Question

If a 3 digit number is 30 times the sum of its digits, and if the digits are consecutive, then how many such three digits
number are possible?
2​

Answer 1

Answer:

you gave the answer in your question?

Answer 2

Solution :-

Let us assume at digit at unit place = x .

As all three digits are consecutive numbers .

So,

→ Digit at ten's place = (x + 1)

→ Digit at hundred's place = (x + 2) .

Than,

→ Required three digit number will be :- {100(x + 2) + 10(x + 1) + x } .

Now, we have given that, the three digit number is 30 times the sum of its digits.

A/q,

→ {100(x + 2) + 10(x + 1) + x } = 30 * { (x + 2) + (x + 1) + x }

→ {100x + 200 + 10x + 10 + x} = 30{3x + 3}

→ 111x + 210 = 90x + 90

→ 111x - 90x = 90 - 210

→ 21x = (-120)

→ x = (-120/21) .

Conclusion :- Negative value of unit digit is not possible. Therefore, Digit at hundred's place must be x.

Now,

→ Digit at hundred's place = x

→ Digit at ten's place = (x + 1)

→ unit digit = (x + 2)

Than,

→ Three digit number = {100x + 10(x+1) + (x+2)}

A/q,

→ {100x + 10(x+1) + (x+2)} = 30{x + x + 1 + x + 2}

→ 100x + 10x + 10 + x + 2 = 30(3x + 3)

→ 111x + 12 = 90x + 90

→ 111x - 90x = 90 - 12

→ 21x = 78

→ x = (78/21) = (26/7) .

Therefore, with this also no such digit is possible.

Hence, we can conclude that, No such three digit is Possible.