Question

Four positive whole numbers add up to 93
One of the numbers is a multiple of 19
The other three numbers are equal.
Write the four numbers in ascending order.

Answer 1

Answer:

Let the numbers be a, b, c, d

${ \tt{a + b + c + d = 93}}$

Let a be the multiple of 19, a = 57

And b = c = d

${ \tt{57+ 3b= 93}} \\ { \tt{3b = 36}} \\ { \tt{b = 12}}$

${ \tt{a = 57}} \\ { \tt{b = c = d = 12}} \\ \\ { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{12}}} \: \: { \boxed{ \rm{57}}}$

Answer 2

Answer:

12, 12, 12, 57

Step-by-step explanation:

Given:

• Four positive whole numbers add up to 93.
• One of the numbers is a multiple of 19.
• The other three numbers are equal.

If one of the four numbers is a multiple of 19 and the other three numbers are equal, then the sum of the other three numbers will be a multiple of 3.

Multiples of 19:  

19, 38, 57, 76, 95, ...

Subtract each multiple of 19 (that is less than 93) from 93:

⇒ 93 - 19 = 74  →  not divisible by 3

⇒ 93 - 38 = 55  →  not divisible by 3

⇒ 93 - 57 = 36  →  divisible by 3

⇒ 93 - 76 = 17  →  not divisible by 3

The only result that is a multiple of 3 is 36.  

Therefore, the multiple of 19 is 57.

To find the other three numbers, simply divide the result of the total less the found multiple of 19 by 3:

⇒ (93 - 57) ÷ 3 = 36 ÷ 3 = 12

Therefore, the four numbers in ascending order are:

12, 12, 12, 57