Q19. A three-digit number was divided successively in order by 4, 5 and 6 leaving out the remainders. The remainders were respectively 2, 3 and 4. How many such three digit numbers are possible?
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lcm = 60
three digit no = 60k + L
now divide byb 4,5,6 taking quotient as -1
as difference becomes 2 which is there 4-2 = 2, 5-3=2,6-4=2
take negative sign of l as quotient is taken as -1
so it will come negative
so three digit no= 6O k -2
take k= 2,3,4,5,6,7,8,9,10,---------15,16,
as 60×17 -2= 1018
so till 16 take
so no of 3 digit no= 15
three digit no = 60k + L
now divide byb 4,5,6 taking quotient as -1
as difference becomes 2 which is there 4-2 = 2, 5-3=2,6-4=2
take negative sign of l as quotient is taken as -1
so it will come negative
so three digit no= 6O k -2
take k= 2,3,4,5,6,7,8,9,10,---------15,16,
as 60×17 -2= 1018
so till 16 take
so no of 3 digit no= 15
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