Maths puzzle: Can you figure this one out? 10 points!?

Question

6 consecutive numbers. E.G 3+4+5+6+7+8 the first 3 add to 12, the second three add to 21 (12 reversed). Can you find 2 more of these sequences?

Answer 1

14 + 15 + 16 + 17 + 18 + 19

14+15+16 = 45

17+18+19 = 54

25 + 26 + 27 + 28 + 29 + 30

25+26+27 = 78

28+29+30 = 87

Hope this helps.

Answer 2

n+n+1+n+2 = 3n+3

n+3+n+4+n+5= 3n+12

3n+3 = 10X +Y

3n+12=10Y+X

.....................

9 = 9Y- 9X

1= Y-X

Y=1+X

...................

3n+3 =11X +1

3n =11X-2

x can be 1,4,7

3,4,5,6,7,8,

14,15,16,17,18,19

25,26,27,28,29,30

Answer 3

reversed couple like 12 and 21 are in fact more or less than the other 9xn like 21-12=9 or 42-24=18

Besides, we will have the second three is the first three plus 3x3 (6-3=3, 7-4=3...)

Let a the first three

We have 3a+9=9xn+3a it is easy to see this is always true with n=1 (with every a or we can say this is true with avery array of numbers)

This also means 2 digits of the sum must have the difference is 1 like 21 12, 23 32, 78 87 etc.

So u just have to find 3 number with the sum have two digit difference is 1 X+X+1+X+2=Yx10+Y+1(for the first three and (Y+1)x10+Y for the second 3 with Y is a 1 digit number)

3X+3=11Y+1 <=> X=(11Y-2)/3 is integer

only Y = 1,4,7 is satisfied

X=3,14,25 (X is the smallest of 6)