How do I find a number which when divided by 4, 5, 6 and 7 gives 3, 4, 5 and 6 as remainders respectively?

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Is there specific set of steps to be followed when solving similar questions?

My attempt:

A number gives remainder 3 when divided by 4. Hence it is of the form 4m+3 where m = 0,1,2…

Now, 4m+3 gives remainder 4 when divided by 5. It means (4m+3)-4 is divisible by 5.

Hence, 4m-1=5n where n is some positive integer. n=(4m-1)/5.

But, 4m+3 gives remainder 5 when divided by 6. It means (4m+3)-5 is divisible by 6.

Hence, 4m-2=6f where f is some positive integer. f=(4m-2)/6.

Again, 4m+3 gives remainder 6 when divided by 7. It means (4m+3)-6 is divisible by 7.

Hence, 4m-3=7l where l is some positive integer. l=(4m-3)/7.

My confusion: What do I do now? Or am I even in the right path?

Please don't solve it using modular arithmetic. I don't know it yet.

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Let $a$ be your number. Then note that $a + 1$ divided by $4, 5, 6 \text{ and } 7$ will give remainders of $0$ in all cases. Thus, the smallest positive $a + 1$ will be the smallest number which $4, 5, 6 \text{ and } 7$ all divide into. Since $6$ and $4$ each have a factor of $2$, you don't need to repeat this factor, giving it to be $4 \times 3 \times 5 \times 7 = 420$, giving that $a = 419$ works.

Note that, in general, $a = 420n - 1$ for any integer $n$.