The units digit of a three-digit number is 5. The sum of its digits is 11. If the units and hundreds digits are reversed, the sum of the new number and the original number is 787
Find the original number.

The original number is 245.

Step-by-step explanation:

Let A be the hundreds digit of the original number, B be the tens digit of the original number, and C be the units digit of the original number.

We know from the first two sentences in the problem that:

C = 5

A + B + C = 11

Substituting the value of C, we get:

A + B + 5 = 11

A + B = 6

In the third sentence, we are told that when the number is added to the reverse of itself, the sum of the two numbers is 787. Expressed mathematically:

100A + 10B + C + A + 10B + 100C = 787

Where (100A + 10B + C) gives the value of the original number and (A + 10B + 100C) gives the value of the number when reversed. Simplified, we get:

101A + 20B + 101C = 787

Rewriting the "A + B = 6" equation from above as B = 6 - A, we substitute (6 - A) into the equation above for B and substitute 5 for C (because we know C = 5):

101A + 20(6 - A) + 101 * 5 = 787

101A + 120 - 20A + 505 = 787

81A + 625 = 787

81A = 162

A = 2

Now that we have A, we can get B using A + B = 6:

2 + B = 6

B = 4

To summarize:

A = 2

B = 4

C = 5

So the original number must be 245.

solution:

→c= a b , that is product of a and b is possible only when when a is m × n matrix and b is n × p matrix.

that is, number of columns of a= number of rows of b.

so, size of matrix , c= a b is , if a = (m × n)matrix

b= (n × p)matrix

then a b = [ (m × n)matrix] ×[   (n × p)matrix ]

→→c= a b= [m × p ] matrix

→→size of product a b will be equal to = number of rows of a × number of columns of b