M is the largest 4-digit number which,when di | vedclass

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(B) The divisors are $4, 5, 6, 7$ and the remainders are $2, 3, 4, 5$. Note that the difference between each divisor and its corresponding remainder i

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(B) The divisors are $4, 5, 6, 7$ and the remainders are $2, 3, 4, 5$. Note that the difference between each divisor and its corresponding remainder is constant: $4-2=2, 5-3=2, 6-4=2, 7-5=2$.Let the number be $N$. Then $N = k 	imes LCM(4, 5, 6, 7) - 2$.$LCM(4, 5, 6, 7) = 420$.So,$N = 420k - 2$.We want the largest $4$-digit number $M$. The largest $4$-digit number is $9999$.$420k - 2 \leq 9999 \implies 420k \leq 10001 \implies k \leq 23.81$.Taking $k = 23$,$M = 420(23) - 2 = 9660 - 2 = 9658$.Now,we find the remainder when $M = 9658$ is divided by $9$.The sum of the digits of $9658$ is $9 + 6 + 5 + 8 = 28$.$28 \div 9$ leaves a remainder of $1$ (since $28 = 9 	imes 3 + 1$).

$M$ is the largest $4$-digit number which,when divided by $4, 5, 6,$ and $7,$ leaves remainders $2, 3, 4,$ and $5$ respectively. What is the remainder when $M$ is divided by $9$?

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