Question

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.
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Answer 1

Answer:

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. ​

Step-by-step explanation:

Answer 2

Given :-

• Sum of 3 natural.numbers A , B and C is 99.
• A has 3 divisor.

To Find :-

• Minimum value of (B + C) .

Solution :-

As we know, if N is a natural number , where

→ N = a^m * b^n * c^p

Than,

• Total number of divisor of N is = (m+1)*(n+1)*(p+1).
• a, b and c are prime factors.

we have given that, A has 3 divisors .

Therefore, A is Perfect square natural number of a Prime number .

{ Reason :- A = n² , total divisor = (2 + 1) = 3 and here n is a prime number.. }

Now, in order to find the minimum value of (B + C) , we have to take Maximum value of A.

So,

• if A = 2² = 4, => B + C = 99 - 4 = 95
• if A = 3² = 9, => B + C = 99 - 9 = 90
• if A = 5² = 25, => B + C = 99 - 25 = 74.
• if A = 7² = 49, => B + C = 99 - 49 = 50.
• if A = 11² = 121 ≠ Sum is greater than 99.

Hence, Minimum value of (B + C) will be 50 when A is 49.

Learn more :-

Target Olympiad

1. Find the values of A and B if the given number 7A5798B8 is divisible by 33.

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