Question

the digits of a two digit number differ by 3.If digits are interchanged and the resulting number is added to the original number, we get 121 .find the original number

Answer 1

Hey MATE !

There can be 2 cases :

1) The number at units place is greater than number at tens place.

So according to the question the the number at tens place is X and the number at units place is (X + 3).

So the original number(ON) is
10X + (X+3) = 11X +3.

Reversing the digits we get new number (NN) => 10(X+3) + X = 11X + 30.

It is given that ON + NN = 121

(11X + 3) + (11 X + 30 ) = 121

22X + 33 = 121

22X = 121 - 33

22X = 88

X = 4

So the number at tens place in original number is X = 4 and at units place is (X +3) ==> (4+3) = 7.

So the number is 47.

Case2 : Number at units place is smaller and number at tens place is bigger so the number formed in this case is 74.


Hope it helps

Hakuna Matata :))

Answer 2

Let us take the two digit number such that the digit in the units place is x. The digit

in the tens place differs from x by 3. Let us take it as x + 3. So the two-digit number is

10 (x + 3) + x = 10x + 30 + x = 11x + 30.

With interchange of digits, the resulting two-digit number will be 10x + (x + 3) = 11x + 3

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

(11x + 30) + (11x + 3) = 11x + 11x + 30 + 3 = 22x + 33

It is given that the sum is 121.

Therefore, 22x + 33 = 121

22x = 121 – 33

22x = 88

x=4

The units digit is 4 and therefore the tens digit is 4 + 3 = 7.

Hence, the number is 74 or 47.