Have you heard of Euclidean Algorithm?
( 4 6 times 7 + 7 + 7 + ⋯ + 7 ) − ( 1 1 times 2 9 + 2 9 + 2 9 + ⋯ + 2 9 ) ( 2 1 times 7 + 7 + 7 + ⋯ + 7 ) − ( 5 times 2 9 + 2 9 + 2 9 + ⋯ + 2 9 ) = 3 = 2
Is it also possible to find positive integers m and n such that ( m times 7 + 7 + 7 + ⋯ + 7 ) − ( n times 2 9 + 2 9 + 2 9 + ⋯ + 2 9 ) = 1 ?
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6 solutions
That's much easier than my idea!
7 46 322 29 11 319 3; 7 21 147 29 5 145 2; = > 7 25 175 29 6 174 1
From Bezout's Lemma it is always possible to find two integers x,y such that ax+by=G.C.M. (x,y) with a,b natural numbers. Since 7 and 29 are coprime numbers and the GCM is 1 the ans is Y E S
Moreover, we can guarantee that $m$, and $n$ are positive since if $(a,b)$ is a pair of Bezóut cofficients, so is $(a+y,b-x)$. We can repeat this until $b$ is negative and $a$ is positive, as required.
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You are right! Now the solution is complete, thanks. Anyway I suggest you to use LaTex intead of Tex.
7x4 = 28.
7x4 - 29x1 = -1.
7(-4) - 29(-1) = 1.
7x29 - 29x7 = 0.
Adding the last two, 7x25 - 29x6 = 1. QED
Can 7m - 29n = 1 ?
7[m - 4n] - n = 1 is possible when m > 4n, n and m pos integers
For m = 4n + 1, n = 6 and m = 25
The equation can be re-written as
7 m − 2 9 n = 1
or
7 m − 2 8 n − n = 1
or
7 ( m − 4 n ) − n = 1
Since n is an integer, 4 n is also an integer. Since m is an integer, ( m − 4 n ) is also an integer. For the sake of simplicity, let us assume k = m − 4 n . The above equation can then be written as
7 k − n = 1
The smallest possible integer values of k and n that satisfy the above equation, given that n is a positive integer are
k = 1 and n = 6
Plugging back the value of k and n in the previous equation k = m − 4 n , we get m = 2 5 A quick cross-check shows that 7 . 2 5 − 2 9 . 6 = 1
4x7 - 1x29 = -1
Multiply by 6
24x7 - 6x29 = -6
Add one more 7
25x7 - 6x29 = 1
Subtract the given equations, which gives us 2 5 ⋅ 7 − 6 ⋅ 2 9 = 1 .