GRE考试

【单选题】

A、sequence of numbers begins 1, 1, 1, 2, 2, 3 and then repeats this pattern forever. What is the sum of the 135th, 136th, and 137th numbers in the sequenceA、3

B、
B、4
C、5
D、6
E、7

网考网参考答案：C
网考网解析：
Since the pattern has six digits, divide 135 by 6. The quotient is 22 and the remainder is 3. Since 22 × 6 = 132, the 132nd number completes the pattern for the 22nd time. So the 133rd, 134th, and {{U}}135{{/U}}th numbers are Is, and the {{U}}136{{/U}}th and {{U}}137{{/U}}th are 2s; and their sum is {{U}}1{{/U}}+{{U}}2{{/U}}+{{U}}2{{/U}}=5, document.getElementById("warp").style.display="none"; document.getElementById("content").style.display="block"; 查看试题解析出处>>