If this condition is satisfied, then the above congruence has exactly $d$ solutions modulo $m$, and that, $$x = x_0 + \frac{m}{d} \cdot t, \quad t = 0, 1, \ldots, d-1.$$. Find all solutions to the linear congruence $5x \equiv 12 \pmod {23}$. is the solution to the initial congruence. This was really helpful. We first note that $(5, 23) = 1$, hence we this linear congruence has 1 solution (mod 23). If b is not divisible by g, there are no solutions. These cookies will be stored in your browser only with your consent. Linear Congruences ax b mod m Theorem 1. It turns out x = 9 will do, and in fact that is the only solution. Observe that Hence, (a) follows immediately from the corresponding result on linear … Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. We need now aplly the above recursive relation: Finally, solutions to the given congruence are, $$x \equiv 61, 61 + 211, 61 \pmod{422} \equiv 61, 272 \pmod{422}.$$. Thus: Hence our solution in least residue is 7 (mod 23). x ≡ (mod )--- Enter a mod b statement . By finding an inverse, solve the linear congruence $31 x\equiv 12 \pmod{24}.$ Solution. However, if we divide both sides of the congru- Here we use the algorithm to solve: 5x−3y=1 (5x≡1 (mod 3), which is easily solved by testing. In the second example, the order is reversed because the coefficient of the x k is smaller than the coefficient of the y. This field is denoted by $\mathbb{Z}_p$. The one particular solution to the equation above is $x_0 = 0, y_0 = -4$, so $3x_0 – 2y_0 = 8$ is valid. We can repeat this process recursively until we get to a congruence that is trivial to solve. Linear Congruence Calculator. We also use third-party cookies that help us analyze and understand how you use this website. For example, 8x ≡ 3 (mod 10) has no solution; 8x is always an even integer and so it can never end in 3 in base 10. To the solution to the congruence $a’v \equiv b’ \pmod{m’}$, where $a’ = \frac{a}{d}, b’ = \frac{b}{d}$ and $m’ = \frac{m}{d}$, can be reached by applying a simple recursive relation: $$v_{-1}= 0, \quad v_0 = 1, \quad v_i = v_{i-2} – q_{i-1}, \quad i= 1, \ldots, k,$$. However, linear congruences don’t always have a unique solution. In general, we may have to apply the algorithm multiple times until we get down to a problem small enough to solve easily. Section 5.1 Solving Linear Congruences ¶ Our first goal to completely solve all linear congruences \(ax\equiv b\) (mod \(n\)). The answer to the first question depends on the greatest common divisor of a and m. Let g = gcd(a, m). Since $\frac{m}{d}$ divides $m$, that by the theorem 6. and that is the solution to the given congruence. With the increase in the number of congruences, the process becomes more complicated. If u 1 and u 2 are solutions, then au 1 b (mod m) and au 2 b (mod m) =)au 1 au 2 (mod m) =)u 1 u In case the modulus is prime, everything you know from linear algebra goes over to systems of linear congruences. Browse other questions tagged linear-algebra congruences or ask your own question. We assume a > 0. You can verify that 7*59 = 413 so 7*59 ≡ 13 (mod 100). This website uses cookies to ensure you get the best experience on our website. That help us the … Since gcd(50, 105) = 5 and 65 is divisible by 5, there are 5 solutions. Then x = (100*4 + 13)/7 = 59. Proposition 5.1.1. Hence -9 can be used as an inverse to our linear congruence $5x \equiv 12 \pmod {23}$. Adding and subtracting rational expressions, Addition and subtraction of decimal numbers, Conversion of decimals, fractions and percents, Multiplying and dividing rational expressions, Cardano’s formula for solving cubic equations, Integer solutions of a polynomial function, Inequality of arithmetic and geometric means, Mutual relations between line and ellipse, Unit circle definition of trigonometric functions, Solving word problems using integers and decimals. In this case, $\overline{v} \equiv v_k \pmod m’$ is a solution to the congruence $a’ \overline{v} \equiv 1 \pmod{m’}$, so $v \equiv b’ v_k \pmod{m’}$ is the solution to the congruence $a’v \equiv b’ \pmod{m’}$. solve the linear congruence step by step. For another example, 8x ≡ 2 (mod 10) has two solutions, x = 4 and x = 9. In the table below, I have written x k first, because its coefficient is greater than that of y. This is a linear Diophantine equation and it has a solution if and only if $d = \gcd(a, m)$ divides $b$. That is, assume g = gcd(a, m) = 1. Let d = gcd(c,m), and choose q, r 2Z such that c = dq and m = d r. If b is a solution to (1), then it is also a Our rst goal is to solve the linear congruence ax b pmod mqfor x. Unfortu-nately we cannot always divide both sides by a to solve for x. Necessary cookies are absolutely essential for the website to function properly. The result is closely related to the Euclidean algorithm. Let x 0 be any concrete solution to the above equation. This reduces to 7x= 2+15q, or 7x≡ … This problem has been solved! Solution to a linear congruence equation is equivalent to finding the value of a fractional congruence, for which a greedy-type algorithm exists. solutions of a linear congruence (1) by looking at solutions of Diophantine equation (2). Let's use the division algorithm to find the inverse of modulo : Hence we can use as our inverse. By the Euler’s theorem, $$a^{\varphi (m)} \cdot b \equiv b \pmod m.$$, By comparing the above congruence with the initial congruence, we can show that, $$x \equiv a^{\varphi (m) -1} \cdot b \pmod m$$. Update: Here are the posts I intended to write: systems of congruences, quadratic congruences. The linear congruence equation ax = b (mod n) may be rewritten as ax1 = b - nx2 where x1, x2 -E- Z. Solve the linear system sa+ tm= 1: Then sba+ tbm= b: So sba b (mod m) gives the solution x= sb. Therefore, solution to the congruence $3x \equiv 8 \pmod 2$ is, $$x = x_0 + 2t, \quad t \in \mathbb{Z},$$. 24 8 pmod 16q. The algorithm can be formalized into a procedure suitable for programming. Theorem 2. There are several methods for solving linear congruences; connection with  linear Diophantine equations, the method of transformation of coefficients, the Euler’s method, and a method that uses the Euclidean algorithm…, Connection with  linear Diophantine equations. The brute force solution would be to try each of the numbers 0, 1, 2, …, m-1 and keep track of the ones that work. Now what if the numbers a and m are not relatively prime? Since 100 ≡ 2 (mod 7) and -13 ≡ 1 (mod 7), this problem reduces to solving 2y ≡ 1 (mod 7), which is small enough to solve by simply sticking in numbers. We find y = 4. Example 1. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. In particular, (1) can be rewritten as We must now see how many distinct solutions are there. A linear congruence  $ax \equiv b \pmod m$ is equivalent to. The CRT is used solve systems of congruences of the form $\rm x\equiv a_i\bmod m_{\,i}$ for distinct moduli $\rm m_{\,i}$; in our situation, there is only one variable and only one moduli, but different linear congruences, so this is not the sort of problem where CRT applies. To the above congruence  we add the following congruence, By dividing the congruence by $7$, we have. The solution of a linear congruence can be found in the Wolfram Language using Reduce[a*x == b, x, Modulus -> m]. the congruences whose moduli are the larger of the two powers. This means that there are exactly $d$ distinct solutions. Solve The Linear Congruence Step By Step ; Question: Solve The Linear Congruence Step By Step . This simpli es to 5t 2 (mod 8), which we solve by multiplying both sides by We can repeat this process recursively until we get to a congruence that is trivial to solve. Construction of number systems – rational numbers. Substituting this into our equation for yields: Thus it follows that , so is the solution t… The complete set of solutions to our original congruence can be found by adding multiples of 105/5 = 21. Lemma. The notion of congruences was first introduced and used by Gauss in his Disquisitiones Arithmeticae of 1801. Let’s talk. Solve the following congruence: Since $\gcd(7, 15) = 1$, that the given congruence has a unique solution. The congruence $ax \equiv b \pmod m$ has solutions if and only if $d = \gcd(a, m)$ divides $b$. 1/15 15 22 31 47 Fermat's Little Theorem is often used in computing large powers modulo n, 1 point under some conditions. Previous question Next question Get more help from Chegg. The algorithm above says we can solve this by first solving 21y ≡ -13 (mod 10), which reduces immediately to y ≡ 7 (mod 10), and so we take y = 7. stated modulo 90, and so the most satisfying answer is given in terms of congruence classes modulo 90. Solve x^11 + x^8 + 5 mod(49) I have a lot of non-linear congruence questions, so I need an example of the procedure. Solving Linear Congruence by Finding an Inverse. Although Bill Cook's answer is completely, 100% correct (and based on the proof of the Chinese Remainder Theorem), one can also work with the congruences successively; we know from the CRT that a solution exists. For example, we may want to solve 7x ≡ 3 (mod 10). Solve the following congruence: Since $\gcd(3, 2) = 1$, that, by the theorem 1., the congruence has a unique solution. linear congruences (in one variable x). Proof. One or two coding examples would’ve been great, though =P, this really helpful for my project. Solve Linear Congruences Added May 29, 2011 by NegativeB+or- in Mathematics This widget will solve linear congruences for you. This entails that a set of remainders $\{0, 1, \ldots, p-1 \}$ by dividing by $p$, whit addition and multiplication $\pmod p$, makes the field. Since 7 and 100 are relatively prime, there is a unique solution. See the answer. Existence of solutions to a linear congruence. Example 3. The result is closely related to the Euclidean algorithm. These cookies do not store any personal information. So we first solve 10x ≡ 13 (mod 21). We can calculate this using the division algorithm. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. Email: donse[email protected] Tel: 800-234-2933; Let $x_0$ be any concrete solution to the above equation. First, let’s solve 7x ≡ 13 (mod 100). most likely will be coming back here in the future, Thank you! The equation 3x==75 mod 100 (== means congruence), input 3x into Variable and Coeffecient, input 100 into modulus, and input 75 into the last box. Thus: Hence for some , . If (a;m) = 1, then the congruence ax b mod mphas exactly one solution modulo m. Constructive. The algorithm says we should solve 100y ≡ -13(mod 7). If the number $m =p$ is a prime number, and if $a$ is not divisible by $p$, then the congruence $ax \equiv b \pmod p$ always has a solution, and that solution is unique. Solve the following congruence: $$x \equiv 5^{\varphi(13) -1} \cdot 8 \pmod{13}.$$, Since $\varphi (13) =12$, that it follows, By substituting it in $x \equiv 3^{11} \cdot 8 \pmod{13}$ we obtain. A linear congruence is the problem of finding an integer x satisfying, for specified integers a, b, and m. This problem could be restated as finding x such that, Two solutions are considered the same if they differ by a multiple of m. (It’s easy to see that x is a solution if and only if x + km is a solution for all integers k.). 1 point In order to solve the linear congruence 15x = 31 (mod 47) given that the inverse of 15 modulo 47 is 22, what number should be multiplied to both sides in the given congruence? Thanks :) Then $x_0 \equiv b \pmod m$ is valid. Now substitute for x in the second congruence: 3(6+7t) 4 (mod 8). Theorem. By subtracting obtained equations we have: It follows: $x – x_0 = 2t, t \in \mathbb{Z}$. first place that I’ve understood it, after looking through my book and all over the internet Example. If b is divisible by g, there are g solutions. 10 15 20 25 30 None of the above 1 point Using the binary modular exponentiation algorithm (as shown in lecture, Algorithm 5 in Section 4.2) to … Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. If this condition is met, then all solutions are given with the formula: $$x = x_0 + \left (\frac{m}{d} \right) \cdot t, \quad y= y_0 \left (\frac{a}{d} \right) \cdot t,$$. Thanks a bunch, Your email address will not be published. Then x 0 ≡ … Solution: We have gcd(42,90) = 6, so there is a solution since 6 is a factor of 12. Find more at https://www.andyborne.com/math See how to solve Linear Congruences using modular arithmetic. We first put the congruence ax ≡ b (mod m) in a standard form. It is mandatory to procure user consent prior to running these cookies on your website. Then the solutions to ax ≡ b (mod m) are x = y + tm/g where t = 0, 1, 2, …, g-1. This category only includes cookies that ensures basic functionalities and security features of the website. But opting out of some of these cookies may affect your browsing experience. Therefore, $x_1$ and $x_2$ are congruent modulo $m$ if and only if $k_1$ and $k_2$ are congruent modulo $d$. The proof for r > 2 congruences consists of iterating the proof for two congruences r – 1 times (since, e.g., € ([m 1,m 2],m 3)=1). For example 25x = 15 (mod 29) may be rewritten as 25x1 = 15 - 29x2. If $d \nmid b$, then the linear congruence $ax \equiv b \pmod m$ has no solutions. Featured on Meta “Question closed” … Multiply the rst congruence by 2 1 mod 7 = 4 to get 4 2x 4 5 (mod 7). The given congruence we write in the form of a linear Diophantine equation, on the way described above. The most important fact for solving them is as follows. $3x \equiv 8 \pmod 2$ means that $3x-8$ must be divisible by $2$, that is, there must be an integer $y$ such that. Since $\gcd(6,8) = 2$ and $2 \nmid 7$, there are no solutions. Let $a$ and $m$ be natural numbers, and $b$ an integer. If y solves this new congruence, then x = (my + b)/ a solves the original congruence. This means that a linear congruence also has infinitely many solutions which are given in the form: $$x = x_0 + \left( \frac{m}{d}\right) \cdot t, \quad t \in \mathbb{Z}.$$. Menu. If not, replace ax ≡ b (mod m) with –ax ≡ –b (mod m). Solving linear congruences is analogous to solving linear equations in calculus. Solving the congruence $ax \equiv b \pmod m$ is equivalent to solving the linear Diophantine equation $ax –  my = b$. Example 4. The linear congruence Solve the following congruence: We must first find $\gcd(422, 186)$ by using the Euclidean algorithm: Therefore, $\gcd(422, 186) = 2$. Recall that since $(31,24)=1$ and $1|12$ there is exactly one incongruent solution modulo $24.$ To find this solution let’s use the definition of congruence, … Suppose a solution exists. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Since $2 \mid 422$, that the given congruence has solutions ( it has exactly two solutions). Which of the following is a solution for x? Get 1:1 help now from expert Advanced Math tutors For this purpose, we take any two solutions from that set: $$x_1 = x_0 + \left( \frac{m}{d}\right) \cdot k_1,$$, $$x_2 = x_0 + \left (\frac{m}{d}\right)  \cdot k_2.$$, $$x_0 + \left( \frac{m}{d} \right) \cdot k_1 \equiv x_0 + \left( \frac{m}{d} \right) \cdot k_2 \pmod m$$, $$\left( \frac{m}{d} \right) \cdot k_1 \equiv \left( \frac{m}{d} \right) \cdot k_2 \pmod m.$$. Solving the congruence 42x ≡ 12 (mod 90) is equivalent to solving the equation 42x= 12+90qfor integers xand q. Let , and consider the equation (a) If , there are no solutions. This says we can take x = (105*7 + 65)/50 = 16. Example 1. Gauss illustrates the Chinese remainder theorem on a problem involving calendars, namely, "to find the years that have a certain period number with respect to the solar and lunar cycle and the Roman indiction." Linear Congruence Calculator. Therefore, if $ax \equiv b \pmod m$ has a solution, then there is infinitely many solutions. So if g does divide b and there are solutions, how do we find them? This is progress because this new problem is solving a congruence with a smaller modulus since a < m. If y solves this new congruence, then x = (my + b)/a solves the original congruence. (b) If , there are exactly d distinct solutions mod m.. If we need to solve a system of three linear congruences with one unknown, then we need first solve a system of two linear congruences, and then see which of the obtained solutions also satisfy the third congruence. Expert Answer . My colleagues and I have decades of consulting experience helping companies solve complex problems involving data privacy, math, statistics, and computing. Then first solve the congruence (a/g)y ≡ (b/g) (mod (m/g)) using the algorithm above. So, we restrict ourselves to the context of Diophantine equations. The algorithm can be formalized into a procedure suitable for programming. 1 point Solve the linear congruence 2x = 5 (mod 9). For instance, solve the congruence $6x \equiv 7 \pmod 8$. So the solutions are 16, 37, 58, 79, and 100. The method of  transformation of coefficients consist in the fact that to the given equation we add or subtract a well selected true congruence. Solving the congruence a x ≡ b (mod m) is equivalent to solving the linear Diophantine equation a x – m y = b. Linear Congruences In ordinary algebra, an equation of the form ax = b (where a and b are given real numbers) is called a linear equation, and its solution x = b=a is obtained by multiplying both sides of the equation by a1= 1=a. For daily tweets on algebra and other math, follow @AlgebraFact on Twitter. If d does divide b, and if x 0is any solution, then the general solution is given by x = x Here, "=" means the congruence symbol, i.e., the equality sign with three lines. Now let’s find all solutions to 50x ≡ 65 (mod 105). First, suppose a and m are relatively prime. Your email address will not be published. This simpli es to x 6 (mod 7), so x = [6] 7 or x = 6 + 7t, where t 2Z. How do I solve a linear congruence equation manually? // Example: To solve € … 12 ( mod 10 ) get more help from Chegg symbol, i.e. the. 2011 by NegativeB+or- in Mathematics this widget will solve linear congruences and how to solve …! Multiples of 105/5 = 21 4 and x = ( 100 * 4 13... The form of a linear congruence equation manually mod 21 ) equation of the.! Example, the process becomes more complicated congruence by $ 7 $, that the given we. \Mid 422 $, we may want to solve use the division to... Congruences whose moduli are the larger of the y coding examples would ’ ve been great, =P... Enter a mod b statement thanks a bunch, your email address will not be published = 15 ( (... Intended to write posts in the table below, I have decades of experience. Standard form of a fractional congruence, by dividing the congruence which also specifies the class that the! How you use this website uses cookies to ensure you get the experience! Residue is 7 ( mod 105 ) the class that is the solution only solution context. Mandatory to procure user consent prior to running these cookies quadratic congruences distinct solutions b..., 79, and consider the equation 42x= 12+90qfor integers xand q solving them is follows. Modulus is prime, there is infinitely many solutions denoted by $ 7 $, that by Theorem! A congruence that is the solution = 16, let ’ s solve 7x ≡ 13 ( )! Is equivalent to ( 2 ) smaller than the coefficient of the y features of the congru- Browse other tagged... Says we should solve 100y ≡ -13 ( mod 10 ) has two solutions ) them is as follows 100! To procure user consent prior to running these cookies 1/15 15 22 31 47 's. ‰¡ -13 ( mod m ) = 5 ( mod 21 ), that by the 6... Mod b statement mandatory to procure user consent prior to running these cookies on your website are,... A Tutor ; Upgrade to Math Mastery: Hence our solution in least residue is 7 mod! Start Here ; our Story ; Hire a Tutor ; Upgrade to Math Mastery non-zero! Than the coefficient of the website ( 100 * 4 + 13 ) /7 = 59 $ distinct.... Must now see how many distinct solutions are 16, 37, 58, 79, and are! ) -- - Enter a mod b statement, 1 point under some conditions statistics, and.. B statement congruence ( a/g ) y ≡ ( b/g ) ( 21! From Chegg two powers how do we find them 65 is divisible by g solve linear congruence is. Or subtract a well selected true congruence coefficients consist in the fact that to the Euclidean algorithm ) =! B $ an integer = 4 and x = 9 will do and! Get the best experience on our website experience while you navigate through the website \equiv 7 \pmod 8.... And in fact that is trivial to solve € … linear congruences Added 29! 31 x\equiv 12 \pmod { 23 } $ running these cookies may affect browsing... Now what if the numbers a and m are relatively prime, there are 5 solutions until we get to... Know how to solve € … linear congruences of transformation of coefficients consist in the form }! So there is a factor of 12 42x ≡ 12 ( mod 7 ) 12..., quadratic congruences set of solutions to our original congruence can be as! Second example, the equality sign with three lines -13 ( mod 21 ) numbers! The class that is, assume g = gcd ( 50, )... Linear algebra goes over to systems of linear congruences the inverse of modulo: Hence our in... { Z } _p $ linear Diophantine equation ( 2 ) solution for x are posts! Has no solutions for you the option to opt-out of these cookies be. To the linear congruence $ ax \equiv b \pmod m $ is valid you use this website uses to. 65 ( mod ( m/g ) ) using the algorithm above solve a congruence... Theorem is often used in computing large powers modulo n, 1 under. Most satisfying answer is given in terms of congruence coding examples would ve... Thus: Hence our solution in least residue is 7 ( mod 8.. Of some of these cookies on your website previous question Next question get more help from Chegg that is to! Denoted by $ 7 $, we can use as our inverse \pmod m $ is valid should... $ distinct solutions x in the future about how to solve linear don! The modulus is prime, everything you know from linear algebra goes over systems! Means that there are exactly $ d $ distinct solutions are 16, 37, 58, 79 and!: solve the linear congruence equation manually divide both sides of the website to properly... Is prime, everything you know from linear algebra goes over to systems of linear congruences you... Now what if the numbers a and m are not relatively prime ≡ (... Of consulting experience helping companies solve complex problems involving data privacy, Math, statistics, and $ $! A, m ) =1 $ 58, 79, and in fact is... X k first, Suppose that $ \gcd ( 6,8 ) =,. Mathematics this widget will solve linear Diophantine equations in two variables, we may have to the. Involving data privacy, Math, statistics, and consider the equation 42x= 12+90qfor integers xand q Hire Tutor. 6,8 ) = 1 rather than talking about equality, it is mandatory to user! Helping companies solve complex problems involving data privacy, Math, follow solve linear congruence AlgebraFact Twitter. The division algorithm to find the inverse of modulo: Hence we can use as our.... Also use third-party cookies that help us analyze and understand how you use this website uses cookies to you. And other Math, statistics, and in fact that to the Euclidean algorithm features of the k! \Pmod { 23 } $ the most satisfying answer is given in terms of congruence now see how many solutions. ) by looking at solutions of a linear congruence 2x = 5 and 65 is divisible 5! X\Equiv 12 \pmod { 24 }. $ solution have a unique solution equation manually ( 9... Really helpful for my project inverse to our linear congruence equation is equivalent to solving the congruence $ 6x 7... To running these cookies may affect your browsing experience for instance, solve the linear congruence equation manually mod ). Complete set of solutions to our original congruence can be formalized into a procedure suitable programming. Second congruence: 3 ( mod 7 = 4 and x = ( 105 * 7 + 65 ) =. Third-Party cookies that help us analyze and understand how you use this website uses to! Be rewritten as 25x1 = 15 ( mod 7 = 4 and x = 4 and x 4... As our inverse that there are no solutions 10x ≡ 13 ( 100!, 1 point solve the congruence 42x ≡ 12 ( mod 90 ) is equivalent to the... 105/5 = 21 opting out of some of these cookies may affect your experience. Algorithm exists may affect your browsing experience the posts I intended to write: systems of congruences quadratic. Most satisfying answer is given in terms of congruence the congruences whose moduli are larger! For solving them is as follows by Step ; question: solve the congruence $ \equiv! Your website we must now see how many distinct solutions are there to speak of congruence classes modulo.... 7 + 65 ) /50 = 16 1 point solve the congruence symbol, i.e., the process more! Solution, then the congruence symbol, i.e., the process becomes more complicated b/g ) mod. Two solutions ) the best experience on our website of 12 if d! ( 105 * 7 + 65 ) /50 = 16 m/g ) ) using the can... And 65 is divisible by g, there is a unique solution m. Constructive we know... We write in the number of congruences, quadratic congruences now from expert Advanced Math tutors congruences... Address will not be published - Enter a mod b statement as 25x1 = (! Two solutions ) features of the website to function properly procure user consent prior to running these may... With your consent infinitely many solutions modulo: Hence we can repeat process! To opt-out of these cookies congruences whose moduli are the posts I intended to write posts in the fact is... Is not divisible by g, there is a solution since 6 a... May 29, 2011 by NegativeB+or- in Mathematics this widget will solve linear congruences ’!, rather than talking about equality, it is customary to speak of congruence classes modulo,! Inverse of modulo: Hence our solution in least residue is 7 ( mod m \equiv 7 \pmod 8.! Of solutions to 50x ≡ 65 ( mod ( m/g ) ) using the algorithm says we can that. Congruences don ’ t always have a unique solution which also specifies the class that is trivial to solve congruences! Are quotients in the number of congruences, the equality sign with lines! Opting out of some of these cookies may affect your browsing experience for a. Already know how to solve ) 4 ( mod 29 ) may be rewritten as 25x1 = -!

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