Difference between revisions of "2008 iTest Problems/Problem 19"
Rockmanex3 (talk | contribs) (Solution to Problem 19) |
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The first few numbers in set <math>B</math> are | The first few numbers in set <math>B</math> are | ||
− | <cmath>6,24,60,120,210,336 \cdots</cmath> | + | <cmath>6,24,60,120,210,336 \cdots</cmath> |
To see if these numbers equal the product of two consecutive numbers, note that <math>n(n+1) = n^2+n</math> and check perfect squares just below these values. After guessing and checking, the first values that equal the product of two consecutive numbers are <math>2 \cdot 3 = 6</math> and <math>14 \cdot 15 = 210</math>, and the sum of these two equal <math>\boxed{216}</math>. | To see if these numbers equal the product of two consecutive numbers, note that <math>n(n+1) = n^2+n</math> and check perfect squares just below these values. After guessing and checking, the first values that equal the product of two consecutive numbers are <math>2 \cdot 3 = 6</math> and <math>14 \cdot 15 = 210</math>, and the sum of these two equal <math>\boxed{216}</math>. | ||
Latest revision as of 18:18, 22 June 2018
Problem
Let be the set of positive integers that are the product of two consecutive integers. Let
be the set of positive integers that are the product of three consecutive integers. Find the sum of the two smallest elements of
.
Solution
The first few numbers in set are
To see if these numbers equal the product of two consecutive numbers, note that
and check perfect squares just below these values. After guessing and checking, the first values that equal the product of two consecutive numbers are
and
, and the sum of these two equal
.
See Also
2008 iTest (Problems) | ||
Preceded by: Problem 18 |
Followed by: Problem 20 | |
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