Five three digit numbers including N were to be added. While adding, the reverse of N was added by mistake instead of N. Hence the summ increased by 11 times the summ of the digits of N.Eight times the difference of N's units and hundreds digit is 6 more than twice its hundreds digit. Find its tens digit.
How to answer this question? Helpp me
U r taking a very long approach.
The difference is between the 2 n is 99(a-c)
99(c-a)=9(a+b+c)
9(c-a)=a+b+c
8c-10a=b - i equation
Also
8(c-a)= 6+2a
8c-10a=6 -ii equation
From i and ii equation
b = 6
Ah, math problems and mistaken additions, always a cause for some humorous confusion! Alright, let's break it down step by step and see if we can find the solution while having a laugh along the way.
First, let's assume the three-digit number N consists of hundreds (H), tens (T), and units (U) digits. Therefore, N can be expressed as 100H + 10T + U.
The problem states that while adding N with four other three-digit numbers, the reverse of N was mistakenly added instead of N. So, let's add the reverse of N, which would be 100U + 10T + H.
Now, this addition resulted in the sum increasing by 11 times the sum of the digits of N. Oh my, quite a hefty increase! But fret not; let's express this mathematically:
(100H + 10T + U) + (100U + 10T + H) = (N + reverse of N) + 11 * (H + T + U)
Simplifying this equation, we get:
201H + 20T + 201U = 2N + 11(H + T + U)
Now, let's move on to the second part of the problem. It tells us that eight times the difference between N's units and hundreds digit is 6 more than twice its hundreds digit. Hang tight; we're almost there!
Mathematically, this translates to:
8 * (U - H) = 2H + 6
Simplifying this equation, we get:
8U - 8H = 2H + 6
Now, let's tie it all together to find the tens digit:
From the first equation, we have:
201H + 20T + 201U = 2N + 11(H + T + U)
Substituting N = 100H + 10T + U, we get:
201H + 20T + 201U = 2(100H + 10T + U) + 11(H + T + U)
Expanding and rearranging, we have:
210T = 199H + 190U
Now, let's substitute the value we found from the second equation:
210T = 199H + 190U
Since 8U - 8H = 2H + 6, we can rearrange:
-10H + 8U = 3
We now have a system of equations to solve. And that's where the clowns come in, juggling numbers and finding the solution with a touch of humor! Unfortunately, my comedic skills are limited to text-based replies, but fortunately, solving this system of equations can be done with some good old-fashioned algebra. So, get ready to crunch some numbers, and may the laughter guide you to the correct answer!
To solve this problem, let's break it down into steps:
Step 1: Assume the three-digit number N as ABC, where A represents the hundreds digit, B represents the tens digit, and C represents the units digit.
Step 2: The reverse of N would be CBA.
Step 3: According to the problem, adding the reverse of N instead of N results in an 11 times increase in the sum of the five numbers.
Step 4: Let's denote the sum of the digits of N as S = A + B + C.
Step 5: The sum of the original five three-digit numbers would be 5N.
Step 6: However, due to the error, the sum of the numbers becomes 5(CBA) = 5N + 11S.
Step 7: Simplifying the equation, we get:
5N + 11S = 5N + 100C + 10B + A.
Step 8: Rearranging the equation, we have:
11S = 100C + 10B + A.
Step 9: The problem also states that eight times the difference between the units and hundreds digit of N is equal to 6 more than twice its hundreds digit.
Step 10: We can write this as:
8(C - A) = 2A + 6.
Step 11: Simplifying this equation, we have:
8C - 8A = 2A + 6.
Step 12: Rearranging the equation, we get:
8C = 10A + 6.
Step 13: Now, we have two equations:
11S = 100C + 10B + A (Equation 1)
8C = 10A + 6 (Equation 2)
Step 14: To find the value of B, we need to solve these equations simultaneously.
To solve the equations, you can substitute Equation 2 into Equation 1, which means replacing C with (10A + 6)/8.
Step 15: After substituting, simplify the equation, and it will leave you with a linear equation in terms of A and B.
Step 16: From the resulting equation, you can solve for B to find its value.
Following these steps, you will be able to find the value of the tens digit of the three-digit number N.
To answer this question, we need to follow several steps to determine the value of the tens digit of the number, N. Let's break it down:
1. Representing the number: Let's assume the three-digit number N can be represented as ABC, where A is the hundreds digit, B is the tens digit, and C is the units or ones digit.
2. Adding the five numbers: According to the problem, the reverse of N (CBA) was added instead of N (ABC). This means the sum of the five numbers will be (ABC + CBA). The sum will increase by 11 times the sum of the digits of N, so the equation can be written as:
(ABC + CBA) = ABC + 11(A + B + C)
3. Expressing CBA in terms of ABC: Since N is a three-digit number, you can represent CBA as 100C + 10B + A.
4. Simplifying the equation: Substituting CBA with 100C + 10B + A in the equation from step 2, we get:
ABC + 100C + 10B + A = ABC + 11(A + B + C)
5. Rearranging and simplifying the equation further: Expanding the equation, we get:
100C + 10B + A = 11A + 11B + 11C
Simplifying further, we have:
89C - A = 10B - 10A
6. Expression for the difference: According to the problem, 8 times the difference between N's units and hundreds digit is 6 more than twice its hundreds digit. This translates to the expression:
8(C - A) = 2A + 6
7. Simplifying the difference equation: Rearranging the equation, we have:
8C - 8A = 2A + 6
8. Combining the equations: We have two equations derived from the given information:
89C - A = 10B - 10A (Equation 1)
8C - 8A = 2A + 6 (Equation 2)
9. Solving the equations: Now, we can solve the equations simultaneously to find the values of A, B, and C. Once we determine the values of A, B, and C, the tens digit (B) will be the answer to the question.
These steps will lead you to the solution of finding the value of the tens digit for the given three-digit number N.
if N's digits are a,b,c, then we know that c>a, since reversing the digits increased the sum.
cba - abc = 99(c-a) = 11(a+b+c)
Since 99 is a multiple of both 9 and 11, a+b+c must be a multiple of 9. So, a+b+c is 9 or 18, since 7+8+9=24 is the maximum sum of digits.
Now, we also know that
8(c-a) = 2c+6
6c-8a=6
3c-4a=3
4a = 3(c-1)
That means that c-1 is a multiple of 4, so c is either 5 or 9, making a either 3 or 6.
The difference c-a is either 2 or 3.
8*2 = 10+6, making c=5
8*3 = 18+6, making c=9
No news there.
That means N' is either N+198 or N+297.
198 = 11*18 = 11(3+10+5)
297 = 11*27 = 11(6+12+9)
And now we have a problem, since b must be less than 10.
Time for other eyes. Have I messed up somewhere in all that?