Arithmetic Operation

Algebra Level 4

Consider the arithmetic operations \blacktriangle and \blacktriangledown defined by

a b = { a if a b b if a < b a b = { a if a b b if a > b \large \begin{aligned} a \blacktriangle b & = \begin{cases} a & \text{ if } \lvert a \rvert \geq \lvert b \rvert \\ b & \text{ if } \lvert a \rvert < \lvert b \rvert \end{cases} \\ a\blacktriangledown b & = \begin{cases} a & \text{ if } \lvert a \rvert \leq \lvert b \rvert \\ b & \text{ if } \lvert a \rvert > \lvert b \rvert \end{cases} \end{aligned}

How many integers k k are there such that ( 20 9 ) ( k 5 ) = 5 ? (-20 \blacktriangle 9) \blacktriangledown (k \blacktriangle 5)=5?

8 11 10 9

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1 solution

Mahindra Jain Staff
Feb 16, 2014

Since 20 9 = 20 -20 \blacktriangle 9=-20 , thus the equation in the problem can be rewritten as ( 20 9 ) ( k 5 ) = 20 ( k 5 ) = 5. (-20 \blacktriangle 9) \blacktriangledown (k \blacktriangle 5)=-20 \blacktriangledown (k \blacktriangle 5) =5. Considering the final equality 20 ( k 5 ) = 5 -20 \blacktriangledown (k \blacktriangle 5) =5 , the value of the left-hand side must be either 20 -20 or ( k 5 ) (k \blacktriangle 5) by the definition of \blacktriangledown . Since the value on the right-hand side is 5 5 , which is not 20 -20 , it must be the case that ( k 5 ) = 5 (k \blacktriangle 5) =5 . So, our goal now becomes finding the number of integers k k that satisfy k 5 = 5. ( 1 ) k \blacktriangle 5 =5. \qquad (1)

If k > 5 \vert k \rvert>5 , then k 5 = k k \blacktriangle 5 =k , which contradicts ( 1 ) (1) .

If k = 5 k=-5 , then k 5 = 5 5 = 5 k \blacktriangle 5= -5 \blacktriangle 5 =-5 , which contradicts ( 1 ) (1) .

If k = 5 k=5 , then k 5 = 5 5 = 5 k \blacktriangle 5= 5 \blacktriangle 5 =5 , which satisfies ( 1 ) (1) .

If k < 5 \lvert k \rvert<5 , then k 5 = 5 k \blacktriangle 5 =5 , which satisfies ( 1 ) (1) .

Thus, the integers k k that satisfy ( 1 ) (1) lie in the interval [ 4 , 5 ] [-4, 5] . Therefore, the number of integers in this interval is 5 ( 4 ) + 1 = 10. 5-(-4)+1=10.

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