What are the last two digits of 72008?
  • a)
    01                                         
  • b)
    21
  • c)
    61                                       
  • d)
    71  
Correct answer is option 'A'. Can you explain this answer?

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Answers

Yashika Mahajan
Mar 27, 2019
See 7^1=7           last two digit =07
        7^2=49         last two digit=49
        7^3=243       last two digit=43
        7^4=1701     lasr two  digit=01
        7^5=8507     last two digit=07
        7^6=59549   last two digit=49
        7^7=              last two digit=43
        7^8=              last two digit=01
We notice that
last two digit in 7^n=07 when n=4k+3
last two digit in 7^n=49 when n=4k+2
last two digit in 7^n=43 when n=4k+1
last two digit in 7^n=01 when n=4k
2008=4*502=>7^2008=7^n=>n=504
Hence last two digit in 7^2008 is 01.

See 7^1=7 last two digit =07 7^2=49 last two digit=49 7^3=243 last two digit=43 7^4=1701 lasr two digit=01 7^5=8507 last two digit=07 7^6=59549 last two digit=49 7^7= last two digit=43 7^8= last two digit=01We notice thatlast two digit in 7^n=07 when n=4k+3last two digit in 7^n=49 when n=4k+2last two digit in 7^n=43 when n=4k+1last two digit in 7^n=01 when n=4k2008=4*502=>7^2008=7^n=>n=504Hence last two digit in 7^2008 is 01.
See 7^1=7 last two digit =07 7^2=49 last two digit=49 7^3=243 last two digit=43 7^4=1701 lasr two digit=01 7^5=8507 last two digit=07 7^6=59549 last two digit=49 7^7= last two digit=43 7^8= last two digit=01We notice thatlast two digit in 7^n=07 when n=4k+3last two digit in 7^n=49 when n=4k+2last two digit in 7^n=43 when n=4k+1last two digit in 7^n=01 when n=4k2008=4*502=>7^2008=7^n=>n=504Hence last two digit in 7^2008 is 01.