Question

In a three-digit number, the digit in the units' place is four times the digit in the hundreds' place. The digit in the hundreds place is one-third of the digit in the tens place. If the digit in the units' place and tens' place are inter-changed, the new number so formed is 18 more than the original number. What is the original number?​

Answer 1

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Answer 2

Answer:

The Original number is 268

Step-by-step explanation:

Let the number be ABC

Then, according to the Question,

Units digit = 4 × Hundreds digit

C = 4A

A = C/4 ------ 1

Also,

Hundreds digit = Tens digit/3

A = B/3 ------ 2

Now, we can solve it in two ways,

Method 1 (Calculations)

Also, given,

If the tens and units digit are interchanged it becomes 18 more than the original

Interchanging the digits we get,

ACB = ABC + 18

ACB - ABC = 18

Now, we know that,

ABC and ACB are 3 digit numbers

So,

ABC = 100A + 10B + C

ACB = 100A + 10C + B

Then,

ACB - ABC = 18

(100A + 10C + B) - (100A + 10B + C) = 18

100A + 10C + B - 100A - 10B - C= 18

9C - 9B = 18

9(C - B) = 18

C - B = 18/9

C - B = 2

C = B + 2 ------ 3

From eq.1 and eq.2 we get,

C/4 = B/3

Cross multiplying we get,

3C = 4B

Now putting in eq.3 we get,

3(B + 2) = 4B

4B = 3B + 6

4B - 3B = 6

B = 6

From eq.2

A = B/3

A = 6/3

A = 2

From eq.1 we get,

C/4 = 2

C = 4 × 2

C = 8

Hence,

The Original number = 268

Method 2 (Reasoning/Logical)

We have,

A = C/4

C = 4A

This tells is that,

Possible outcomes are A = (1 or 2) and C = (4 or 8)

A can't be 3 or higher as C will become 12 or higher which will not give us a 3 digit number.

Again,

A = B/3

B = 3A

This tells us that,

Possible outcomes are A = (1 or 2 or 3) and B = (3 or 6 or 9)

We need single digit numbers only then we can make a 3 digit number and we know that C is a single digit.

But from the 1st observation, we know that,

A can only be 1 or 2

Then, in 2nd Observation,

C = 3 or 6

So,

A = 1 or 2

B = 3 or 6

C = 4 or 8

Now, we know that,

ABC and ACB are 3 digit numbers

So,

ABC = 100A + 10B + C

ACB = 100A + 10C + B

Then,

ACB - ABC = 18

(100A + 10C + B) - (100A + 10B + C) = 18

100A + 10C + B - 100A - 10B - C= 18

9C - 9B = 18

9(C - B) = 18

C - B = 18/9

C - B = 2

C = B + 2 ------ 3

Here, we must do trial and error,

Let B = 3

From eq.3

C = 3 + 2 = 5

But C can only be 4 or 8

Then,

Let B = 6

C = 6 + 2

C = 8

Here both B and C is True

Thus,

B = 6

C = 8

From eq.1 we get,

A = 8/4

A = 2

Hence,

A = 2

B = 6

C = 8

Hence,

The Original number is 268.

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