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Age 11 to 14

Challenge Level

- Problem
- Student Solutions

For 4 numbers, the first times the last is always 2 less than the product of the middle numbers.

For 5 numbers, the first times the last is always 3 less than the second times the fourth.

For $n$ numbers, the first times the last is always $n-2$ less than the second times the second-last.

**Why?**

Let the first number be $a$

__4 numbers__

$a, (a+1), (a+2), (a+3)$

First times last: $a(a+3) = a^3 + 3a$

Second times third: $(a+1)(a+2) = a^2 + 3a + 2$

Difference: $2$

__5 numbers__

$a, (a+1), (a+2), (a+3), (a+4)$

First times last: $a(a+4) = a^3 + 4a$

Second times fourth: $(a+1)(a+3) = a^2 + 4a + 3$

Difference: $3$

__$n$ numbers__

$a, (a+1), (a+2), ..., \left(a+(n-2)\right), \left(a+(n-1)\right)$

First times last: $a(a+n-1) = a^2 +an-a$

Second times second last: $(a+1)(a+n-2) = a^2+an-a+n-2$

Difference: $n-2$