please help with math?!?!?!?!:( and please show solutions:D?

You need to use a little logic here:

The only possible numbers would be 16, 27, 38, or 49. nothing else fits the 1st requirement.

So, you divide 16 by 7, 27 by 9, 38 by 11, and 49 by 13 to figure out which one would give you 3 with a remainder of 5, which would be 38.

Let: x = the ones digit

y = the tens digit

Let us translate each statement:

"The units digit of a 2-digit number exceeds its tens digit by 5"

In symbols, it is x = y + 5

"If the number is divided by the sum of its digits, the quotient is 3 and the remainder is 5"

In symbols, it is

((10y+x)/(x+y)) = 3 + (5 / x+y)

multiply both sides by (x + y),

10y + x = 3(x + y) + 5

10y + x = 3x + 3y + 5

10y - 3y + x - 3x = 5

7y - 2x = 5

We have formed two linear equations:

x = y + 5

7y - 2x = 5

Substituting the first equation to the second equation,

7y - 2(y + 5) = 5

7y - 2y - 10 = 5

5y - 10 = 5

5y = 5 + 10

5y = 15

y = 3

To find x,

x = y + 5

x = 3 + 5

x = 8

Therefore,

the number is 38.

tens digit = x unit digit = x+5

no. = 10x + x+5 = 11x+5 sum of digits = 2x+ 5

11x+5/2x+5 = 3+ (5/2x+5

x=3 number is 38

Let a and b be the digits. Then the number N = 10a + b

The relations are:

a + 5 = b

(a+b)*3 + 5 = 10a +b

or 7a - 2b = 5

Solving these equations, you get 7a - 2a - 10 = 5, or a = 3 and b = 8

The number is 38.

N = 10x + y

y = x + 5

[10x + y] / [x + y]

[10x + (x + 5)] / [x + (x + 5)]

[11x + 5] / [2x + 5] = 3 + 5/[2x + 5]

11x + 5 = 3[2x + 5] + 5

11x = 6x + 15

5x = 15

x = 3

y = x + 5 = 3 + 5 = 8

N = 10x + y

N = 10*3 + 8

N = 38

b = a + 5

(10a+b)/(a+b) = 3 + 5/(a+b)

multiply across by (a + b)

10a + b = 3(a + b) + 5

10a + b = 3a + 3b + 5

7a - 2b = 5

7a - 2(a + 5) = 5

7a - 2a - 10 = 5

5a = 15

a = 3, so b = 8

Answer is 38