Question

If we divide a two digit number by sum of its digits we get 4 as quotient and 3 as remainder.Now if we divide two digit number by the product of digits we get 3 as quotient and 5 as remainder.Then the two digit number is:

Answer 1

Suppose the two-digit number is 10x+y, where x is at the place of tens and y is at the place of ones.

According to the question,

10x+y=4(x+y)+3 and 10x+y=3xy+5

Taking first equation,

10x+y=4(x+y)+3

Solving further we get,

2x-y=1  

Now, we can assume the value of y thus get the value of x  

Putting y=1 we get x=1

Now, verify these values by putting in second equation i.e.

10x1+=3x1x1+5

11=8 which is not true.

Putting y=2 we get x=3/2 which is not possible.

Now, put y=3 we get x=2

Now, verify these values by putting in second equation i.e.

10x2+3=3x2x3+5

23=23

Thus, the two-digit number will be 23

Answer 2

Answer:

23

Step-by-step explanation:

Let n(x,y) be the two digit number with digits x & y.

Then, n(x,y) = 10x+y

As per first statement,

=> 10x+y = 4(x+y)+3

=> 6x-3y-3 = 0

=> y = 2x-1  --- eq(1)

As per second statement,

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

=> 10x+(2x-1) = 3x(2x-1) + 5 ∵ y = 2x-1 from eq(1)

=> 12x-1 = 6$x^{2}$ - 3x + 5

=> 6$x^{2}$ - 15x + 6 = 0

=> 2$x^{2}$ - 5x +2 = 0

=> 2$x^{2}$ - 4x -x +2 = 0

=> 2x(x-2) -1(x-2) = 0

=> (2x-1)(x-2) = 0

As x & y are single digits in the range of 0 to 9, x=2 is the only valid solution.

So, by eq(1), y = 2x-1 = 3

∴ The two digit number must be 23.