Elitmus
Exam
Category
What are last two digits of 7^2008
A.
21
B.
61
C.
01
D.
41
Read Solution (Total 9)
-
- C.01
7^1 last two digits will be 07
7^2 last two digits will be 49
7^3 last two digits will be 43
7^4 last two digits will be 01
this pattern will be reeating after every 4 terms i.e
7^5 last two terms will be 07 again & so on
hence for 7^2008
since the power is completely cancelled out from 4 it will be similar to that of
7^4 i.e. 01 - 10 years agoHelpfull: Yes(17) No(0)
- C. 01
7^2008 = (7^4)^502
= 2401^502
=...01^502
= ....01 - 10 years agoHelpfull: Yes(7) No(0)
- 01
2008,last two digit is a multiple of 4.so whatever last two digits of 7*7*7*7=2401,it will become the last two digits of the question.so C is final answer. - 10 years agoHelpfull: Yes(2) No(0)
- ((7)^502)^4)....
means
7^1=7
7^2=49
7^3=343
7^4=2401.....
07,49,43,01.....are repeated a 7 power
hence correct answer is 01....(c) - 10 years agoHelpfull: Yes(2) No(0)
- B.21
(7^3)^669*7=3*7=21 - 10 years agoHelpfull: Yes(0) No(7)
- @Pooja How can u explain ??
- 10 years agoHelpfull: Yes(0) No(1)
- [0*8][7^8]
01
- 10 years agoHelpfull: Yes(0) No(2)
- i think its 01 ...can anyone please explain the answer ??
- 10 years agoHelpfull: Yes(0) No(1)
- C.01 divide the no. by 100 and find the remainder
- 10 years agoHelpfull: Yes(0) No(1)
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