Can anyone guide on how I can solve the following problem without using trial-and-error:
A number when divided by 5 leaves a remainder of 2 and when divided by 7, leaves a remainder of 4. What is the remainder when the same number is divided by 5×7?
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You see, let us take the no. to be X. And € represent congruence. X € 2 (mod 5) and X € 4 (mod 7). X € 2 (mod 5)= X € 2-5 (mod 5)=X € -3 (mod 5) & similarly,
X € 4 (mod 7)= X € 4-7 (mod 7)=X € -3 (mod 7). Now, X € -3 (mod 5×7)= X € -3 (mod 35)=X € -3 +35 (mod 35)=X € 32 (mod 35). So, the remainder left when divided by 35 is 32.
You see let the number be X. Then X is of the form 5a+2 or 5b-3. Similarly, X is of tge form, 7c+4 or 7d-3. So, when divided by 35 , we can say it will leave remainder -3 which is common both cases. So, X = 35e -3 which corresponds to a remainder of 35-3 = 32. You can do it in congruent modulo concept too.
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Chinese Remainder Theorem.
It is 32.
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I got that answer by trial-and-error. But, may you please explain the mathematical way to solve it?
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You see, let us take the no. to be X. And € represent congruence. X € 2 (mod 5) and X € 4 (mod 7). X € 2 (mod 5)= X € 2-5 (mod 5)=X € -3 (mod 5) & similarly, X € 4 (mod 7)= X € 4-7 (mod 7)=X € -3 (mod 7). Now, X € -3 (mod 5×7)= X € -3 (mod 35)=X € -3 +35 (mod 35)=X € 32 (mod 35). So, the remainder left when divided by 35 is 32.
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You see let the number be X. Then X is of the form 5a+2 or 5b-3. Similarly, X is of tge form, 7c+4 or 7d-3. So, when divided by 35 , we can say it will leave remainder -3 which is common both cases. So, X = 35e -3 which corresponds to a remainder of 35-3 = 32. You can do it in congruent modulo concept too.