Question

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

Answer 1

Answer:

here is your answer

Step-by-step explanation:

Answer

The sum of three prime numbers can only be even if one of the prime numbers is even, the only prime number that is even is 2.

So, the sum of the other two prime numbers must be 98.

x+x+36=98

⇒2x=62

⇒x=31

Numbers are 2,31 and 67.

The highest among them is 67.

hope it is helpful

Answer 2

Answer:

✩ According to the Question :

Sum of three non - zero prime numbers is 100. 100 is an Even Number , and all Prime Numbers are Odd and 3 Odd Numbers can't make 100.

⠀⠀⠀⠀⠀⠀⠀Hence , we will take First Prime Number 2 as one of the non - zero prime numbers.

⠀

Let the Second non - zero number be n, and Third non - zero number be (n + 36) i.e. 36 Exceeds the Other Number.

⠀

$\begin{tabular}{|c|c|}\cline{1-2}\sf 1st Number&\sf 2\\\cline{1-2}\sf2nd Number&\sf n\\\cline{1-2}\sf 3rd Number &\sf (n + 36)\\\cline{1-2}\end{tabular}$

$\rule{90}{0.8}$

☯ $\underline{\boldsymbol{According\: to \:the\: Question :}}$

$\dashrightarrow\sf\:\:1st\:No.+2nd\:No.+3rd\:No.=100\\\\\\\dashrightarrow\sf\:\:2+n+(n+36)=100\\\\\\\dashrightarrow\sf\:\:2 + n + n + 36 = 100\\\\\\\dashrightarrow\sf\:\:38 + 2n = 100\\\\\\\dashrightarrow\sf\:\:2n = 100 - 38\\\\\\\dashrightarrow\sf\:\:2n = 62\\\\\\\dashrightarrow\sf\:\:n = \dfrac{62}{2}\\\\\\\dashrightarrow\sf\:\:n = 31$

⠀⠀⠀$\rule{160}{1.5}$

☢ $\underline{\textsf{All three non - zero prime numbers :}}$

$\bullet\:\:\textsf{1st No. = \textbf{2}}\\\bullet\:\:\textsf{2nd No. = n = \textbf{31}}\\\bullet\:\:\textsf{3rd No. = (n + 36) = \textbf{67}}$

⠀

$\therefore\:\underline{\textsf{Hence, Largest Number will be C) \textbf{67}}}.$