Find the least number which when divided by 5,6,7 & 8 leaves a remainder 3, but when divided by 9 leaves no remainder.
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1683 is the number.............
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devided by 5 6 7 8 it leaves 3
so it is 3 more than multiples ot LCM of 5 6 7 8
now LCM of 5 6 7 8
factorisee
5 = 5
6 = 2*3
7 = 7
8 = 2* 2* 2 = 2^3
LCM = 5*2^3*3*7 = 840
now it has to be of the form 840n + 3
this is divisible by 9
840n + 3 = 9m
3 = 9m - 840n
now using extended euler algorithm
we need to find GCD of 840 and 9
840 = 9*93 + 3
3= 840*1 - 9*93
so n = -1 or 2 as it is mod 3
so 840*n +3 = 1683 divisible by 9
hence ans is 1683
so it is 3 more than multiples ot LCM of 5 6 7 8
now LCM of 5 6 7 8
factorisee
5 = 5
6 = 2*3
7 = 7
8 = 2* 2* 2 = 2^3
LCM = 5*2^3*3*7 = 840
now it has to be of the form 840n + 3
this is divisible by 9
840n + 3 = 9m
3 = 9m - 840n
now using extended euler algorithm
we need to find GCD of 840 and 9
840 = 9*93 + 3
3= 840*1 - 9*93
so n = -1 or 2 as it is mod 3
so 840*n +3 = 1683 divisible by 9
hence ans is 1683
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