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8. Find the last two digits of (1941^3881)*(1961^4181)
Read Solution (Total 9)
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- I think last two digits will be 01.
- 12 years agoHelpfull: Yes(19) No(1)
- taking last two digits of first term i.e. (41)^3881 now , taking (41)^1 = 41,
(41)^2=1681, (41)^3=68921, (41)^4=2825761, (41)^5=115856201.......so on... Do not write whole terms as i written take only last two digits we get a sequence of last digits-> 41,81,21,61,01, again 41,...... now by observation 81st(^3881) term is: 41, Similarly for second term(1961^4181) the sequence of last digits is-> 61,21,81,41,01, again 61,2...... now again by observation 61 is the 81st (^4181)now multiply 41 and 61 we get last two digits as 01. - 12 years agoHelpfull: Yes(16) No(1)
- ( 41 ^81 ) * ( 61^81 )
= 41 * 61
= 01 last two digit - 12 years agoHelpfull: Yes(8) No(0)
- According to my calculation i think the answer must be 01.
- 12 years agoHelpfull: Yes(1) No(1)
- last digit is definitely 1.
so answer can not have options as 12,22,42,44.
I think Mp is right.
01 will be last two digits. - 12 years agoHelpfull: Yes(1) No(0)
- let us take the last 2 digits in the given numbers 1941 i.e, 41 & 1961 i.e,61 so 41^3881=41 & 61^4181=61..now will multiply 41 & 61 so final solution for this is "01"
- 12 years agoHelpfull: Yes(1) No(0)
- but options are 12,22,42,44
- 12 years agoHelpfull: Yes(0) No(2)
- (1941^3881) gives 4*1=4 ie 41 at last digits
(1961^4181) gives 6*1=6 ie 61 as last digit
so total product will give 41*61= ..01 as last digits - 12 years agoHelpfull: Yes(0) No(0)
- it is coming 01
- 12 years agoHelpfull: Yes(0) No(0)
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