Question

Consider three natural numbers a, b, c such that 2 ≤ a ≤ b ≤ c and whenever the product of any two divided by the third leaves the remainder 1. Find the value of (a + b + c + abc)​

Answer 1

Answer:

a=2, b=1, c=1

=6 is the answer

Answer 2

Given:

Three natural numbers a, b, and c satisfying the condition2$\leq$a$\leq$b$\leq$c, product of any two of the three natural numbers when divided by the third natural number leaves remainder 1.

To Find:

The value of (a + b + c + abc)​.

Solution:

1. The given condition for the three natural numbers is 2$\leq$a$\leq$b$\leq$c.

• Using trial and error methods can be helpful in such cases.
• Since the remainder is non-zero, all the numbers must be different else, the remainder will be zero if any of the two numbers are identical.
• Consider values as a = 2, b = 3, and c = 5.

=> a x b = 6, 6 when divided by 5 leaves remainder 1.

=> b x c = 15, 15 when divided by 2 leaves remainder 1.

=> a x c = 10, 10 when divided by 3 leaves remainder 1.

• Hence, the set of numbers are satisfying the given conditions. Therefore, the values of a, b, and c are 2, 3, and 5 respectively.

• The value of (a + b + c + abc) is,

=> 2 + 3 + 5 + (2x3x5),

=> 10 + 30,

=> 40.

Therefore, the value of (a+b+c+abc) is 40.