# Which Numbers Replace the Question Marks?

Can you find out what the numbers below have in common? Can you figure out what comes next?

I = 9,9,9,8,8,8,2,7,?

II = 6,6,7,5,7,5,1,3,?

III = 4,8,4,8,4,8,5,3,?

IV = 5,5,5,6,6,6,6,8,?

V = 2,8,1,4,7,2,8,6,?

VI = 9,8,7,5,4,3,3,5,?

VII = 2,0,2,0,2,0,2,0,?

VIII = 3,6,1,2,2,4,0,1,?

• Are the 8 sequences related? Or are they separate? – Dr Xorile Nov 12 '18 at 22:53
• They are separate. :) – danimoth Nov 13 '18 at 3:40
• Mmm. That makes this extremely difficult. You might find it gets closed as too broad. It becomes a "guess what's in my head" if there's no connecting theme. – Dr Xorile Nov 13 '18 at 4:13
• I believe the OP means that the eight sequences follow the same algorithm, but in different cases – Omega Krypton Nov 13 '18 at 7:39
• Hmmm, I wonder what the 7th one is? I know! 5! (just joking) – Yout Ried Jan 5 '19 at 5:48

I. 9,9,9,8,8,8,2,7,12

II. 6,6,7,5,7,5,1,3,11

III. 4,8,4,8,4,8,5,3,16

IV. 5,5,5,6,6,6,6,8,7

V. 2,8,1,4,7,2,8,6,1

VI. 9,8,7,5,4,3,3,5,-8

VII. 2,0,2,0,2,0,0

VIII. 3,6,1,2,2,4,0,1,5

I. (9+9+9)-(8+8+8)=3 (8+8+8)-(2+7+12)=3

II. (6+6+7)-(5+7+5)=2 (5+7+5)-(1+3+11)=2

III. (4+8+4)-(8+4+8)=-4 (8+4+8)-(5+3+16)=-4

IV. (5+5+5)-(6+6+6)=-3 (6+6+6)-(6+8+7)=-3

V. (2+8+1)-(4+7+2)=-2 (4+7+2)-(8+6+1)=-2

VI. (9+8+7)-(5+4+3)=12 (5+4+3)-(3+5-8)=12

VII. (2+0+2)-(0+2+0)=2 (0+2+0)-(0)=2

VIII. (3+6+1)-(2+2+4)=2 (2+2+4)-(0+1+5)=2

I guess the 1. solution is...

$$8 * 3 = 24$$, because there are three times a 9 and $$9 * 3 = 27$$

The 4. answer could be...

$$68 + 20 = 88$$, because if you erase the commas, you get $$55 + 1 = 56, 56 + 10 = 66, 66 + 2 = 68$$

The 5. answer could be...

$$3$$, because if you erase the commas, all numbers are multiples of $$7$$, the only 2 digit number starting with $$6$$ and being a multiple of $$7$$ is $$63$$

The 6. answer could be...

$$98 - 75 = 23$$, $$75 - 43 = 32$$, $$43 - 35 = 8$$, you can see, that the difference of the first pairs is mirrored, so maybe the next number is $$35 - 8 = 27$$

The 7. answer could be...

$$3$$, because there are $$2^2 = 4$$ times the number 2, separated of $$0$$. So my idea is, that the sequence continues with $$3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,3,0,4,0,...$$

The 8. sequence could work...

like the 5. If all numbers are multiples of $$3$$, the last number could be $$5$$ or $$8$$, if all numbers are multiples of $$6$$, the last number could be $$8$$