Some Of Multiples

Level pending

Alex had a habit of viewing Natural numbers in groups of 10.

Group 0 : 1 to 10; Group 1 : 11 to 20; Group 2 : 21 to 30; and so on .

One day he observed, that multiplying any number in 'Group 0' with 9, the results have sum of digits equal to 9. If he multiplied any number in 'Group 1' with 9, all the results except 11*9, has sum of digits equal to 9.

He kept on doing this until he figured 'Group m'. When, we multiply each number of 'Group m' with 9, the sum of digits of each of the digits are greater than 18.

If 'm' is the minimum value of such a group fulfilling this property,

If 'n' is the number formed by multiplying 9 with smallest number in group 'm'. Find the leftmost 3 digits of 'n'


The answer is 990.

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1 solution

Jaidip Palit
Sep 7, 2016

We can see that 11*9 yields us a number whose sum of digits is equal to 18.So if we make it 111 we will get sum of digits greater than 18.But if we take the next number we are getting sum of digits less than 18.Therefore according to the conditions given we cannot take 111.Now let us take 1101( P.S. HERE THE KEY IS THE NUMBER 11 AND WE NEED TO USE IT. ) We see that all the 10 numbers satisfies the condition.The group m is 112 and the lowest number obtained by multiplying it by 9 is 9909. So the answer is 990

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