Question

find the smallest number of five digits which leaves 8 as remainder when divided by the number 54,63 and 81​

Answer 1

Answer:

11111  is the of this question

Answer 2

Answer: 10214 is the right answer

Step-by-step explanation:

The smallest number of 5 digits shall leave remainder of 8 when divided by 3 different numbers.

Ex. Consider number 240 it is divisible by 30, 20, 40.

Number 243 will leave a remainder of 3 when divided by 30 or 20 or 40.

In other words 240 is a common multiple of 30, 20, 40. (Make note I used word Common multiple instead of Lowest common multiple; LCM of 30, 20, 40 being 120)

Going back to question

Step 1 find out LCM of 54,63,81 = 9x6x7x9 = 3402

If 5 digit answer is not mentioned, answer would have been 3402 + 8 = 3410. Since 3410 will leave a remainder of 8 when divided by 54, 63 or 81.

But our answer requires a 5 digit number.

Hence we mutiply 3402 by 3 to get lowest 5 digit multiple = 10206.

This number, 10206 is perfectly divisible by 54, 63 and 81 you can check.

Next it should leave a remainder of 8 so just add 8 to it which makes 10214 the right answer to this question