Latest TCS Aptitude Question SOLUTION: Last two digits of 87^474?

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(#M40152873) TCS Numerical Ability question Keep an EYE

Last two digits of 87^474?

Asked In TCS Nakshatra Mukhopadhyay (8 years ago)

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Soumita Ranasingha  (8 years ago)

(87^4)^118*(87)^2
=(61)^118*(7569)
=481*7569
89
last two digits are 89

Is this solution Helpfull? Yes (9) | No (2)  

sandipan  (8 years ago)

(87^4)^118 * 87^2
=(87^2 * 87^2)^118 * 87^2
now 87^2=7569 we will take only last two digit number i.e. 69
=(69 * 69)^118 *87^2
69 *69 gives unit digit only=01 and 87^2 gives unit digit 9
01 * 9= 09
so, last two digit= 09

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SUKANYA GHOSH  (8 years ago)

From the cyclicity of 7,It repeats after 4 intervals. now 474/4 makes remainder 5.
so last digit in thw extreme rigit is 5.
again from the cyclicity of 87 ,it repeats after 2 intervals. but 474 /2 makes the remainder 0.
so we takes as 87^2---- last digit 9.
ultimately the answerwill be 95

Is this solution Helpfull? Yes (1) | No (7)  

priyadharshini  (8 years ago)

http://www.slideshare.net/nickyswetha/number-systems-theory-for-cat-2009-quant please have a luk at this link for clear solution

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