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what will be the last two digit of (5306)^214
Read Solution (Total 6)
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- last two digits of (06)2 is 36.
last two digits of (06)3 is 16.
last two digits of (06)4 is 96.
last two digits of (06)5 is 76.
last two digits of (06)6 is 56.
last two digits of (06)7 is 36.
last two digits of (06)8 is 16.
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after 6th power last two digits is repeated.
when 214 is divided by 6 the remainder is 4.
Ans is last two digits of (06)214 is 96 same as last two digits of (06)4 is 96.
- 11 years agoHelpfull: Yes(24) No(2)
- If the given number is expanded using binomial theorem, it will be (5300 +6)^ 214
The terms will be 53006 214 + 214 * 5300^213 * 6 + ..............+ 214 * 5300 * 6 ^213 + 6 ^ 214
Other than the last term, every other term ends with two zeroes at least.
So, the last two digits of the netire expression depends on the last two digits of 6 ^ 214
Lets see the pattern of 6 powers
6 power 1-- 06 : 6^2----36; 6^3----16; 6^4---- 96; 6^5----76; 6^6---56
6^7----36; 6^8----16 and so on
It is following a cycle of 5, starting from power two
So, 6 ^ 212 will be again 36
6^213 will be 16
6^214 will be 96.
So, answer is 96
- 11 years agoHelpfull: Yes(4) No(2)
- last two digits will be 96
on sixth power the pattern of last two digit is repeating.
so it is unidentical upto 5th power.
Now 214/5 gives remainder of 4.
on its 4th power the last two digits are 96.
So 96 will be the required answer
- 11 years agoHelpfull: Yes(4) No(0)
- 5306^214=6^53*6^2
=6^13*6^3
=6^3*6^4
=6^7
=6^3
=216
so the last two digits are 16 - 11 years agoHelpfull: Yes(1) No(3)
- Just focus on units digit of power given i.e '4'
Now , (6) ^ any digit = will always end with 6, hence unit's digit of answer is confirmed to be six.
Now focus on (06)^4 : 6^4 = 1296, 6 will be kept and 9 will be transferred to the ten's place.
Ten's place is obtained by (0)^4 + carry over from unit's digit( which is 9) = 0 + 9 = 9
Hence last two digit's of given expression will be 96 - 8 years agoHelpfull: Yes(1) No(0)
- ans will be 96
- 11 years agoHelpfull: Yes(0) No(1)
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