Question

the sum of the digits of the three digit number is 11.if the digits are reversed the new number is 46 more than five times the former number. if the hundreds digit plus twice the tens digit is equal to the units digit then find the original three digit number?​

Answer 1

Answer:

Hello mtps students parsauniya.

Step-by-step explanation:I am going to re-write the problem to what I think you meant:

:

The sum of the digits of a three-digit number is 11. If the digits are reversed,

the new number is 46 more than five times the old number. If the hundreds digit

plus twice the tens digit is equal to the units digit, then what is the number,

:

Let the number be: 100x + 10y + z

:

Let's try solving this using only the 1st and last statements:

"The sum of the digits of a three-digit number is 11."

x + y + z = 11

:

"the hundreds digit plus twice the tens digit is equal to the units digit,"

x + 2y = z

x + 2y - z = 0

:

Add to the 1st equation to eliminate z

x + y + z = 11

x +2y - z = 0

--------------

2x + 3y = 11

:

2x = 11 - 3y

x = %28%2811-3y%29%29%2F2

Only two values for y will give a positive integer value for x, namely 1 and 3:

From the 2nd statement we know that x is a low value, therefore 3 seems likely:

x = %28%2811-3%283%29%29%29%2F2

x = %28%2811-9%29%29%2F2

x = 2%2F2

x = 1 when y = 3

Find z

1 + 3 + z = 11

z = 11-4

z = 7

;

Our number is 137

:

See if that makes the 2nd statement true.

"If the digits are reversed, the new number is 46 more than five times the old number"

5(137) + 46=731

685 + 46=731 ANSWER

Answer 2

• The three digit number is 137

Step-by-step explanation:

Let,

The three digit numbers be x, y & z

• The sum will be = x + y + z = 11 ------ (1)

Original number,

• 100x + 10y + z

(As because x is in the hundredth place, y is in the tense place and z is in ones place)

Reversing,

• 100z + 10y + x

According to the question,

It is given that the new number is 46 more than 5 times the former number (original number),

⟶ 100z + 10y + x = 46 + [5 (100x + 10y + z)]

⟶ 100z + 10y + x = 46 + (500x + 50y + 5z)

⟶ 100z + 10y + x - 500x - 50y - 5z = 46

⟶ (x - 500x) - (50y - 10y) + (100z - 5z) = 46

⟶ (-499x) - (40y) + 95z = 46

⟶ - 499x - 40y + 95z = 46 -------- (2)

Hundred digit plus twice the tense digit is equal to the unit digit,

⟶ x + (2y) = z

⟶ x + 2y = z

⟶ x + 2y - z = 0 ---------- (3)

Solving (1) & (2),

x + y + z = 11 (+)

x + 2y - z = 0

_________

2x + 3y = 11 --------- (4)

_________

(By adding (+z) & (-z) will get cancelled)

Solving (2) & (3),

To make the 'z' value same in both (2) & (3), we will multiply (4) × 95,

-499x - 40y + 95z = 46 (+)

95x + 190y - 95z = 0

_________________

-404x + 150y = 46 ---------- (5)

_________________

(By adding (+95) & (-95) will get cancelled)

Solving (4) & (5),

To make the 'y' value same in both (4) & (5), we will multiply (4) × 50,

-404x + 150y = 46 (-)

100x + 150y = 550

______________

-504x = -504

______________

⟶ x = -504 / 504

⟶ x = 1

Substituting 'x' in (4),

⟶ 2x + 3y = 11

⟶ 2 (1) + 3y = 11

⟶ 2 + 3y = 11

⟶ 3y = 11 - 2

⟶ 3y = 9

⟶ y = 9 / 3

⟶ y = 3

Substituting 'x' & 'y' value in (1),

⟶ x + y + z = 11

⟶ 1 + 3 + z = 11

⟶ 4 + z = 11

⟶ z = 11 - 4

⟶ z = 7

• Hence, the original three digit number is 137