While adding ten two-digit numbers the digits of one of the numbers were interchanged. As a result the sum of all the ten numbers increased by a value which was four less than that number. Three times the sum of the digits of the original number is ten less than the number. What is the product of the digits of that number?

Please help how to obtain that number. and also explain (While adding ten two-digit numbers the digits of one of the numbers were interchanged. As a result the sum of all the ten numbers increased by a value which was four less than that number ) this sentence

a1 ÷ a10 = your numbers.

You can write 10th digit in form:

a10 = 10 x + y

Original sum:

S = a1 + a2 +...+ a9 + 10 x + y

If 10th digit be interchanged new number will be 10 y + x

New sum wil be:

S1 = a1 + a2 +...+ a9 + 10 y + x

The sum of all the ten numbers increased by a value which was four less than that number mean:

S1 - S = 10 x + y - 4

S1 - S = a1 + a2 +...+ a9 + 10 y + x - ( a1 + a2 +...+ a9 + 10 x + y )

S1 - S = ( a1 + a2 +...+ a9 ) + 10 y + x - ( a1 + a2 +...+ a9 ) - 10 x - y

S1 - S = 10 y + x - 10 y + x

S1 - S = 9 y - 9 x

So:

S1 - S = 10 x + y - 4

9 y - 9 x = 10 x + y - 4

9 y - y = 10 x + 9 x - 4

8 y = 19 x - 4

Three times the sum of the digits of the original number is ten less than the number mean:

3 ( x + y ) = 10 x + y - 10

3 x + 3 y = 10 x + y - 10

3 y - y = 10 x - 3 x - 10

2 y = 7 x - 10

Now you must solve system:

8 y = 19 x - 4

2 y = 7 x - 10

The solutions are:

x = 4 , y = 9

The product of the digits:

x ∙ y = 4 ∙ 9 = 36

Suppose the two digits are a and b. Then the value of the number is 10a+b.

So, the value after the digit swap is 10b+a
I assume "that number" refers to the number whose digits were swapped.

The amount of increase in the sum is (10b+a)-(10a+b)=9b-9a
Now we know that
9b-9a = 10a+b-4
3(a+b) = 10a+b-10

19a-8b = 4
7a-2b = 10
a=4 b=9

ab = 36

I don't understand the entire problem

To solve this problem, let's break it down step by step:

1. Let's assume the original number is represented by the two digits "xy" (where x is the tens digit and y is the ones digit).
2. When the digits are interchanged, the new number becomes "yx".
3. The sum of all ten numbers is increased by a value which was four less than that number means that the difference between the sum after interchanging the digits and the sum before interchanging the digits is 4. Let's call the sum after interchanging the digits S'.

Step 1: Find the sum before interchanging the digits:
Since we are adding ten two-digit numbers, the sum before interchanging the digits can be represented as 10xy.

Step 2: Find the sum after interchanging the digits:
The new number formed after interchanging the digits is represented as 10yx. The sum of all ten numbers after interchanging the digits can be represented as 10yx + (S' - 4), where (S' - 4) is the increased value.

Step 3: Formulate the equation based on the given information in the question:
According to the question, three times the sum of the digits of the original number is ten less than the number. In mathematical terms, it can be written as:
3(x + y) = 10x + y - 10

Step 4: Solve the equation:
Expand the equation:
3x + 3y = 10x + y - 10

Rearrange the equation:
3x - 10x + 3y - y = -10

Simplify:
-7x + 2y = -10

Step 5: Find the value of x and y:
To solve this equation, we need another equation. Looking at the given information, we know that the sum after interchanging the digits increased by four less than the number, so we can write another equation:
10yx + (S' - 4) = 10xy

Now let's combine the equations:
10yx + (S' - 4) = 10xy
10yx + 10x + 10y = 10xy
yx + x + y = xy

Simplify:
x + y = xy - yx
x + y = (10x + y) - (10y + x)
x + y = 9x - 9y

Rearrange the equation:
10x - 8y = 0

Now we have a system of equations:

-7x + 2y = -10
10x - 8y = 0

Step 6: Solve the system of equations:
Using any method you prefer (substitution, elimination, or matrix), solve for x and y. In this case, let's use the elimination method to solve the system.

Multiply the first equation by 5 and the second equation by 7 to eliminate y:

-35x + 10y = -50
70x - 56y = 0

Add the equations together:

-35x + 10y + 70x - 56y = -50
35x - 46y = -50

Divide both sides by 23 to simplify:

x - 2y/23 = -50/23

Since the solution to the system is not an integer, let's shift our focus. The original question asks for the product of the digits of the original number. Since the solution to the system is not providing us with the values of x and y directly, we need to devise another strategy.

Step 7: Deduce the relationship between x and y:
Looking at the second equation of the system (10x - 8y = 0), we can deduce that y must be a multiple of 10 and x must be a multiple of 8 for the equation to be satisfied. Since we are dealing with two-digit numbers, we can focus on values that fit this pattern.

Step 8: Find the possible values for x and y:
Based on the previous step, let's check the possible combinations of two-digit numbers where y is a multiple of 10 and x is a multiple of 8:
y=10, x=16
y=20, x=8
y=30, x=24
y=40, x=16
y=50, x=40
y=60, x=32
y=70, x=56
y=80, x=48
y=90, x=64

Now, let's plug in these values of x and y into the equation from step 3 (3(x + y) = 10x + y - 10) and see if any of them satisfy the equation.

Step 9: Check the values of x and y in the equation:
y=10, x=16:
3(16 + 10) = 10(16) + 10 - 10
3(26) = 160
78 = 160 (not satisfied)

y=20, x=8:
3(8 + 20) = 10(8) + 20 - 10
3(28) = 80
84 = 80 (not satisfied)

y=30, x=24:
3(24 + 30) = 10(24) + 30 - 10
3(54) = 230
162 = 240 (not satisfied)

y=40, x=16:
3(16 + 40) = 10(16) + 40 - 10
3(56) = 160 + 30
168 = 190 (not satisfied)

y=50, x=40:
3(40 + 50) = 10(40) + 50 - 10
3(90) = 400 + 50
270 = 450 (not satisfied)

y=60, x=32:
3(32 + 60) = 10(32) + 60 - 10
3(92) = 320 + 50
276 = 370 (not satisfied)

y=70, x=56:
3(56 + 70) = 10(56) + 70 - 10
3(126) = 560 + 60
378 = 620 (not satisfied)

y=80, x=48:
3(48 + 80) = 10(48) + 80 - 10
3(128) = 480 + 90
384 = 570 (not satisfied)

y=90, x=64:
3(64 + 90) = 10(64) + 90 - 10
3(154) = 640 + 80
462 = 720 (not satisfied)

None of the combinations satisfy the equation. Therefore, it seems there is no solution to the problem as stated.

Hence, we cannot find the product of the digits of that number based on the given information.

Why (10b+a)-(10a+b) this step can you explain

cmon. That is the difference between the original number and with its digits reversed. Better reread what I wrote.