Find the smallest number which when divided by 24 30 48 and 60 leaves a remainder 2 in each case
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LCM - LCM of two or more numbers is the smallest number which is divisible by all the given numbers.
LCM of two or more numbers= Product of the greatest power of each prime factor involved in the numbers with highest power.
SOLUTION :
To find the smallest number, we first find LCM ( least common multiple) of 24,30,48 & 60 and then add remainder 2 in the LCM.
LCM by Prime factorization method :
24 = 2³ × 3¹
30 = 2¹ × 3¹ × 5¹
48 = 2⁴ × 3¹
60 = 2² × 3¹ × 5¹
Least Common Multiple(LCM) = 2⁴ × 3¹ × 5¹ =240
Required Number = 240 + 2 = 242
Hence, the smallest number which when divided by 24 ,30 ,48 and 60 leaves a remainder 2 in each case is 242.
HOPE THIS WILL HELP YOU….
LCM of two or more numbers= Product of the greatest power of each prime factor involved in the numbers with highest power.
SOLUTION :
To find the smallest number, we first find LCM ( least common multiple) of 24,30,48 & 60 and then add remainder 2 in the LCM.
LCM by Prime factorization method :
24 = 2³ × 3¹
30 = 2¹ × 3¹ × 5¹
48 = 2⁴ × 3¹
60 = 2² × 3¹ × 5¹
Least Common Multiple(LCM) = 2⁴ × 3¹ × 5¹ =240
Required Number = 240 + 2 = 242
Hence, the smallest number which when divided by 24 ,30 ,48 and 60 leaves a remainder 2 in each case is 242.
HOPE THIS WILL HELP YOU….
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