Elitmus
Exam
Numerical Ability
Number System
How many positive number less than 1000 have sum of there digit 17?
Read Solution (Total 13)
-
- digits between 1-100 sum 17 =only 2 (89,98)
digits between 101-200 sum 17 =3 (188,179,197)
digits between 201-300 sum 17 = 4 (296,287,278,269)
and so on.....
digits between 301-400 sum 17 =5
digits between 401-500 sum 17 =6
digits between 501-600 sum 17 =7
digits between 601-700 sum 17 =8
digits between 701-800 sum 17 =9
digits between 801-900 sum 17 =10
but in
digits between 901-999 sum 17 =9 (980,971,962,953,944,935,926,917,908)
so total number is
2+3+4+5+6+7+8+9+10+9=63
- 8 years agoHelpfull: Yes(15) No(2)
- I think answer is 75.
digits between 1-100 sum 17 = 2 (89,98)
digits between 101-200 sum 17 =3 (188,179,197)
digits between 201-300 sum 17 = 4
and so on
we have
2+3+4+5+6+7+8+9+10+11=65
I might be wrong. please correct me!
- 8 years agoHelpfull: Yes(3) No(1)
- I think answer is 65.
digits between 1-100 sum 17 = 2 (89,98)
digits between 101-200 sum 17 =3 (188,179,197)
digits between 201-300 sum 17 = 4
and so on
we have
2+3+4+5+6+7+8+9+10+11=65
I might be wrong. please correct me! - 8 years agoHelpfull: Yes(3) No(2)
- i think it will be 75.
because 2 digit numbers whose sum is 17 are 89 and 98.
in 3 digit numbers whose sum is 17 starts from (1,8,8) which can be 3 ways
and(1,9,7) in 6 ways.
similarly,(2,8,7),(2,9,6) --------12 numbers can be formed ,
(3,9,5)(3,8,6)(3,7,7)-----------15 numbers can be formed,
(4,9,4),(4,8,5),(4,7,6)---------15 numbers can be formed,
(5,9,3),(5,8,4),(5,7,5),(5,6,6)-----18 numbers can be formed,
(9,8,0),(9,0,8)(8,9,0)(8,0,9)------4 numbers can be formed
- 8 years agoHelpfull: Yes(2) No(5)
- Answer will be 65
In my case 1 sorry I say 5 no bw 100 to. 200 ....it will be three only
See My previous answer - 8 years agoHelpfull: Yes(2) No(2)
- i guess 64
- 8 years agoHelpfull: Yes(1) No(1)
- I think answer is 75.
digits between 1-100 sum 17 = 2 (89,98)
digits between 101-200 sum 17 =3 (188,179,197)
digits between 201-300 sum 17 = 4
and so on
we have
2+3+4+5+6+7+8+9+10+11=75
I might be wrong. please correct me! - 8 years agoHelpfull: Yes(1) No(6)
- In the question it's not mentioned only 3 digit number ,so you have to consider 98 & 89 too. Now,let the three digits are a,b and c . Basically you have to find the number of values of a,b and c,such that a+b+c=17
So there are 19c2 ways, in this we have to subtract the case in which a,b,c can be zero. So, 19c2-(9c2*3)=63
So total numbers less than 1000 having sum of their digits as 17 is 63+2=65. This is a right answer. - 8 years agoHelpfull: Yes(1) No(1)
- my ans is
1330
- 8 years agoHelpfull: Yes(0) No(4)
- i guess its 48....bcz the smallest 2 digit num that sum 17 is 98.....so the no of combinations of their sum is 48....
- 8 years agoHelpfull: Yes(0) No(1)
- No of 2 digits=2
No of 3 digits=fix first hundreds digit
When it is 1,we have 5 ways
I.e 187,188,178,197,179....
Similarly
For 2. 4
For 3. 7
For 4. 6
For 5. 7
For 6. 8
For 7. 9
For 8. 10
For 9. 9
Total = 65+2= 67 - 8 years agoHelpfull: Yes(0) No(4)
- x+y+z=17
so no of possible solutions is given by
N+R-1 C R-1
N= 17 R= 3
so ans = 171 - 7 years agoHelpfull: Yes(0) No(0)
- Ans is 171
- 7 years agoHelpfull: Yes(0) No(0)
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