Question

100pts The digits of a two digit number differ by 3. If the digits are interchanged and the two numbers are added, their sum is 77. Find the numbers

Answer 1

Let the digit at units place be x

Let the digit at tens place be x + 3

10(x + 3) + x

When digits interchanged, the number becomes 10x + x + 3

According to the question,

10(x + 3) + x + 10x + x + 3 = 77

10x + 30 + 12x + 3 = 77

22x + 33 = 77

22x = 77 - 33

22x = 44

Answer 2

The digits of a two-digit number differ by ⇒ 3

When the gits are interchanged and the two numbers are added, their sum is ⇒ 77

Let's assume that the digits are 'a' and 'b'.

So according to the question,

the digit of the two-digit number differs by 3 ⇒ $\sf b-x=3$

⇒ $\sf b-a+a=3+a$

⇒ $\sf b=a+3$ .......(i)

From the question, we know that the digits are interchanged and the two numbers are added giving the sum 77. So now let's proceed with this information.

⇒ $\sf 10a+b+10b+a = 77$

⇒ $\sf 11a+11b = 77$

⇒ $\sf 11\times(a+b) = 77$

⇒ $\sf a+b = 77 \div 11$

⇒ $\sf a+b = 7$

⇒ $\sf a+a+3=7$

⇒ $\sf 2a+3=7$

⇒ $\sf 2a = 4$

⇒ $\sf a = \frac{4}{2}$

⇒ $\sf a=2$

Now we have to put the value of 'x' in (i).

⇒ $\sf b = 2+3$

⇒ $\sf b=5$

∴ The original value is ⇒ (10×5)+2 = 50+2 = 52

∴ The interchanged value is ⇒ (10×2)+5 = 20+5 = 25