Find all 3-digit numbers which are 19 times the sum of their digits.

The 3 digit numbers are;

171, 152, 133, 114, 285, 266, 247, 228, 209, 399

Step-by-step explanation:

Let the 3 digit number be xyz

Since x is the Hundreds digit, y is the tens digit and z is the unit digit, we have;

100x, 10y and z

We are told the sum of this is equal to 19 times the sum of their digits.

Thus;

100x + 10y + z = 19(x + y + z)

100x + 10y + z = 19x + 19y + 19z

This gives;

81x - 9y - 18z = 0

Divide through by 9 to get;

9x - y - 2z = 0

x = (2z + y)/9

Now, we will need to select values of z and y between 0 to 9 in such a way that (2z + y) is divisible by 9 and in such a way that x > 0 but ≤ 10

The possible solutions are;

At z = 1, y = 7 and x = 1

At z = 2, y = 5 and x = 1

At z = 3, y = 3 and x = 1

At z = 4, y = 1 and x = 1

At z = 5, y = 8 and x = 2

At z = 6, y = 6 and x = 2

At z = 7, y = 4 and x = 2

At z = 8, y = 2 and x = 2

At z = 9, y = 0 and x = 2

At z = 9, y = 9 and x = 3

Compiling all these, the 3 digit numbers are;

171, 152, 133, 114, 285, 266, 247, 228, 209, 399

option 2 is correct.

step-by-step explanation:

we are given two inequality equation

x+y> 5 and -3x+2y< -2

first we draw the graph of both function using t-table method

equation: x+y> 5

for x=0, y=5

for x=-5, y=0

table:

x   :   0     5

y   :   5     0

test point (0,0) , 0> 5

false

equation: -3x+2y< -2

for x=0, y=-1

for x=2, y=4

table:

x   :   0     2

y   : -1       4

test point (0,0) , 0< -2

true

now we draw the graph. see the attachment for graph.

common region would be solution set.

option 2 is correct.