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How many different integers can be expressed as the sum of three distinct numbers from the set {3,8,13,18,23,28,33,38,43,48}
Read Solution (Total 10)
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- minimum sum of 3 no.=24
maximum sum =129
now the series 24,29,34,....................,129
no of terms ={(129-24)/5}+1=22 - 9 years agoHelpfull: Yes(29) No(4)
- can anybody send me TCS placement papers of last 10 years in order.please mail me at piyushbag44@gmail.com
- 9 years agoHelpfull: Yes(21) No(6)
- 10c3=120
now
(33+18+8)=(38+8+3)=49
(43+13+3)=(48+8+3)=59
(43+18+13)=(48+23+3)=74
so these are not distinct ,so deduct these three from 120
(120-3)=117 - 9 years agoHelpfull: Yes(11) No(11)
- all except 3
- 9 years agoHelpfull: Yes(3) No(1)
- Minimum sum=24 maximum sum=129. so the no=24,29,34......129. so total =129-24/5=21
- 9 years agoHelpfull: Yes(2) No(4)
- From the above discussion, minimum number = 24 and maximum number = 129. So there exist many numbers in between these two numbers, with common difference of 5. All these numbers can be expressed as a sum of 3 different integers from the given set.
Total numbers = l−ad+1=129−247+1 = 22. - 8 years agoHelpfull: Yes(2) No(2)
- All except 3
Remaining all are separated in 3 distinct number
- 9 years agoHelpfull: Yes(1) No(3)
- 10c3 because of selected 3 different numbers
- 9 years agoHelpfull: Yes(1) No(6)
- can anybody send me TCS placement papers of last 10 years in order.please mail me at
Read more at sumancse.jis@gmail.com - 9 years agoHelpfull: Yes(1) No(6)
- ans:22
minimum sum of 3 no.=24
maximum sum =129
now the series 24,29,34,....................,129 so,these are in ap
we have forula to find no,of terms ={l-a}/d +1
no of terms ={(129-24)/5}+1=22 - 6 years agoHelpfull: Yes(0) No(0)
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