorite]
Response by poster: Yes, looking at Brent Parker's spreadsheet it would appear "798" would be first. The spreadsheet is the only kind of solution we saw but it reeks of counting on your fingers.
posted by uncaken at 7:54 AM on January 28, 2014 [1 favorite]
posted by uncaken at 7:54 AM on January 28, 2014 [1 favorite]
Can be accomplished using code. For example, in VBA (and printing to a cell in an Excel worksheet):
Sub whatNumberAmI()
Dim i As Integer
For i = 1 To 3000
If i Mod 32 = 30 And i Mod 58 = 44 Then
Sheets("Sheet2").Cells(1, 3) = i
End If
Next
End Sub
The code declares a variable as an integer, and goes thu every integer from 1-3000 checking if the remainder when the variable is divided by 32 is 30 AND the remainder when divided by 58 is 44, and if some number satisfies that it puts that number in a certain cell in an Excel worksheet.
This code would only show the largest number that satisfies the conditions (i.e., if there were a smaller number that also satisfied the conditions, it would be overwritten by the larger number - 2654). Would probably be simple enough to list all, and I could figure that out if you want (not a VBA expert).
posted by 3FLryan at 7:56 AM on January 28, 2014
Sub whatNumberAmI()
Dim i As Integer
For i = 1 To 3000
If i Mod 32 = 30 And i Mod 58 = 44 Then
Sheets("Sheet2").Cells(1, 3) = i
End If
Next
End Sub
The code declares a variable as an integer, and goes thu every integer from 1-3000 checking if the remainder when the variable is divided by 32 is 30 AND the remainder when divided by 58 is 44, and if some number satisfies that it puts that number in a certain cell in an Excel worksheet.
This code would only show the largest number that satisfies the conditions (i.e., if there were a smaller number that also satisfied the conditions, it would be overwritten by the larger number - 2654). Would probably be simple enough to list all, and I could figure that out if you want (not a VBA expert).
posted by 3FLryan at 7:56 AM on January 28, 2014
Best answer: Telegraph, maybe I miscalculated, but using your notation I would have gotten y=(58x+14)/32. No?
Uncaken, no offense to your granddaughter's teacher, but I can't think what he or she was trying to teach the students with this problem. How to brute-force an answer? I can't see where there is enough information for an even remotely-grade-appropriate single-value algebraic solution--it's quite a "stretch" indeed.
posted by spelunkingplato at 7:56 AM on January 28, 2014 [1 favorite]
Uncaken, no offense to your granddaughter's teacher, but I can't think what he or she was trying to teach the students with this problem. How to brute-force an answer? I can't see where there is enough information for an even remotely-grade-appropriate single-value algebraic solution--it's quite a "stretch" indeed.
posted by spelunkingplato at 7:56 AM on January 28, 2014 [1 favorite]
Best answer: A lot of 5th grade math problems are most easily solved by enumeration AKA brute force. It's an oddly common issue with math around that level from what I saw of my kids' math homework.
Maybe they're trying to get kids to be clever but really, if you've seen one enumeration problem, you've seen one enumeration problem.
(that was a joke)
posted by GuyZero at 8:04 AM on January 28, 2014 [14 favorites]
Maybe they're trying to get kids to be clever but really, if you've seen one enumeration problem, you've seen one enumeration problem.
(that was a joke)
posted by GuyZero at 8:04 AM on January 28, 2014 [14 favorites]
First write out what you know:
x/32 = a + 30
x/58 = b + 44
0 ≤ x < 3000 (I'm going to assume the number is nonnegative)
You can rearrange so that the equalities are all in terms of x:
x = 32a + 960
x = 58b + 2252
Therefore:
32a + 960 = 58b + 2252
0 ≤ 32a + 960 < 3000
You can plug that equation into Wolfram Alpha, which says there are three integer solutions:
a = -14, b = -30
a = 15, b = -14
a = 44, b = 2
Plugging these values back into the equalities in terms of x you get three solutions: x ∈ {512, 1440, 2368}.
posted by grouse at 8:08 AM on January 28, 2014 [2 favorites]
x/32 = a + 30
x/58 = b + 44
0 ≤ x < 3000 (I'm going to assume the number is nonnegative)
You can rearrange so that the equalities are all in terms of x:
x = 32a + 960
x = 58b + 2252
Therefore:
32a + 960 = 58b + 2252
0 ≤ 32a + 960 < 3000
You can plug that equation into Wolfram Alpha, which says there are three integer solutions:
a = -14, b = -30
a = 15, b = -14
a = 44, b = 2
Plugging these values back into the equalities in terms of x you get three solutions: x ∈ {512, 1440, 2368}.
posted by grouse at 8:08 AM on January 28, 2014 [2 favorites]
To give you an idea of how much stretching of thinking this problem can inspire, this is basically an application of the Chinese Remainder Theorem, a topic that's covered many college mathematics classes.
posted by RonButNotStupid at 8:12 AM on January 28, 2014 [3 favorites]
posted by RonButNotStupid at 8:12 AM on January 28, 2014 [3 favorites]
Grouse: 44*58=2552, not 2252.
(edited, haha - made a mistake here, too!)
posted by 3FLryan at 8:13 AM on January 28, 2014
(edited, haha - made a mistake here, too!)
posted by 3FLryan at 8:13 AM on January 28, 2014
D'oh! It's too late to edit my comment. Here is a corrected version:
First write out what you know:
x/32 = a + 30
x/58 = b + 44
0 ≤ x < 3000 (I'm going to assume the number is nonnegative)
You can rearrange so that the equalities are all in terms of x:
x = 32a + 960
x = 58b + 2552
Therefore:
32a + 960 = 58b + 2552
0 ≤ 32a + 960 < 3000
You can plug that inequality system into Wolfram Alpha, which says there are four integer solutions:
a = -30, b = -44
a= -1, b = -28
a = 28, b = -12
a = 57, b = 4
Plugging these values back into the equalities in terms of x you get four solutions: x ∈ {0, 928, 1856, 2784}. Still don't have the same answers as Brent Parker. It's possible I messed something else up.
posted by grouse at 8:18 AM on January 28, 2014
First write out what you know:
x/32 = a + 30
x/58 = b + 44
0 ≤ x < 3000 (I'm going to assume the number is nonnegative)
You can rearrange so that the equalities are all in terms of x:
x = 32a + 960
x = 58b + 2552
Therefore:
32a + 960 = 58b + 2552
0 ≤ 32a + 960 < 3000
You can plug that inequality system into Wolfram Alpha, which says there are four integer solutions:
a = -30, b = -44
a= -1, b = -28
a = 28, b = -12
a = 57, b = 4
Plugging these values back into the equalities in terms of x you get four solutions: x ∈ {0, 928, 1856, 2784}. Still don't have the same answers as Brent Parker. It's possible I messed something else up.
posted by grouse at 8:18 AM on January 28, 2014
Best answer: Let's see if there's a pattern!
The first possible candidate is 44.
44 divided by 32 has a remainder of 12
Well that's not right, so let's add 58 to it and try again
102 divided by 32 has a remainder of 6
Next!
160 divided by 32 has a remainder of 0.
12, 6, 0? That looks like it might be going down by six each time, but what happens after 0? It can't go negative... Let's keep going!
218 divided by 32 has a remainder of 26 Okay ... does the pattern continue?
276 divided by 32 has a remainder of 20 Whew! Okay well we can just follow the pattern and do easy math (adding 58 and subtracting 6) instead of hard math (dividing big numbers by 32)
334 -> 14
392 -> 8
460 -> 2 (If we don't trust our pattern we can do the division on any of these to check it)
... Now what? If we haven't figured it out to deal with this we can just do the math on 508 and see what happens.
508 divided by 32 has a remainder of 28. Fine. Let's run through our pattern a bit and add 58 and subtract 6 some more
566 -> 22
624 -> 16
682 -> 10
740 -> 4
Okay we'll divide this one out again, unless we've noticed that when it goes negative you just add 32
798 -> 30 Hey that's it! Done!
posted by aubilenon at 8:21 AM on January 28, 2014 [5 favorites]
The first possible candidate is 44.
44 divided by 32 has a remainder of 12
Well that's not right, so let's add 58 to it and try again
102 divided by 32 has a remainder of 6
Next!
160 divided by 32 has a remainder of 0.
12, 6, 0? That looks like it might be going down by six each time, but what happens after 0? It can't go negative... Let's keep going!
218 divided by 32 has a remainder of 26 Okay ... does the pattern continue?
276 divided by 32 has a remainder of 20 Whew! Okay well we can just follow the pattern and do easy math (adding 58 and subtracting 6) instead of hard math (dividing big numbers by 32)
334 -> 14
392 -> 8
460 -> 2 (If we don't trust our pattern we can do the division on any of these to check it)
... Now what? If we haven't figured it out to deal with this we can just do the math on 508 and see what happens.
508 divided by 32 has a remainder of 28. Fine. Let's run through our pattern a bit and add 58 and subtract 6 some more
566 -> 22
624 -> 16
682 -> 10
740 -> 4
Okay we'll divide this one out again, unless we've noticed that when it goes negative you just add 32
798 -> 30 Hey that's it! Done!
posted by aubilenon at 8:21 AM on January 28, 2014 [5 favorites]
grouse - you've still got your first two equations wrong. If x/32 gives a remainder of 30, that means x/32 = (some integer)/32 + 30/32. Those equations should read
x-30 = 32a
x-44 = 58b
posted by spielzebub at 8:22 AM on January 28, 2014 [2 favorites]
x-30 = 32a
x-44 = 58b
posted by spielzebub at 8:22 AM on January 28, 2014 [2 favorites]
This is what's given:
x / 32 = N + 30
x / 58 = M + 44
x < 3000
This is how I was solving it:
x = 58 * M + 58 * 44 = 58 * M + 2552 (used calculator to multiply 58 by 44)
58 * M + 2552 < 3000
58 * M < 448
This gives us possible values for M : 1..7
And possible values for x: 2610, 2668, 2726, 2784, 2842, 2900, 2958
Checking each possible value of x against the very first equation x / 32 = N + 30, we'll find the only possible answer 2784
posted by ringu0 at 8:22 AM on January 28, 2014
x / 32 = N + 30
x / 58 = M + 44
x < 3000
This is how I was solving it:
x = 58 * M + 58 * 44 = 58 * M + 2552 (used calculator to multiply 58 by 44)
58 * M + 2552 < 3000
58 * M < 448
This gives us possible values for M : 1..7
And possible values for x: 2610, 2668, 2726, 2784, 2842, 2900, 2958
Checking each possible value of x against the very first equation x / 32 = N + 30, we'll find the only possible answer 2784
posted by ringu0 at 8:22 AM on January 28, 2014
Best answer: I don't think you can solve this without brute force. Manually trying out different values, plugging into Wolfram Alpha, or writing up VBA code are all brute force solutions.
For a fifth grader, I would not recommend writing code -- I think there's value in playing with the numbers and getting a sense for any patterns that arise. There are underpinnings of basic algebra here, as well as number theory. Math isn't just about coming up with an answer, it's also about the journey you take to find the answer. Brute force is a perfectly acceptable starting point, especially if you have nothing else to go on. Once you have some solutions, you can see how they are all related to each other (there's basically "one" solution; once you have that solution, you can repeatedly add or subtract 928 = least common multiple of 32, 58 to get more solutions) and if you wanted to do further stretch work you can play around with the parameters of the problem and discover the cases where there are no solutions at all. Doing this will build up intuition for the ideas of least common multiple and greatest common denominator and develop number sense in general.
posted by leopard at 8:24 AM on January 28, 2014 [3 favorites]
For a fifth grader, I would not recommend writing code -- I think there's value in playing with the numbers and getting a sense for any patterns that arise. There are underpinnings of basic algebra here, as well as number theory. Math isn't just about coming up with an answer, it's also about the journey you take to find the answer. Brute force is a perfectly acceptable starting point, especially if you have nothing else to go on. Once you have some solutions, you can see how they are all related to each other (there's basically "one" solution; once you have that solution, you can repeatedly add or subtract 928 = least common multiple of 32, 58 to get more solutions) and if you wanted to do further stretch work you can play around with the parameters of the problem and discover the cases where there are no solutions at all. Doing this will build up intuition for the ideas of least common multiple and greatest common denominator and develop number sense in general.
posted by leopard at 8:24 AM on January 28, 2014 [3 favorites]
OK, if I actually put the correct equations into WolframAlpha I get the same answers as Brent Parker.
posted by grouse at 8:25 AM on January 28, 2014
posted by grouse at 8:25 AM on January 28, 2014
Not sure why you guys are getting 2784 as an answer, that is divisible by both 32 and 58 (no remainder).
posted by 3FLryan at 8:27 AM on January 28, 2014 [1 favorite]
posted by 3FLryan at 8:27 AM on January 28, 2014 [1 favorite]
You can even put in a single entry that will cause WolframAlpha to do everything for you:
x = 32a + 30 = 58b + 44, 0 ≤ x < 3000
I'd be willing to bet that the person who posed this question did not realize there was more than one answer to it.
posted by grouse at 8:32 AM on January 28, 2014 [2 favorites]
x = 32a + 30 = 58b + 44, 0 ≤ x < 3000
I'd be willing to bet that the person who posed this question did not realize there was more than one answer to it.
posted by grouse at 8:32 AM on January 28, 2014 [2 favorites]
I'll second GuyZero that these "stretch your thinking" problems for the kids generally seem to only be do-able to them with lots of enumeration looking for how things change / patterns. To that end, aubilenon really has the best process for this, and how I would have explained it to my kids (all the while thinking to myself WTF, how much time do they expect the kids to work on this (without calculators of course), and are they trying to kill any appreciation for math).
posted by nobeagle at 9:11 AM on January 28, 2014
posted by nobeagle at 9:11 AM on January 28, 2014
Here's a non-brute-force (or at least less brute force) method:
We know that the mystery number, X = 32k + 30 = 58j + 44 for some integers, j,k.
32k + 30 = 58j + 44 --> 16k + 15 = 29j +22 --> (16k - 29j) = 7
If we choose the pair (2, 1) [that is, k=2, j = 1], we see that 16k - 29j = 3. In other words, taking our function as 16k - 29j, (2,1) maps to 3.
(2,1) maps to 3 means that 3x(2,1) = (6,3) maps to 9.
It's also true that (3, 2) maps to -10.
Adding these two, we see (6,3) + (3,2) maps to -1 --> (9,5) maps to -1.
Once we have an expression for -1, getting solutions is easy.
--> (6,3) + 2x(9,5) maps to 7 --> (24, 13) maps to 7. And this is what we were looking for: a pair of numbers such that (16k - 29j) = 7. And we see that 16 x 24 - 29 x 13 = 384 - 377 = 7.
This gives us a k and j, and we can plug back into the original question:
x= 32k + 30 = 58j + 44 --> x = 32 x 24 + 30 = 58 x 13 + 44 = 798
Any combination that maps to 7 will work. For example, that (1,0) maps to 16, so (1,0) + 9x(9,5) = (82,45) maps to 7, therefore 32*82 + 30 = 58*45 + 44 = 2654 also works.
It's natural for someone who's seen the Chinese Remainder Theorem before... not so natural for a fifth grader.
posted by cgs06 at 9:34 AM on January 28, 2014 [5 favorites]
We know that the mystery number, X = 32k + 30 = 58j + 44 for some integers, j,k.
32k + 30 = 58j + 44 --> 16k + 15 = 29j +22 --> (16k - 29j) = 7
If we choose the pair (2, 1) [that is, k=2, j = 1], we see that 16k - 29j = 3. In other words, taking our function as 16k - 29j, (2,1) maps to 3.
(2,1) maps to 3 means that 3x(2,1) = (6,3) maps to 9.
It's also true that (3, 2) maps to -10.
Adding these two, we see (6,3) + (3,2) maps to -1 --> (9,5) maps to -1.
Once we have an expression for -1, getting solutions is easy.
--> (6,3) + 2x(9,5) maps to 7 --> (24, 13) maps to 7. And this is what we were looking for: a pair of numbers such that (16k - 29j) = 7. And we see that 16 x 24 - 29 x 13 = 384 - 377 = 7.
This gives us a k and j, and we can plug back into the original question:
x= 32k + 30 = 58j + 44 --> x = 32 x 24 + 30 = 58 x 13 + 44 = 798
Any combination that maps to 7 will work. For example, that (1,0) maps to 16, so (1,0) + 9x(9,5) = (82,45) maps to 7, therefore 32*82 + 30 = 58*45 + 44 = 2654 also works.
It's natural for someone who's seen the Chinese Remainder Theorem before... not so natural for a fifth grader.
posted by cgs06 at 9:34 AM on January 28, 2014 [5 favorites]
Another brute force solution, this time for Unix/Linux shell:
posted by scruss at 9:45 AM on January 28, 2014 [1 favorite]
$ comm -12 <(seq -w 30 32 3000) <(seq -w 44 58 3000) | tail -1 2654The seq commands produce two lists of numbers: (0030, 0062, …¸2974) and (0044, 0102, …, 2944), formatted with leading zeroes so that comm command would happily pick out common lines. I used tail to output only the highest number (= the last line).
posted by scruss at 9:45 AM on January 28, 2014 [1 favorite]
It won't help you learn how to do it, but on problems like this, if you just want to check your answer, the easiest instruction to give Wolfram Alpha is to use the built-in modulo (remaninder) function and just give it the problem: x mod 32 = 30, x mod 58 = 44, 0 < x < 3000. Pops out the same 798, 1726, and 2654 given by other methods.
posted by Partial Law at 10:08 AM on January 28, 2014 [1 favorite]
posted by Partial Law at 10:08 AM on January 28, 2014 [1 favorite]
My approach arose out of me screwing up my brain to try to figure out what kind of solution a teacher could realistically hope for from a 5th grader, who hasn't had algrebra yet.
Honestly, learning how to use excel (or google spreadsheets or whatever) is a lot more useful.
And actually by the time I was in fifth grade I'd had some exposure to BASIC and Logo. I imagine schools are not less computery than they were 30 years ago. This would be a pretty good exercise for someone with a "simplest introduction to programming possible" level of expertise.
posted by aubilenon at 10:32 AM on January 28, 2014 [1 favorite]
Honestly, learning how to use excel (or google spreadsheets or whatever) is a lot more useful.
And actually by the time I was in fifth grade I'd had some exposure to BASIC and Logo. I imagine schools are not less computery than they were 30 years ago. This would be a pretty good exercise for someone with a "simplest introduction to programming possible" level of expertise.
posted by aubilenon at 10:32 AM on January 28, 2014 [1 favorite]
Yeah I take back my comment about programming. Honestly if you wanted to use this exercise to develop number sense, you could do perfectly fine with smaller numbers where the solutions take less endurance to find. I was just trying to say that there's nothing wrong with manually searching for a solution by brute force, you can learn a lot that way.
The size of the numbers here reminds me of the possibly apocryphal story of Gauss's primary school teacher assigning him the task of adding up the numbers from 1 to 100. Such a task doesn't really have any value and is meant to just take up some of the student's time (except in Gauss's case he figured out a trick that allowed him to answer the question in seconds).
posted by leopard at 11:54 AM on January 28, 2014
The size of the numbers here reminds me of the possibly apocryphal story of Gauss's primary school teacher assigning him the task of adding up the numbers from 1 to 100. Such a task doesn't really have any value and is meant to just take up some of the student's time (except in Gauss's case he figured out a trick that allowed him to answer the question in seconds).
posted by leopard at 11:54 AM on January 28, 2014
So at the risk of digressing here... You can plug that inequality system into Wolfram Alpha...
This is fifth grade math. I too have been down this path. I once tried to do a closed-form solution to an equality of two integral equations to solve a 3rd grade math problem[1]. Do not assume that primary school math textbook authors really know 100% what the heck they're doing. Sometimes math problems are not easily solvable given the skills at hand of the students involved.
[1] - "A shape has a perimeter of 14. What is its area?" This was a for-reals 3rd (maybe 2nd?) grade math problem my son had.
posted by GuyZero at 5:59 AM on January 30, 2014
This is fifth grade math. I too have been down this path. I once tried to do a closed-form solution to an equality of two integral equations to solve a 3rd grade math problem[1]. Do not assume that primary school math textbook authors really know 100% what the heck they're doing. Sometimes math problems are not easily solvable given the skills at hand of the students involved.
[1] - "A shape has a perimeter of 14. What is its area?" This was a for-reals 3rd (maybe 2nd?) grade math problem my son had.
posted by GuyZero at 5:59 AM on January 30, 2014
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Then list the multiples of 58 and add 44 to those numbers.
One of those numbers should match.
posted by Brent Parker at 7:41 AM on January 28, 2014 [5 favorites]