Difference between revisions of "2008 iTest Problems/Problem 19"

(Solution to Problem 19)
 
m (Solution)
 
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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 17:18, 22 June 2018

Problem

Let $A$ be the set of positive integers that are the product of two consecutive integers. Let $B$ be the set of positive integers that are the product of three consecutive integers. Find the sum of the two smallest elements of $A \cap B$.

Solution

The first few numbers in set $B$ are \[6,24,60,120,210,336 \cdots\] To see if these numbers equal the product of two consecutive numbers, note that $n(n+1) = n^2+n$ and check perfect squares just below these values. After guessing and checking, the first values that equal the product of two consecutive numbers are $2 \cdot 3 = 6$ and $14 \cdot 15 = 210$, and the sum of these two equal $\boxed{216}$.

See Also

2008 iTest (Problems)
Preceded by:
Problem 18
Followed by:
Problem 20
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