2 + 6+ 10+ 14 + 18 + 22 + 26+ 30+ 34 + 38 = 200 To inme se kisi 5 ko Total kar ke 100 banaao. Agar dimaag me dum hai to reply karo. It's A Challenge For You !!
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23 Answers

thats integer sequens with start=2, step size=4, num terms=100

me dont hav a klue bout wot tung yu yuze, so yer werds dont meen nuthun

sum av sequens=20,000

last number=398

averaej=200
by
That question didnot say that there is restricted to addition. 6+10*2+30+38=100. This is d answer. Reasoning hai
by

38+26+24+10+2=100

by
There are ten numbers adding up to 200. If you could take out five numbers that add up to 100, then automatically, the other five must also add up to 100. Ten numbers add up to 200, means that the average value is 20. Five numbers add up to 100, also anaverage of 40. Once you pick one number to remove (n), then the other four must have an average of (100 - n)/4 It would be nice if you could pick out a number divisible by 4... and there is not one. --- Another way is to look at "modulo 4" arithmetic. A "modulo" is the remainder, when you divide a number by the base. Here, we make the base 4. Every number in the list is congruentto 2 modulo 4. Meaning that when you divide each number by 4, you get a remainder of 2. Modulo arithmetic has a few simple rules, one of them being that if you add two numbers, the modulo of the sum will be equal to the sum of the individual modulo. For example 14 18 = 32 14 is congruent to 2 (mod 4) 18 is congruent to 2 (mod 4) therefore their sum must be congruent to 4 (mod 4) which is thesame as 0 (mod 4) because 4 would leave a remainder of 0, when divided by 4 32 divided by 4 = 8, with a remainder of 0. If you add 2 to itself an even number of times (like 10 times), the final number will be divisible by 4. 200 is congruent to 0 (mod 4): it is divisible by 4. If you take an ODD number of numbers, all congruent to 2, their sum CANNOT be divisible by 4. For example, take the last three numbers: 30 34 38 = 102 102 is congruent to 2 (mod 4), as promised by the sum of three numbers congruent to 2 (mod 4). Yet, 100 is divisible by 4. Therefore, it is impossible to make these numbers add up to 100, by taking five of them (or three or seven or any odd combination). The ONLY way to get 100 is to take the sum of an EVEN number of numbers.
by
200-(10+30+26+34)=100
by
14+6*2+38+22=100.. total means does nt mean to add...
by
Kinhi 5 ko totalkrke 100 bnana hai Kinhi 5 ka sum 100 ho Aisa nhi likha So 2+98=100 6+94=100 10+90=100 14+86=100 18+82=100 I think so
by
[(22/2=11)] (11+14) * (22-18)=100
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--> 2+6+10+14+18+22 +26 +30 +34 +38 =200 --> 2+6 +10 +14 +18 +22 +26 +30 +34 +38 - 200=0 Hence, (- 200)+38+22+30+10=100 
by
(38+22)+(6-2)*(10)=60+40=100
by
26+38+24+10+2=100
by


2 + 6+ 10+ 14 + 18 + 22 + 26+ 30+ 34 + 38 = 200
To in me se kisi 5 ko total kar ke 100 banao

by
38+26+18+10+6+2=100
by
1 to 100
by Level 1 User (140 points)

2+6+10+14+18+22+26+30+34+38=200

by Level 9 User (42.6k points)

20,000 is your answer

by Level 1 User (460 points)
Yeah the answer is 100. 6 + 10 × 2 + 30 + 38 = 100

 

Hope this helps solve your math problem! :)
by Level 3 User (2.7k points)
Average = 200

 

So your answer must be 200, right? Look, I'm only in the 7th grade i'm not good at this stuff
by Level 1 User (240 points)
6+14+18+26+36=100
by
2,6+10+26+38=
by
the answer is 100. 6 + 10 × 2 + 30 + 38 = 100

hope this solves it !
by Level 1 User (840 points)

thats integer sequens with start=2, step size=4, num terms=100

me dont hav a klue bout wot tung yu yuze, so yer werds dont meen nuthun

sum av sequens=20,000

last number=398

averaej=200

by
reshown by
2,6,10,14,18,22,26,30,34,38 pach achaer Ka 100 Lana hai
by

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