If the sum of 3 natural numbers a,b, and c is 99 and a has 3 divisors then what will be the minimum value of b+c.
Answers
Answer:
a+b+c=99
a=96 (96 is the highest number before 99 which is divisible by 3)
b+c=3
Given : sum of 3 natural numbers a,b, and c is 99 and a has 3 divisors then
To Find : what will be the minimum value of b+c.
Solution:
A has 3 Divisors
only prefect square has odd number of divisors
A has 3 Divisors hence A must be perfect square
A = 1 * A
A = √A * √A
=>√A must be a prime
√A = 2 ,, 3 , 5 , 7 , 11
A = 4 , 9 , 25 , 49 , 121
sum of 3 natural numbers a,b, and c is 99
=> A < 99
minimum value of b+c. if A is maximum
A = 49
=> B + C = 99 - 49 = 50
If the sum of 3 natural numbers a,b, and c is 99 and a has 3 divisors then what will be the minimum value of b+c is 50
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