Elitmus
Exam
Numerical Ability
Number System
2) P=(101^X-99^X) IS AN EXPRESSION where X>=10,and X is a natura number ,how many different
values can the last digit of place ?
A)1 B)10 C)4 D)2
Read Solution (Total 9)
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- Ans- 2
unit digit of 101^x = 1 (always)
and unit digit of 99^x= 1 or 9 ( all value of x )
so at last place after difference 2 values are possible . - 10 years agoHelpfull: Yes(61) No(0)
- Ans: 2
1^anything 1
9^anything gives 1 or 9
So and is. 2 - 10 years agoHelpfull: Yes(6) No(0)
- ans is 2.. d option
- 10 years agoHelpfull: Yes(5) No(0)
- two values 0and 8
- 10 years agoHelpfull: Yes(3) No(5)
- answer is 2 and last digits values are 0 and 2.
- 10 years agoHelpfull: Yes(3) No(0)
- 2
as 9*9=1 and1*9=9(as9 has 2 cycles i.e repeat after two cycle) and 1*1=1(no cycle) always
so there are two values of last digit is possible. - 10 years agoHelpfull: Yes(2) No(0)
- answer is A)1
- 10 years agoHelpfull: Yes(0) No(7)
- d option is correct
- 10 years agoHelpfull: Yes(0) No(0)
- 2
becoz 101^x=1 and 99^x=1or 9
- 8 years agoHelpfull: Yes(0) No(0)
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