Question

the sum of the digits of a two digit number is 4 the number got by interchanging the digits is 18 less than the original number what is the number​

Answer 1

Answer:

31

Step-by-step explanation:

x+y=4

10x+y- 18= 10y+x

9x - 18=9y

________divide by 9_______

x-y=2

x+y=4

x-y=2

thus, 2x=6, or x= 3

x-y=2

3-y=2

y=1

so, number is 10(3) + 1 or 31

check-

13( digits interchanged) is 18 less than 31

hence solved.

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Answer 2

The original number is $31$

Step-by-step explanation:

Let the number at ten's place be $y$ and number at one's place be $x$

The original number is $10y+x$

According to the given condition that the sum of the digits of the number is 4.

Therefore, we have

$x+y=4\\y=4-x\cdots(1)$

Now, on interchanging the place values of the original number we have the two-digit number as $10x+y$

Also, according to the given condition we have,

$10y+x-(10x+y)=18$

Solving it further we have,

$10y+x-10x-y=18\\9y-9x=18\\9(y-x)=18\\y-x=\frac{18}{2} \\y-x=2\cdots(2)$

Now, using equation (1) in equation (2) we have,

$y-x=2\\(4-x)-x=2\\4-2x=2\\-2x=2-4\\-2x=-2\\x=1$

Now using the value of $x$ we have,

$y-1=4\\y=3$

Hence the original number is,

$10y+x=10(3)+1\\=31$