Math, asked by swarupmajee, 11 months ago

Find the number of such number
of multiples of 9 less than 2012 and having sum of the digits as 18.

Answers

Answered by RitaNarine
10

The number of multiples of 9 less than 2012 and having sum of digits as 18 is 118.

We have the multiples of 9.

The last multiple of 9 less than 2012 is 2007.

Also digits should add up to 18.

For 2 digit number,

  • only 99 adds up to 18.

For 3 digit numbers ,

  • Starting with 1 : 198 and 189 add up to 18 : 2 nos.
  • Starting with 2 : 297, 288, 297 ad up to 18 : 3nos.
  • Similarly,
  • 3 : 4 nos.
  • 4 : 5 nos.
  • 5 : 6 nos.
  • 6 : 7 nos.
  • 7 : 8 nos.
  • 8 : 9 nos.
  • 9 : 10 nos.
  • Total = 1 + 2+ 3+ ... 10 = 55

For 4 digit number only 1000s are possible.

Second digit starting with ,

  • 0 : 1098 , 1098 =2 nos.
  • 1 : 3 nos
  • 2 : 4 nos.
  • 3 : 5 nos
  • 4 : 6 nos
  • 5 : 7 nos
  • 6 : 8 nos
  • 7 : 9 nos
  • 8 = 10 nos
  • 9 = 9 nos

Therefore,

total  = 2 + 3 + 4 + ...10 + 9 = 55 -1 + 9 = 63.

Therefore total number of multiples less than 2012 and having sum of digits as 18 is 118.

Answered by krishika2086
3

Answer:

109

Step-by-step explanation:

2 digit number:- 99 :-1 number

3digit number:-

189, 198 :-2 numbers

279,297,288:-3 numbers

369, 396, 387, 378:-4 numbers

similarly, 477:- 5 numbers

567:-6numbers

666:- 7 numbers

765.:- 8

So, 1+2+3+4+5+6+7+8+9+10=55numbers

4digit numbers:-

1098, 1089= 2 numbers

1179, 1197= 2 numbers

1269, 1296 , 1287, 1278 = 4 numbers

1359, 1395= 5 numbers

So, 2+2+4+5+6+7+8+9+10= 54 numbers

Total numbers:- 55+54= 109

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