Difference between revisions of "1967 AHSME Problems/Problem 39"
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== Solution == | == Solution == | ||
| − | <math>\fbox{B}</math> | + | The last element of the 21st set is <math>\frac{21\times 22}{2}=231</math>, and hence the first element of the 21st set is <math>211</math>. |
| + | So <math>S_{21}=\frac{231\times 232}{2}-\frac{210\times 211}{2}=4641</math>, hence our answer is <math>\fbox{B}</math> | ||
== See also == | == See also == | ||
| − | {{AHSME box|year=1967|num-b=38|num-a=40}} | + | {{AHSME 40p box|year=1967|num-b=38|num-a=40}} |
[[Category: Intermediate Algebra Problems]] | [[Category: Intermediate Algebra Problems]] | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Latest revision as of 00:40, 16 August 2023
Problem
Given the sets of consecutive integers 
,
,
,
,
, where each set contains one more element than the preceding one, and where the first element of each set is one more than the last element of the preceding set. Let 
be the sum of the elements in the nth set. Then 
equals:
Solution
The last element of the 21st set is 
, and hence the first element of the 21st set is 
.
So 
, hence our answer is 
See also
| 1967 AHSC (Problems • Answer Key • Resources) | ||
| Preceded by Problem 38 |
Followed by Problem 40 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
