If a Gareth sequence begins 10,8 , then the 3 rd number in the sequence is $10-8=2$, the 4 th is $8-2=6$, the 5 th is $6-2=4$, the 6 th is $6-4=2$, the 7 th is $4-2=2$, the 8 th is $2-2=0$, the 9 th is $2-0=2$, the 10 th is $2-0=2$, and the 11 th is $2-2=0$.

Thus, the resulting sequence is $10,8,2,6,4,2,2,0,2,2,0, \ldots$.

The first 5 numbers in the sequence are 10,8,2,6,4, the next 3 numbers are $2,2,0$, and this block of 3 numbers appears to repeat.

Since each new number added to the end of this sequence is determined by the two previous numbers in the sequence, then this block of 3 numbers will indeed continue to repeat. (That
is, since the block repeats once, then it will continue repeating.)

The first 30 numbers of the sequence begins with the first 5 numbers, followed by 8 blocks of $2,2,0$, followed by one additional 2 (since $5+8 \times 3+1=30$ ).

The sum of the first 5 numbers is $10+8+2+6+4=30$.

The sum of each repeating block is $2+2+0=4$, and so the sum of 8 such blocks is $8 \times 4=32$. Thus, the sum of the first 30 numbers in the sequence is $30+32+2=64$.

ANSWER: (E)
A Gareth sequence is a sequence of numbers in which each number after the second is the non-negative difference between the two previous numbers. For example, if a Gareth sequence begins 15,12 , then

- the third number in the sequence is $15-12=3$,
- the fourth number is $12-3=9$,
- the fifth number is $9-3=6$,

and so the resulting sequence is $15,12,3,9,6, \ldots$ If a Gareth sequence begins 10,8 , what is the sum of the first 30 numbers in the sequence?
(A) 40
(B) 72
(C) 34
(D) 56
(E) 64