Units place in math expression
Many exams try to make you feel that you cannot solve the problem by making question as big as possible like units place of (186937)^178909 but solving this problem is as simple as solving 9%4
Before solving these kind of problems, understand the great concept called Cyclicity of numbers which states that any digit when multiplied by itself repeatedly, generates a cycle of digits.
points to remember:
Number powers Series/cycle(units place of powers column) [cyclicity of number]
2 2,4,8,16,32 2,4,8,6 [4]
3 3,9,27,81,243 3,9,7,1 [4]
4 4,16,64 4,6 [2]
7 7,49,343,2401 7,9,3,1 [4]
8 8,64,512,4096 8,4,2,6 [4]
9 9,81,729 9,1 [2]
Numbers 0,1,5,6 generates same numbers when multiplied y themselves, so the cyclicity of numbers for 0,1,5,6 is [1] .
--------------------------------------------------------------------------------------------------------------------------
Algorithm for finding units place
Example 1:What is the units place of (186937)^178909
solution:
As units place alone generates units place,this is equivalent to units place of (7)^178909
Cyclicity of number 4 is 4,
R=178909%4=1
1st number of cycle is 7.
Example 2: what is the units place of ((188888888888888)*18888888888888888888885)^19979
solution:
5*8=40
units place=0
The given question can be reframed as (0)^19979
Answer is 0.
Example 3:what is the units place of 129*189*123*1773*87127*999
Solution:
As units place alone is responsible to generate units place,
answer = 9*9*3*3*7*9
=81 * 9 * 63 [again take units place]
= 1* 9* 3
=27
units place = 3
Before solving these kind of problems, understand the great concept called Cyclicity of numbers which states that any digit when multiplied by itself repeatedly, generates a cycle of digits.
points to remember:
- Only units place will be responsible for generating units place
- Every number after applying successive multiplications,repeats the cycle of digits.
So try to figure out a common series in powers of digits
for example units(3)=3,units(3*3)=9,units(9*3)=7,units(7*3)=1,units(1*3)=3(we get a cycle of multiplication)
Number powers Series/cycle(units place of powers column) [cyclicity of number]
2 2,4,8,16,32 2,4,8,6 [4]
3 3,9,27,81,243 3,9,7,1 [4]
4 4,16,64 4,6 [2]
7 7,49,343,2401 7,9,3,1 [4]
8 8,64,512,4096 8,4,2,6 [4]
9 9,81,729 9,1 [2]
Numbers 0,1,5,6 generates same numbers when multiplied y themselves, so the cyclicity of numbers for 0,1,5,6 is [1] .
--------------------------------------------------------------------------------------------------------------------------
Algorithm for finding units place
- Reframe the entire number by discarding all digits except units place.
- Find the cycle count of the units place.
- find the remainder when power is divided by cycle count(Exponent%cyclecount) =R
- R th number in the cycle is the answer.
Example 1:What is the units place of (186937)^178909
solution:
As units place alone generates units place,this is equivalent to units place of (7)^178909
Cyclicity of number 4 is 4,
R=178909%4=1
1st number of cycle is 7.
Example 2: what is the units place of ((188888888888888)*18888888888888888888885)^19979
solution:
5*8=40
units place=0
The given question can be reframed as (0)^19979
Answer is 0.
Example 3:what is the units place of 129*189*123*1773*87127*999
Solution:
As units place alone is responsible to generate units place,
answer = 9*9*3*3*7*9
=81 * 9 * 63 [again take units place]
= 1* 9* 3
=27
units place = 3
Comments
Post a Comment