Difference between revisions of "2007 Cyprus MO/Lyceum/Problem 29"
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==Problem== | ==Problem== | ||
| − | The minimum value of a positive integer <math>k</math>, for which the sum <math>S=k+(k+1)+(k+2)+\ldots+(k+10)</math> is a perfect square, is | + | The minimum value of a positive integer <math>k</math>, for which the sum <math>\displaystyle S=k+(k+1)+(k+2)+\ldots+(k+10)</math> is a [[perfect square]], is |
<math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}</math> | <math> \mathrm{(A) \ } 5\qquad \mathrm{(B) \ } 6\qquad \mathrm{(C) \ } 10\qquad \mathrm{(D) \ } 11\qquad \mathrm{(E) \ } \mathrm{None\;of\;these}</math> | ||
==Solution== | ==Solution== | ||
| − | <math>k+(k+1)+(k+2)+\ldots+(k+10)=11k+55=11(k+5)</math> | + | <math>\displaystyle k+(k+1)+(k+2)+\ldots+(k+10)=11k+55=11(k+5)</math> |
| − | + | Thus, <math>k+5</math> must be divisible by <math>11</math>, and the minimum for <math>k</math> is <math>6\Longrightarrow\mathrm{ B}</math>. | |
==See also== | ==See also== | ||
| − | + | {{CYMO box|year=2007|l=Lyceum|num-b=28|num-a=30}} | |
| − | + | [[Category:Introductory Algebra Problems]] | |
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| − | |||
Latest revision as of 17:12, 6 May 2007
Problem
The minimum value of a positive integer 
, for which the sum 
is a perfect square, is
Solution
Thus, 
must be divisible by 
, and the minimum for is 
.
See also
| 2007 Cyprus MO, Lyceum (Problems) | ||
| Preceded by Problem 28 |
Followed by Problem 30 | |
| 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 | ||
