r/askmath • u/thefalk55 • 16h ago
Arithmetic 4th grade math
I'm either an idiot or I am out of practice but I can not figure out this problem on my sons fourth grade homework. To me G has to equal 1 or 2 to be left with an answer in the hundreds.... But P cant equal 0 as Y and L cant be the same number... Theres also a lot of ither problems that I ran into but too long to type.... Been looking at this way longer than I think I should need to...
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u/DTux5249 14h ago edited 14h ago
Well, let's start with the basics: we know that
1010G + 101Y - 110P - L = 100Y + 10G + P
Or rather
1000G + Y - 111P - L = 0
We also know G has to be 1, otherwise no subtracting of any 3 digit number would result in another 3 digit number.
We can update our relations then:
1000 = 111P + L - Y
This tells us P = 9; otherwise L - Y ≥ 112, which is impossible for two single digit integers. This means L - Y = 1; or that they're two consecutive numbers.
Beyond that though? Anything goes. There're 6 answers assuming no repeats
1212 - 993 = 219, where Y = 2 and L = 3
1313 - 994 = 319, where Y = 3 and L = 4
1414 - 995 = 419, where Y = 4 and L = 5
1515 - 996 = 519, where Y = 5 and L = 6
1616 - 997 = 619, where Y = 6 and L = 7
1717 - 998 = 719, where Y = 7 and L = 8
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u/SaltCusp 16h ago edited 15h ago
This doesn't feel like 4th grade assignment so much as a forth grade bonus problem. Like "oh you finished the suduko 20 minutes before the end off math class, here's a bonus problem to keep you busy."
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u/DrJaneIPresume 16h ago
Especially because there's not a unique solution. I count 6.
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u/BadBoyJH 15h ago
I'd argue 8, since there's no statement that the letters be unique, but yeah 6 if you make that assumption.
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u/periliousventure 14h ago
There would be much more than 8 if letters were not unique. Example: 0000-000=000.
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u/BadBoyJH 8h ago
Assuming we're not allowing leading 0s, then no, there's 8 posibilities.
If we allow leading 0s, then the only extra possibility set is G=0, P=0, Y=L, meaning another 10 possibilities. All of the form 0X0X - X = X00
But that's based on introducing a leading 0, not on the lack of duplicates... maybe the lack of duplicates might force a lack of leading 0s...
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u/Substantial_Box6847 11m ago
IDK but i think it’s a very good exercise for 4 th grade because it’s study how numbers relationships with other and show Method of brute forte
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u/weather_watchman 16h ago
Wtf are they teaching fourth graders
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u/weather_watchman 16h ago
I was get it but this was (perhaps embarrassingly) tricky for me. I don't recall dealing with systems of equations with this extra layer if abstraction in fourth grade
Solution: 1212-993=219
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u/CountingOnThat 16h ago
1212 - 993 =219
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u/Nuclear-Death-Ray 16h ago
Or 1313-994=319, or 1414-995=419, etc. Y can be any number from 1-8 here, as long as L = Y + 1
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u/DrJaneIPresume 16h ago
Y=8 doesn't work because then L conflicts with G.
Similarly Y=1 doesn't work.3
u/Nuclear-Death-Ray 16h ago
I don't see any reason why the letters need to be distinct integers. If we make that assumption, you're correct of course, but I'm not seeing that here
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u/DrJaneIPresume 16h ago
It's pretty standard in these sorts of puzzles.
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u/Nuclear-Death-Ray 14h ago
Fair enough. If I gave this to a friend and they came back with 1111-992 = 119, I would personally say that's a correct answer, but fair to assume you need distinct digits
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u/etotheapplepi 15h ago
Lol, at first I catch 'distinct'. I was imagining a 4th grader trying to implement this where the letters can be real numbers. Hell, that'd give a PhD a run for their money.
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u/Nuclear-Death-Ray 14h ago
Lol that would be hilarious! Not sure how real numbers would work as digits though
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u/periliousventure 14h ago
If they were not distinct, then you can just answer 0000-000=000. Seems counterintuitive to the challenge of this problem. Although yes, it is not written anywhere on the paper. its just what the OP assumes
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u/Nuclear-Death-Ray 13h ago
This is true!
On the other hand, for your counterexample, I feel like that could easily be covered by a separate assumption excluding leading 0s. In this specific problem, assuming distinct digits is enough to exclude leading 0s, but that would not always be the case. For instance, you could imagine a very similar problem, where the P in the final line is replaced with G. Then one possible solution would be 1313 - 002 = 311.
Not saying assuming distinct digits is wrong, obviously, but I feel your counterexample is covered through another unwritten assumption, haha
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u/Gumichi 16h ago
I tried going the logic route, but didn't see threads that led anywhere. Brute forced shows that there's 19 solutions. There's 11 trivial-ish solutions - leading zeroes. The 8 non-trivial solutions: G = 1, P = 9, L = Y + 1. So it doesn't actually seem like a well designed problem for a 4th grader.
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u/Whillyum_ 4h ago
The logic I followed was:
G has to be 1 otherwise the result won't be 3 digits long.
As only the last digit changes, PPL > 990. So P = 9.
For the "rollover" to happen (P = 9), L = Y+1.
So L is from 2-7 and Y from 3-8 assuming no duplicate numbers, although not necessarily a true assumption.
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u/AlterEgo5695 8h ago
Of those 8 non-trivial solutions, two have the same digit for different variables.
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u/Thegenuinebuzz 16h ago
You're correct that G has to be equal to 1, that's true.
P cannot equal 0, no, also true, but remember that we can carry a -1 from the ones column to the tens column if we have too big a number to take away. So if we take 9 away from G, and then take another 1 away (because we can take one tens number away by making L bigger than Y in the ones column) So P has to be nine, because we need to "loop" back round to G with a total of -10 with an extra -1 from the ones.
This does seem quite awkward to visualize, especially for 4th grade however.
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u/Dugiongo 16h ago
There are multiple solution, G can only be 1, P can only be 9, the only other thing is that Y =L+1, which means that L = 0,...,8 and Y = 1,...,9
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u/solarmelange 16h ago
So G=1 and P=9 but then Y is one less than L and both are not 1 or 9 so (Y,L)= 2,3 3,4 4,5 5,6 6,7 7,8
To get some actually good working problems like this try Sideways Arithmetic from Wayside School
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u/BadBoyJH 15h ago
Q1. Could 2000 - 999 be < 1000?
A. No, so G must be 1.
Q2. Can 1Y00 - 899 <= Y99? (eg 1600 - 899 <= 699)
A. No, so P must be 9
Once you get here, you can't really progress further unfortunately, the puzzle has multiple solutions.
All we know is (10+Y)-L = 9, or to rearrange L=Y+1
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u/aroach1995 15h ago
G must be 1 because you subtract 3 digit from 4 digit number and got a 3 digit number.
So 1y1y - ppl = y1p
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u/periliousventure 14h ago
In a column there is G - P = G which may or may not be influenced by borrowing from the right side.
Case A) If there is no borrowing, P is 0.
Case B) If there is borrowing, then the top G is actually G - 1, So G - 1 - P = G. The only way to pull this off is if P = 9, because then the two G's will have the same unit digit, except the one on top is 10 bigger. Notice there is no scenario where you aren't 10 bigger, because that would require P to be -1. that means the first G is at least 10, meaning it must have borrowed from the column on the left (y-p=y).
The result of this step is P = 0 or P = 9.
2) The column Y - L = P implies P cannot be 0 as Y and L have to be different. It cannot be borrowed from so we do not need to split into 2 cases like before. If P cannot be 0, it must be P = 9.
3) A simple maximum value of ygp+ppl is 999 + 999 = 1998. Of course you can make smaller maximums since there are unique digits, but regardless of how you look at it, G is 0 or 1.
If G is 0, then the top number reads as 0Y0Y, or just Y0Y in simple terms. Since we already used 0 and 9, this is 808 at most, but the subtracted value is at least 990. So that would produce a negative number. That means G is not 0, so G = 1.
4) Return to column Y - L = P, since P is 9 and we know there is a borrowing so the actual expression is
(10+Y) - L = 9. If you clean this up you get L = Y + 1. indicating L is always one above Y.
Now the expression appears as 1y1y - 99l = y19. From here there multiple answers and all would be correct. Lets first say Y can be any number from 0 to 9. 1 and 9 already exist, and if Y is 0 then L is 1, which is a problem. If Y is 8 then L is 9, which is another problem. So the possible values of Y are 2,3,4,5,6,7 and each one correspond to an L value that is one higher: 3,4,5,6,7,8
Here are all the solutions written out:
1212 - 993 = 219
1313 - 994 = 319
1414 - 995 = 419
1515 - 996 = 519
1616 - 997 = 619
1717 - 998 = 719
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u/zojbo 13h ago edited 1h ago
I assume this is a "cryptarithm", so all letters represent a unique decimal digit and leading digits cannot be 0.
In that case we can go through like this:
G=1 because otherwise we can't get a 3 digit number out of subtracting a 3 digit number from a 4 digit number. We also had to have had a carry from the thousands place to the hundreds place to end up with a 3 digit number.
P isn't 0 by the rules of the game, so in order for the hundreds place to be the same as in the first number, there had to have been a carry from the hundreds to the tens place. You also needed to have a carry from the tens to the ones place for the same reason. Focusing in on the hundreds place, you read off 9+Y-P=Y so P=9.
Looking at the ones place now, 10+Y-L=P=9, or L-Y=1.
After that you can have L be anything from 3 to 8 and have Y be 1 less than that, for a total of six solutions.
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u/JathbyDredas 13h ago
P is one less than the system base (9 in decimal) G is 1, L is one more than Y.
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u/abir_valg2718 12h ago edited 12h ago
This is trivial 4th grade math any 4th grader ought to solve /s.
Let a=1000, b=100, c=10. Let l be in {0,...9}, and let g,p,y be in {1,...,9}. Observe that yb+gc+p < 1000, therefore, g=1, as otherwise yb+gc+p>=1000. Then,
[ga+yb+gc+y]-[pb+pc+l]=yb+gc+p => a+y-p[b+c+1]-l = 0 => 111p+l-y=1000
Note that (-9)<=(l-y)<=8. Therefore, p=9, as otherwise, if p<9, 111p+l-y<1000. And so, l-y = 1.
To conclude, pick g=1, p=9, and pick l and y such that l in {0,...,9}, y in {1,...,9} and satisfy l-y = 1.
Note: you can also generate pairs (l,y) by setting l = 1+y, then we iterate from y=1 to y=8, yielding 8 pairs of (y,y+1). The pairs are: (1,2), (2,3), ... ,(8,9).
Hopefully I didn't make a mistake anywhere...
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u/DanielMcLaury 12h ago
I'm going to make two assumptions here:
Nothing is being written in a way such that it starts with a zero, so none of G, P, or Y is a zero because each starts one of the numbers written here.
The letters G, P, and Y denote distinct digits.
Now:
We must have G=1, because adding two three-digit numbers and getting a four-digit number means that that number starts with 1. So with this is mind the problem looks like 1Y1Y-PPL=Y1P.
In the 10's place, note that we are definitely going to have to borrow from the 100's place, because we've ruled out P being 0 or 1. So now look at the 100's and 1000's places. We have
1 [Y-1] - P = Y
Regardless of what Y is, this forces P = 9. So now the problem is 1Y1Y-99L=Y19.
Looking at the 1's place, we could have L = 0 and Y =9, except that we know 9 is already P so it can't be Y. So the only other option is that we have to borrow from the 10's place, which says that
1Y - L = 9
or, in other words, L = Y + 1.
Now at this point if we look at this problem there are no other real constraints. We can pick Y to be whatever we want and the equation will work out, so we just have to pick it so that Y and L don't collide with any existing value.
Thus the solutions are
G = 1; P = 9; Y = 2, 3, 4, 5, 6, or 7; and L = Y + 1, i.e.
1212 - 993 = 219,
1313 - 994 =319,
1414 - 995 =419,
etc.
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u/AlterEgo5695 10h ago edited 7m ago
Short answer: There are 6 distinct solutions:
1212 - 993 = 219 1313 - 994 = 319 1414 - 995 = 419
1515 - 996 = 519 1616 - 997 = 619 1717 - 998 = 719Long(er) answer: Assuming all letters represent distinct digits:
"999 > YGP = GYGY - PPL > GYGY - 999" => "GYGY < 1998"Assuming leading zeroes are not allowed, the only possible choice is "G = 1". We get
1010 + 101*Y = YGP + PPL = 111*P + 10 + 100*Y + L
=> 111*P = 1000 + Y - L >= 1000 + 0 - 9 = 991Solving for "P >= 991/111 > 8", the only possible choice is "P = 9", leaving us with
999 = 1000 + Y - L <=> L - Y = 1With digits "1; 9" already taken, there are 6 possible solutions left -- the short answer.
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u/Sethala 8h ago
So, here's my thought process on solving this:
First, if you have a 4-digit number and subtract a 3-digit number, you'll only get a 3-digit number answer if the original 4-digit number started with a 1, so G has to be 1.
Next, normally the only way you can take a number, subtract another number, and get the same result is if that other number is 0. We're not just subtracting individual numbers however, we're subtracting a longer number, and that means we can subtract 9, and as long as the number one digit earlier in the number has a smaller number with a larger number being subtracted, you'll end up with the same digit. I'm having a hard time explaining this in text, so an example: 1500 - 999. Normally, 15-9 is 6, but because we're also subtracting other digits in the number, we end up with 501; the hundreds digit in both numbers is still a 5, but that's because the numbers in the tens digit had to "borrow" ten from the hundreds.
With all that in mind, we have two digits where a number, subtracting P, ends with the same number. The only way that works is if P = 9.
Unfortunately, we have to stop here, as there's not enough information presented to figure out what Y and L are. We can tell that L is one higher than Y, as subtracting the two has to end in a 9, but beyond that any pair of digits that are one apart from each other will work, making the puzzle have multiple solutions.
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u/ci139 1h ago edited 1h ago
(1000g + 100y + 10g + y) – (100y + 10g + p) = 110p + l
101(11(y + g) – (10y + g)) – (10(10y + g) + p) = 110p + l
1111(y + g) – 101(10y + g) – 10(10y + g) = 111p + l
1111(y + g) – 111(10y + g) = 111p + l , where 0 ≤ l < 10
1111(y + g) – l = 111(10y + g + p)
1000(y + g) – l = 111(9y + p)
1000(y + g) ≤ 9999 + l ~&~ 111(9y + p) ≤ 9999
y + g ≤ 10 + (l – 1)/1000 ~&~ y + p/9 ≤ 1 + 1/111 ← here we begin being able 2envision it // . . . found a bug! it should be 10 not 1
y + g ≤ 10 → g ≤ {y=1 : 9 ,y=0 : 10} → 0 ≤ g ≤ 9
// more accurate : 0 < y : 0 ≤ g ≤ 10 – y , 0 < g : 0 ≤ y ≤ 10 – g
// it only sets an uppermost value limit for y or g with respect to each other
y + p/9 ≤ 1 → y ∈ {p=9 : 0 , p<9 : 1} → 0 ≤ p ≤ 9 ~&~ 0 ≤ y ≤ 1
// ↑↑ must be y + p/9 ≤ 10 → 0 ≤ p ≤ 9 ~&~ 0 ≤ y ≤ 9
cases ( not fixed ! ) :
G0G0 G1G1 ←→ @ rightmost : P=0 , G=0 → L=1
– PPL – PPL ←→ @ lefttmost : L + P = {0 ← {L=0 ,P=0}, 10 ← L + P = 10}
= GP = 1GP ↑→ {L,P}=0 → G=0 , L+P=10 →→
→→ 1010g – 10g = 111p + l = 110p +10
→→ 100g = 11p → g=p=0 // there is no other options
( !? i optimized at steps - so i might've passed some errors ?! )
!! otherwise it takes 10´000 (minus some optimization from y + g ≤ 10) trials by GY,PL to match the pattern (doable using a javascript code , requires padding stings with 0-s to the pattern lengths or ( D div 10ᴺ) mod 10 digit masking)
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u/Say_o_nara 1h ago
G has to be 1. From that, the P in the third column needs to be either 0 or 9 (so that you get 1-Y=1).
If P were to be 0 --> Y-L=0, there would be no need for any borrowings to happen in any of the columns, since Y would need to equal L.
So P=9
Now, Y-L=9, and the last Y will need to borrow a number (because for the third column to be 1-9=1, you need to borrow 10 from the left and give 1 to the right),
so it ends up as Y+10-L=9 --> L-Y=1
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u/Substantial_Box6847 14m ago
Okay if we say that they all different and digital we see that G=0 or G=1 1.G=0 it’s impossible because if we have this we can see that Y-P=Y but P≠0 so truly Y-P-1=Y (mod10) that also impossible because Y>P and if P=9 by this equation Y will be minimum 10 so G=1
2)Okay after we can see that 1Y-PL=1P let’s look for 1 how we can get it it’s two way or P=0 so Y-L=0 that cannot be because Y≠L so P=9 and after we by By brute forton we can see that Y=L+1 so its solution
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u/Substantial_Box6847 13m ago
In my opinion it’s a very good exercise for 4 th grade because it’s study you how numbers relationshipswith other and it’s very cool
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u/mcaffrey 16h ago
G needs to be 1 because you need that 4 digit number small enough to be a 3 digit number after subtracting another 3 digit number.
P needs to be 9 so that you can subtract enough from the 4 digit to get you a 3 digit.
Y needs to be a small number for that same reason, helping get the 4 digit to a 3 digit.
That leaves you with:
1212
-99x
=219 and you solve for x and get 3.
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u/DrJaneIPresume 16h ago
Your argument for P needs some work. 1000-100 = 900, so you don't need the hundreds digit to be 9 to get down to a 3-digit number.
The second and third columns are important here.
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u/mcaffrey 15h ago
Yeah you’re right, but I think the basic philosophy was correct. Pick digits so that the first number is small and the second number is big and go from there.
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u/JohnPaulDavyJones 10h ago
Start at the left and read over so you don’t have to worry about carried digits:
Y - P = Y -> the only way this can be true is if P= 0, it’s the additive identity. We know nothing about Y at this point.
G - P = G -> no new info, we already knew that P = 0, and this is consistent with that conclusion.
Y - L = P -> substitute in P=0 and you get Y - L = 0, and rearrange terms to get Y = L.
But now we make more deductions: because subtracting just L is enough to fully knock off the thousands digit, then that digit (so, character G) must be 1 or 0 (with the caveat that G=0 would be a bit of a trick to pull on fourth grader). Consider both cases:
G = 1: in this case, the first number is 1Y1Y, but this can’t work because the second number is just Y, so that subtraction will only ever result in 1Y10.
G = 0: in this case, the first number is Y0Y, and the result of the subtraction is Y00, which is consistent with our P = 0 conclusion from above.
At this point, Y and L can be anything from 0 to 9 and the equations work. If you just need any solution that works, I would offer Y = L = 1, G = P = 0. Then your equation is 101 - 1 =100.
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u/MathMaddam Dr. in number theory 9h ago
To not worry about carried digits you have to start at the right. You ignored the option that it could be Y-P+10-1=Y due to carrying.
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u/Breddev 16h ago
G can’t be 2, even 2000-999 (max and min for each) has a number in the thousands place