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.
let the 10th digit be interchanged and in form 10x+y
therefore original sum
=> (a1+a2+...+a9) + 10x + y = S
but the sum obtained was::: (a1+a2+...+a9) + 10y + x = S' bcoz the digits were interchanged
"the sum of all the ten numbers increased by a value which was four less than that number"
therefore!!
S' - S = (10x+y) - 4
the a1, a2...a9 gets cancelled out & what is left is
=> 9(y-x)=10x+y-4 => 8y-19x=(-4)-----------------------------...
"three times the sum of digits of that original number is ten less than the number"
therefore
=> 3(x+y) = 10x+y - 10 => 2y-7x = -10-------------------------------------...
by (1) & (2), the y can be cancelled out by multiplying (2) with -4 & adding (1) & (2)
the result we get is
=> 9x = 36 => x=4
therefore by (2)
2y -7.4= -10 => y=9
therefore x.y = 36
Well, let's call the original number AB, where A represents the tens digit and B represents the ones digit.
We are told that when the digits are interchanged, we get the number BA. So, the sum of the ten numbers will increase by (BA - AB).
We are also told that three times the sum of the digits of the original number is ten less than the number. In equation form, that would be:
3(A + B) = 10(A + B) - 10
Simplifying this equation, we get:
2(A + B) = 10
Now, let's focus on the first piece of information. The sum of all the ten numbers increased by (BA - AB), which is four less than the number. In equation form, that would be:
10(A + B) + (BA - AB) = (10A + B) - 4
Simplifying this equation, we get:
9(A + B) = 4
So, we have two equations:
2(A + B) = 10---------(1)
9(A + B) = 4---------(2)
Let's solve these equations simultaneously to find the values of A and B!
From equation (1), we can rewrite it as:
A + B = 5
Now, we can substitute this value into equation (2):
9(5) = 4
Oh dear, we have a contradiction! 45 is definitely not equal to 4.
So, it seems there is no solution to this problem. It's a bit of a mathematical mystery! Though, I must admit, it is a rather amusing puzzle.
Let's solve this step by step:
Step 1: Let's assume the original two-digit number as 10x + y, where x is the tens digit and y is the units digit.
Step 2: According to the problem, the sum of all ten numbers increased by a value which was four less than that number. This means that the sum increased by (10x + y) - 4.
Step 3: The sum of all ten numbers can be expressed as the sum of the digits of the original number (x + y) multiplied by 10, plus the original number itself. So, the sum of all ten numbers is (x + y) * 10 + (10x + y).
Step 4: From step 2 and step 3, we can write the equation:
(x + y) * 10 + (10x + y) = (x + y) * 10 + (10x + y) - 4.
Step 5: Simplifying the equation from step 4, we get:
10x + y = 4.
Step 6: According to the problem, three times the sum of the digits of the original number is ten less than the number. This can be expressed as:
3(x + y) = 10x + y - 10.
Step 7: Simplifying the equation from step 6, we get:
-7x + 2y = -10.
Step 8: Solving the equations from step 5 and step 7 simultaneously, we get:
x = 1 and y = 6.
Step 9: Therefore, the original number is 10x + y = 10 * 1 + 6 = 16.
Step 10: The product of the digits of the number 16 is 1 * 6 = 6.
Therefore, the product of the digits of that number is 6.
To solve this problem, let's break it down step by step:
Step 1: Let's assume the original number is a two-digit number with digits AB, where A represents the tens digit and B represents the ones digit.
Step 2: According to the problem, the digits of the number were interchanged, so the new number becomes BA. The value of the new number can be calculated using the formula: (10 * B) + A.
Step 3: The sum of all ten numbers can be represented as the sum of the original number and the new number, which is (10A + B) + (10B + A) = 11A + 11B = 11(A + B).
Step 4: According to the problem, the sum of all ten numbers increased by a value that was four less than the original number. This can be written as:
11(A + B) = (10A + B) + (10B + A) + (10B + A) - 4.
Step 5: Simplifying the equation, we get:
11(A + B) = 21A + 21B - 4.
Step 6: Distribute the value of 11 on the left side, and simplify the equation further:
11A + 11B = 21A + 21B - 4.
Step 7: Rearrange the equation to isolate one of the variables:
10A = 10B - 4.
Step 8: Divide both sides of the equation by 10:
A = B - 0.4.
Step 9: Substitute the value of A from Step 8 into the equation from Step 2 to get the original number:
Original number = (10 * (B - 0.4)) + B = 10B - 4.
Step 10: According to the problem, three times the sum of the digits of the original number is ten less than the number, which can be represented as:
3(A + B) = (10B - 4) - 10.
Step 11: Simplify the equation:
3(A + B) = 10B - 14.
Step 12: Substitute the value of A from Step 8 into the equation from Step 11:
3((B - 0.4) + B) = 10B - 14.
Step 13: Simplify and solve the equation:
6B - 1.2 = 10B - 14.
4B = 12.8.
B = 3.2.
Step 14: Substitute the value of B back into the equation from Step 9 to calculate the original number:
Original number = 10(3.2) - 4 = 28.
Step 15: The original number is 28. Now, let's find the product of the digits:
Product of digits = 2 * 8 = 16.
Therefore, the product of the digits of the original number is 16.