If x and y are the number of possibilities that A can assume such that the unit digit of A and A3 are same and the unit digit of A2 and A3 are same respectively, then the value of x – y is (where A is a single digit number)
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Answers
Answer:
Explanation:
Given, A is a single digit number.
if A = 1 then, A³ = 1³ = 1 , here you can see unit digit of A and A³ are same.
if we take A = 2 then, A³ = 2³ = 8 , unit digits are not same.
if we take A = 3 => A³ = 27 ( × )
if we take A = 4 => A³ = 64 ( ✓ )
if we take A = 5 => A³ = 125 (✓)
if we take A = 6 => A³ = 216 (✓)
if we take A = 7 => A³ = 343 ( × )
if we take A = 8 => A³ = 512 ( × )
if we take A = 9 => A³ = 729( ✓)
hence, there are five possible solution where unit digits of A and A³ are same.
so, x = 5
again, if we take A = 1 => A² = 1 and A³ = 1 (✓)
if we take A = 2 => A² = 4 and A³ = 8 ( × )
if we take A = 3 => A² = 9 and A³ = 27 ( × )
if we take A = 4 => A² = 16 and A³ = 64 ( × )
if we take A = 5 => A² = 25 and A³ = 125( ✓ )
if we take A = 6 => A² = 36 and A³ = 216 ( ✓)
if we take A = 7 => A² = 49 and A³ = 343 ( × )
if we take A = 8 => A² = 64 and A³ = 512 ( × )
if we take A = 9 => A² = 81 and A³ = 729 ( × )
here we can see , there are three possible solution where unit digits of A² and A³ are same. so, y = 3
hence, x - y = 5 - 3 = 2
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Answer:
Heya mate here is ur answer.....
Given :
Given :A = A3
Given :A = A3 A3 - A = 0
Given :A = A3 A3 - A = 0A (A2 - 1) = 0
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0 A2(A - 1) = 0
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0 A2(A - 1) = 0A2= 0 or (A - 1) = 0
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0 A2(A - 1) = 0A2= 0 or (A - 1) = 0Therefore A = 0 or A = 1
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0 A2(A - 1) = 0A2= 0 or (A - 1) = 0Therefore A = 0 or A = 1
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0 A2(A - 1) = 0A2= 0 or (A - 1) = 0Therefore A = 0 or A = 1 y = A = {1, 0}
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0 A2(A - 1) = 0A2= 0 or (A - 1) = 0Therefore A = 0 or A = 1 y = A = {1, 0}
Given :A = A3 A3 - A = 0A (A2 - 1) = 0A = 0 or A - 1 = o or A + 1 + 0Therefore x = A = {-1, 1, 0}Next,A2 = A3A3 - A2= 0 A2(A - 1) = 0A2= 0 or (A - 1) = 0Therefore A = 0 or A = 1 y = A = {1, 0} x - y = 3 - 2 = 1
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