Question

The sum of three prime numbers is 100. If one of them exceeds another by 36, then one of the numbers is

A) 17 B) 29 C) 43 D) none

Answer 1

Answer:

(D) none

Step-by-step explanation:

Given that,

Sum of three prime numbers is 100.

Also,

One of them exceeds another by 36.

To find ane of them.

Let, one prime number is x.

Therefore, another one = x+36

Let, the third one is y.

Therefore, we have,

=> x + x+36 + y = 100

=> 2x + y = 100-36

=> 2x + y = 64

=> 2x + y = 62+2

=> 2x + y = 2(31) + 2

On Comparing both sides, we get,

=> x = 31

=> y = 2

Therefore, we get,

=> x +36 = 31+36 = 67

Thus, the prime numbers are 2, 31 and 67.

Hence, the correct answer is (D) none.

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

• Sum of three prime numbers =100
• One of them exceeds another by =36

${\bf{\blue{\underline{To\:Find}}}}$

• other number =?

${\bf{\blue{\underline{Now:}}}}$

• Let the three prime numbers be a ,a+36 , and b

$\dagger \: \underline {\frak{according \: to \: the \: question:}}$

$: \implies{\sf{ a + a + 36 + b = 100}}$

$: \implies{\sf{ 2a + 36 + b = 100}}$

$: \implies{\sf{ 2a + b = 100 - 36}}$

$: \implies{\sf{ 2a + b = 64}}$

From above it is clear that 2a is a even prime number and to get sum 100 ,b should also be an even prime number.

So put b =2

$: \implies{\sf{ 2a + 2 = 64}}$

$: \implies{\sf{ 2a = 64 - 2}}$

$: \implies{\sf{ 2a = 62}}$

$: \implies{\sf{ a = \frac{62}{2} }}$

$: \implies{\sf{ a = 31}}$

__________________________________

Thus , the prime numbers are 31, 67,2.

Option (D) is correct!