Question

The digits of a two-digit number differ by 3. If the digits are interchanged,
and the resulting number is added to the original number, we get 143. What can be the
original number?​

Answer 1

Explanation:

x= first digit and y= second digit

the digits are differ by 3 ie.

[x-y]=3

the sum of number and the number when its digit are interchanged is 143

[10×+y] + [10y+x]= 143

ie

x+y=13

if you take x-y=3 then

x=8 and y=5 and the number will be 85

Answer 2

Solve:-

According to question ❓:-

Take , for example , a 2 digit number , say, 56.

Take , for example , a 2 digit number , say, 56. It can be written as 56 = (10 × 5) + 6.

~If the digits in 56 are interchanged , We get 65 , which can be written as (10×6) + 5.

• Let us take the two digit number such that the digit in the unit place is b.

• The digit in the tens place is different from b by 3.

• Let us take it as b + 3.

• So the two digit number is 10 (b+3) + b

= 10b + 30 + b

= 11b + 30

With interchange the digits , the resulting two number will be

= 10b + (b+3) = 11b + 3

If we add these two two digit numbers , their sum is

(11b + 30) + (11b + 3) = 11b + 11b + 30 + 3

= 22b + 33

It is given that the sum is 143.

Therefore , 22b + 33 = 143

• 22b = 143 - 33

• 22b = 110

• b = 110/22

• b = 5

Now,

Unit place = b

The value of b is 5 .

So, The unit place is 5.

Unit place = 5

Tens place = b + 3

So , We have to sum the both numbers.

Value of b = 5

So ,

b + 3

= 5 + 3

= 8

hence,

the number is 85.

Answer is verified.

Answering check ✅ => On interchange of digits the number we get is 58.

The sum of 85 and 58 is 143 are given.

Hope it helps you ❤️