Question

The sum of three non-zero prime numbers is 100. One of them exceeds the other by 36, then, the largest number is
A. 73
B. 91
C. 67
D. 57

Answer 1

Hint: We will use the property that to get an even number as a sum of three distinct numbers, either one or all 3 should be even. Also, we know that 2 is the only prime number which is even. So, we will suppose the number to be variable and by using the above property we will get the required value.

Complete step-by-step answer:
We have been given the sum of three non-zero-prime numbers of 100. One of them exceeds the other by 36.
We know that the sum of 3 odd numbers is always odd, it can never be 100. So, to get an even number i.e. 100 as sum of 3 distinct numbers, either one or all 3 should be even.
Let the number be x, \[{\rm{x}} + {\rm{36}}\] and y. According to the question,
\(x + x + 36 + y = 100\)
Also, we know that 2 is the only prime number which is even. Thus, 2 must be one of the 3 numbers.
If we add 36 to 2, we get a composite number, so, y must be equal to 2.
\(\begin{array}{l}x + x + 36 + 2 = 100\\2x = 100 - 2 - 36\\2x = 100 - 38\\2x = 62\\x = \dfrac{{62}}{2}\\x = 31\end{array}\)
Now, we have x = 31. So, we can get the second number as 31 + 36 = 67. We already have y = 2. So, the prime numbers are 31, 67 and 2.
Hence, the largest number is 67.
Therefore, the correct option is C.

Note: It is very important to understand that the value of 'x' cannot be equal to 2 because \[\left( {{\rm{x}} + {\rm{36}}} \right)\] equals to 38 which is a composite number but it is given that all the three numbers are prime numbers.
Also, remember the fact that the sum of three odd numbers is always odd and to get an even number as a sum of 3 numbers either one or all 3 should be even.