Math, asked by sumanyawale00, 11 months ago

The consecutive multiples of the number nine are 9,18,27,36,45,54,63,72,81,90,99,108,117,........The sum of the digits in these multiples are respectively 9,9,9,9,9,9,9,9,9,9,9,18,9...Find the least number which when multiplied by 9 gives a number the sum of whose digits is 27.

Answers

Answered by amitnrw
4

Given :  The consecutive multiples of the number nine are 9,18,27,36,45,54,63,72,81,90,99,108,117,........The sum of the digits in these multiples are respectively 9,9,9,9,9,9,9,9,9,9,9,18,9...

To Find :  least number which when multiplied by 9 gives a number the sum of whose digits is 27.

Solution:

Least number to have sum of digits as 27

can be 999

(∵ max digit is 9 hence 2 digit number maximum possible sum = 9 + 9 = 18

3 digit maximum possible sum  9 + 9 + 9 = 27 )

Divide the number by 9 to get the least number which when multiplied by 9 gives a number the sum of whose digits is 27.

999/9  = 111

111 is the least number which when multiplied by 9 gives a number the sum of whose digits is 27.

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Answered by freefirehacker73356
1

Answer:

define M, M, M, R in 36,23,18,18,45,54,9,18,&18

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