integers ?

Question

sum of the digits of a two-digit integer is 12   The value of the integer is 13 times the tens digit - what is the integer?

Answer 1

This is a simultaneous equation. The sum of two digits, say a and b, must be 12. So, a+b=12. The tens digit will be a in this problem to preserve order. Therefore, 13 times a is the integer, say c, so 13a=c. Because a is also a tens value, we know that 10 times the tens digit plus the ones digit will also equal the integer, or 10a+b=c. Since we need the value of the integer, we will solve the last equation plugging in values for both a and b that are represented in terms of c. From the first equation, we know that a is 12-b. Plugging this into the second equation, we can write b as a function of c. 156-13b=c, or 12-c/13=b. From the second equation, we also know that a=c/13. So, plugging into the third equation, we have 10(c/13)+12-c/13=c. Simplifying, 12=4c/13. c=39. You'll notice the sum of the digits is 12, and 13 times 3 is 39, satisfying the conditions set forth in the problem. There may be a simpler way to do this, but I'm not sure. 

Answer 2

for this kind of problem you need to recall how we form number in lower classes.

for example :if tens digit is 5 and ones digit is 2 what will be the number.yes it is 52 but in lower classes we learn that we need to multiply digits with their place value and then add to get the number

i.e.(5*10)+(2*1) = 50 +2 = 52  

 

Now let tens place digit is x and ones place digit is y in the two digit number,so the number will be :

Number :    (x*10) + (y*1) = 10x +y

sum of digits : x+y= 12  ==> y=12-x (this value of y we will use in another equation)

 

 and also,  value of integer = 13 times tens digit 

i.e.               10x +y = 13 x

                    10 x + (12-x) = 13x      (putting y=12-x)

                    10x + 12 - x = 13 x

                     9x + 12 = 13 x 

                            12   = 13x - 9x 

                            12   =  4x

                             x = 12/4 = 3

 

So tens place digit = x = 3

once place digit = y = 12-x = 12- 3 = 9

 

 number or integer = 10x + y = 10 *3 + 9 = 30 + 9 = 39

Answer 3

Let m be the tens digit and n be the ones digit.  Then what you are given translates into the following equations.

 

m+n=12                         (1)

 

10m+n=13m                   (2)

 

Equation (2) is equivalent to

 

3m-n=0                          (3)

 

Now add Equations (1) and (3).  Note that the n cancels out.

 

4m=12

 

m=3

 

Substitute this in Equation (1)

 

3+n=12

 

n=9

 

The number is 39

 

 

 

Answer 4

ab is the number

a+b=12

10a+b=13a

b=13a-10a

b=3a

substitute into a+b=12

a+3a=12

4a=12

a=3

b=9

39 is the number