Mod equation
For all real numbers, is defined as the absolute value of ; for example and . Given that and are integer, how many different solutions does the equation have?
This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try
refreshing the page, (b) enabling javascript if it is disabled on your browser and,
finally, (c)
loading the
non-javascript version of this page
. We're sorry about the hassle.
From ∣ x ∣ + 2 ∣ y ∣ = 1 0 0 , we note that the range of y is − 5 0 ≤ y ≤ 5 0 . The solutions to the equation 2 ∣ y ∣ + ∣ x ∣ = 1 0 0 are as follows:
⇒ 2 ∣ 0 ∣ + { ∣ + 1 0 0 ∣ ∣ − 1 0 0 ∣ , ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 2 ∣ + 1 ∣ + { ∣ + 9 8 ∣ ∣ − 9 8 ∣ 2 ∣ − 1 ∣ + { ∣ + 9 8 ∣ ∣ − 9 8 ∣ , ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 2 ∣ + 2 ∣ + { ∣ + 9 6 ∣ ∣ − 9 6 ∣ 2 ∣ − 2 ∣ + { ∣ + 9 6 ∣ ∣ − 9 6 ∣ , ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 2 ∣ + 3 ∣ + { ∣ + 9 4 ∣ ∣ − 9 4 ∣ 2 ∣ − 3 ∣ + { ∣ + 9 4 ∣ ∣ − 9 4 ∣ , . . . ⎩ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎧ 2 ∣ + 4 9 ∣ + { ∣ + 2 ∣ ∣ − 2 ∣ 2 ∣ − 4 9 ∣ + { ∣ + 2 ∣ ∣ − 2 ∣ , { 2 ∣ + 5 0 ∣ + ∣ 0 ∣ 2 ∣ − 5 0 ∣ + ∣ 0 ∣
Therefore the total number of solutions = 2 + 4 9 × 4 + 2 = 2 0 0