Elitmus
Exam
Numerical Ability
Algebra
If P=(102 power x) - (98 power x) and x is a natural number greater than zero. how many values exist at units place of P?
(a) 4 (b)5 (c)3 (d) 8
Read Solution (Total 6)
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- (102^x - 98^x)
to calculate first digit we only need firs digit of no.
to check power
if x=1
then |2^1-8^1|=6
if x=2
then |2^2-8^2|=|4-64|=4-4=0(only take first digit)
if x=3
then |2^3-8^3|= |8-2 |= 6 (only take first digit)
if x=4
then 2^4-8^4= 6-6=0
after x=4 it is repeat
x = 3 ans - 10 years agoHelpfull: Yes(18) No(0)
- (c) 3
calculate by putting x=1,2,3,4 you can observe that only 0,4,6 are repeated - 10 years agoHelpfull: Yes(5) No(1)
- 102^x - 98^x
to calculate unit digit just consider the unit digit 102^x and 98^x individually.
so 2^1 - 8^1= 2 - 8 =12 - 8= 4
2^2 - 8^2= 4 - 4 = 0
2^3 - 8^3= 8 - 2= 6
2^4 - 8^4= 6 - 6= 0
2^5 - 8^5= 2 - 8= 4
2^6 - 8^6= 4 - 4= 0
2^7 - 8^7= 8 - 2= 6
thus, cycle of 4 having unit digits, 4,0 6, and 0
we have to find different values that exists at unit place. As 0 comes two times in the cycle. so there ill be 3 different values say, 4, 0 and 6
- 9 years agoHelpfull: Yes(3) No(1)
- thank u kumar
- 10 years agoHelpfull: Yes(0) No(0)
- ans is (c)3
- 10 years agoHelpfull: Yes(0) No(0)
- correct ans is 3
- 9 years agoHelpfull: Yes(0) No(0)
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