Okay, let's see. The problem is asking for the smallest possible value of |m - n| where m and n are integers satisfying the equation 3m + 4n = 100. Hmm, so I need to find integers m and n such that when you plug them into 3m + 4n, you get 100, and among all those possible pairs, the one where the absolute difference between m and n is the smallest. Got it.

First, maybe I should express one variable in terms of the other. Let's try solving for m in terms of n. So, starting with 3m + 4n = 100. Subtract 4n from both sides: 3m = 100 - 4n. Then divide both sides by 3: m = (100 - 4n)/3. Since m has to be an integer, (100 - 4n) must be divisible by 3. So, 100 - 4n ≡ 0 mod 3. Let's simplify that. 100 mod 3 is 1 because 3*33=99, so 100 - 99 =1. So 1 - 4n ≡ 0 mod 3. Which means -4n ≡ -1 mod 3. Multiply both sides by -1: 4n ≡ 1 mod 3. Since 4 mod 3 is 1, this simplifies to n ≡ 1 mod 3. So n must be congruent to 1 modulo 3. Therefore, n can be written as n = 3k + 1 for some integer k.

Substituting back into m = (100 - 4n)/3. Let's plug n = 3k + 1 into that. So m = (100 - 4*(3k +1))/3 = (100 -12k -4)/3 = (96 -12k)/3 = 32 -4k. So m = 32 -4k and n =3k +1. Therefore, all integer solutions are of the form m =32 -4k, n=3k +1 where k is an integer. Now, we need to find the value of k that minimizes |m -n|. Let's compute |m -n| in terms of k.

Compute m -n: (32 -4k) - (3k +1) =32 -4k -3k -1=31 -7k. Therefore, |m -n|=|31 -7k|. So we need to find integer k such that |31 -7k| is minimized.

Since 7k is a multiple of 7, we can think of 31 divided by 7. 7*4=28, 31-28=3. So 31 is 7*4 +3. Therefore, 31 -7k =0 when 7k=31, but k is an integer, so the closest integers around 31/7 are 4 and 5. Let's check k=4: 31 -28=3, absolute value 3. k=5: 31 -35= -4, absolute value 4. So k=4 gives a smaller value. Let's check if there are other k's that might give a smaller value. Since 31 is between 28 (7*4) and 35 (7*5), the closest multiple is 28. So the minimal |31 -7k| is 3. Wait, but let's make sure. Let's check k=4: |31 -28|=3. k=3: |31 -21|=10. k=5: |31 -35|=4. So yes, k=4 gives the minimal value of 3. So the minimal |m -n| is 3. But wait, is that the case? Let's check.

But wait, maybe there's a mistake here. Because when we have the equation 3m +4n=100, and we expressed m and n in terms of k, but we need to ensure that m and n are integers. However, since we derived n=3k+1 and m=32-4k, these expressions already ensure that m and n are integers for any integer k. So the minimal |31 -7k| is indeed 3 when k=4. Therefore, the minimal |m -n| is 3.

Wait, but let me verify by plugging back k=4 into m and n. So m=32 -4*4=32-16=16. n=3*4 +1=13. Then |16 -13|=3. Check 3*16 +4*13=48 +52=100. Correct. So that works. Let's check the next closest k=5: m=32 -20=12, n=16. |12 -16|=4. That's correct. And k=3: m=32-12=20, n=10. |20-10|=10. So yes, 3 is the minimal. But wait, hold on. Let's check if there's a smaller value. Suppose we take k=4, which gives |m -n|=3. Is there any other k that gives a smaller difference? Let's see. Since 7k is linear, the minimal difference occurs at the closest integer to 31/7. 31 divided by 7 is approximately 4.428. So the closest integers are 4 and 5. We've checked both, and 4 gives 3, which is smaller than 5's 4. So 3 is the minimal. Therefore, the answer should be 3.

But wait a second, maybe there's another approach here. Suppose we don't parameterize n in terms of k but instead express m in terms of n. Let's see. From 3m +4n=100, m=(100 -4n)/3. Since m must be integer, 100 -4n must be divisible by 3. As before, 100 ≡1 mod3, so 4n ≡1 mod3. Since 4≡1 mod3, so n≡1 mod3. So n=3k+1. Then m= (100 -4*(3k+1))/3=(100 -12k -4)/3=(96 -12k)/3=32 -4k. So same result. Therefore, same as before. So the minimal difference is 3.

Alternatively, maybe we can think of this as a linear Diophantine equation. The equation 3m +4n=100. The general solution is given by m= m0 +4t, n= n0 -3t, where t is an integer, and m0, n0 are particular solutions. Let's find a particular solution. Let's set t=0. Then m0=32 -4k, n0=3k +1. Wait, maybe not. Let's see. Wait, the general solution for equations like ax + by =c is x =x0 + (b/d)t, y=y0 - (a/d)t, where d is gcd(a,b). Here, a=3, b=4, so gcd(3,4)=1. Therefore, the general solution is m = m0 +4t, n= n0 -3t. Let's find a particular solution. Let's set t=0. Let's try to find some m0, n0 such that 3m0 +4n0=100. Let's try m0=16, n0=13. Then 3*16 +4*13=48 +52=100. So yes, that's a particular solution. Therefore, the general solution is m=16 +4t, n=13 -3t. So any solution is of that form. Therefore, m=16+4t, n=13-3t. Then |m -n|=|16+4t - (13 -3t)|=|16 +4t -13 +3t|=|3 +7t|. Therefore, |m -n|=|7t +3|. We need to find integer t such that |7t +3| is minimized. Let's solve 7t +3=0. Then t= -3/7. So the closest integers are t=-1 and t=0. Let's check t=-1: |7*(-1)+3|=|-7+3|=4. t=0: |0 +3|=3. So the minimal value is 3 when t=0. Therefore, the minimal |m -n| is 3. So that's consistent with the previous approach. Therefore, the answer is 3.

But wait, the problem is asking for the smallest possible value of |m -n|. So according to both methods, the minimal is 3. But let me check another way. Suppose we think of m and n as variables. The equation is 3m +4n=100. We can think of this as a straight line in the m-n plane. The problem is to find integer points on this line that are closest to the line m =n, i.e., the line with slope 1. The minimal |m -n| would correspond to the closest point on the line 3m +4n=100 to the line m=n. But since we are dealing with integers, the minimal |m -n| is the minimal distance between the two lines at integer points.

Alternatively, maybe we can use the formula for the distance from a point to a line. Wait, but since we are restricted to integer solutions, maybe that's not directly applicable. But perhaps we can use the concept of minimizing |m -n| given 3m +4n=100. Let's express m in terms of n: m=(100 -4n)/3. Then |m -n|=|(100 -4n)/3 -n|=|(100 -4n -3n)/3|=|(100 -7n)/3|. So we need to minimize |(100 -7n)/3|. Let's write this as |100 -7n|/3. So to minimize this, we need to minimize |100 -7n|. Let's think about 100 divided by 7. 7*14=98, so 100=7*14 +2. So 100 ≡2 mod7. Therefore, 100 -7n ≡2 -0=2 mod7. So 100 -7n is congruent to 2 mod7, meaning that 100 -7n is not divisible by 7, so |100 -7n| can't be zero. The minimal value of |100 -7n| is the minimal distance from 100 to a multiple of 7. Since 7*14=98 and 7*15=105, 100 is 2 away from 98 and 5 away from 105. So the minimal |100 -7n| is 2, achieved when n=14. Therefore, the minimal |100 -7n| is 2, so |100 -7n|/3=2/3. But since n must be integer, and m must be integer, we need |100 -7n| to be a multiple of 3? Wait, no. Wait, |100 -7n|/3 must be an integer because |m -n| is an integer. Wait, but |100 -7n| must be a multiple of 3? Wait, no. Wait, from the previous steps, we have |m -n|=|31 -7k|, which is an integer. So when we have |100 -7n|/3, since m and n are integers, |m -n| must be an integer. Therefore, |100 -7n| must be divisible by 3. So we need to find n such that 100 -7n ≡0 mod3. Let's check 100 mod3 is 1, so 1 -7n ≡0 mod3. 7n ≡1 mod3. Since 7≡1 mod3, so 1*n ≡1 mod3. Therefore, n≡1 mod3. Which is consistent with our previous result. Therefore, n=3k +1. Then 100 -7n=100 -7*(3k +1)=100 -21k -7=93 -21k=3*(31 -7k). Therefore, |100 -7n|=3*|31 -7k|. So |m -n|=|31 -7k|. Therefore, to minimize |31 -7k|, which is the same as before. So minimal |31 -7k| is 3 when k=4. Therefore, this approach also confirms that the minimal |m -n| is 3.

Alternatively, maybe we can consider solving for n in terms of m. Let's try that. From 3m +4n=100, 4n=100 -3m, so n=(100 -3m)/4. For n to be integer, 100 -3m must be divisible by 4. So 100 -3m ≡0 mod4. 100 mod4 is 0, so -3m ≡0 mod4. Multiply both sides by -1: 3m ≡0 mod4. Since 3 and 4 are coprime, this implies m≡0 mod4. Therefore, m=4k for some integer k. Then n=(100 -3*(4k))/4=(100 -12k)/4=25 -3k. Therefore, m=4k, n=25 -3k. Then |m -n|=|4k - (25 -3k)|=|7k -25|. So we need to minimize |7k -25|. Let's compute 25/7≈3.571. So the closest integers are k=3 and k=4. Let's check k=3: |21 -25|=4. k=4: |28 -25|=3. So the minimal is 3, achieved at k=4. Therefore, m=16, n=13. So |16 -13|=3. That's the same result again.

So all these methods lead to the conclusion that the minimal |m -n| is 3. Therefore, the answer should be 3. But let me make sure there's no other pair (m,n) that gives a smaller difference. Let's check for k=4, which gives m=16 and n=13. |16 -13|=3. Let me check k=5: m=12, n=16. |12 -16|=4. k=3: m=20, n=10. |20 -10|=10. So yes, 3 is indeed the minimal. Therefore, the answer is 3.

But wait, just to be thorough, let's check if there's any other solution where |m -n| is 1 or 2. Let's see. Suppose |m -n|=1. Then m =n ±1. Let's substitute into the equation 3m +4n=100. If m =n +1: 3(n +1) +4n=100 →3n +3 +4n=100 →7n +3=100 →7n=97 →n=97/7≈13.857. Not integer. If m =n -1: 3(n -1) +4n=100 →3n -3 +4n=100 →7n -3=100 →7n=103 →n=103/7≈14.714. Not integer. So no solutions where |m -n|=1.

How about |m -n|=2? Then m =n ±2. Substitute into 3m +4n=100. If m =n +2: 3(n +2) +4n=100 →3n +6 +4n=100 →7n +6=100 →7n=94 →n=94/7≈13.428. Not integer. If m =n -2: 3(n -2) +4n=100 →3n -6 +4n=100 →7n -6=100 →7n=106 →n=106/7≈15.142. Not integer. So no solutions for |m -n|=2 either.

Therefore, the minimal possible value is indeed 3.
Therefore, the smallest possible value of $\left| m - n \right|$ is $\boxed{3}$.
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