Difference between revisions of "2003 AMC 10B Problems/Problem 7"
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The first three values in the sum are equal to <math>1,</math> the next five equal to <math>2,</math> the next seven equal to <math>3,</math> and the last one equal to <math>4.</math> For example, since <math>2^2=4</math> any square root of a number less than <math>4</math> must be less than <math>2.</math> Sum them all together to get | The first three values in the sum are equal to <math>1,</math> the next five equal to <math>2,</math> the next seven equal to <math>3,</math> and the last one equal to <math>4.</math> For example, since <math>2^2=4</math> any square root of a number less than <math>4</math> must be less than <math>2.</math> Sum them all together to get | ||
− | <cmath>3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{\ | + | <cmath>3\cdot1 + 5\cdot2 + 7\cdot3 + 1\cdot4 = 3+10+21+4 = \boxed{\textbf{(B) \ } 38}</cmath> |
==See Also== | ==See Also== | ||
{{AMC10 box|year=2003|ab=B|num-b=6|num-a=8}} | {{AMC10 box|year=2003|ab=B|num-b=6|num-a=8}} |
Revision as of 18:55, 26 November 2011
Problem
The symbolism denotes the largest integer not exceeding
. For example,
and
. Compute
Solution
The first three values in the sum are equal to the next five equal to
the next seven equal to
and the last one equal to
For example, since
any square root of a number less than
must be less than
Sum them all together to get
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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 | ||
All AMC 10 Problems and Solutions |