Latest TCS Aptitude Question SOLUTION: find last two digits of (1941^3843)+(1961^4181)

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(#M40012193) TCS question tcs Keep an EYE

find last two digits of (1941^3843)+(1961^4181)

Asked In TCS manish jaiswal (11 years ago)

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PARAG KAMRA  (11 years ago)

1941^3843
first take last two digit 41

first last digit is 1 ..............(1)
then take the 4 (1941) and multiply with 3 (3843)
4*3=12,take last digit from this is 2 ...........(2)
now (1)(2)
so last two digit is 21

now same for 1961^4181
first take last two digit 61

first last digit is 1 ..............(1)
then take the 6 (1961) and multiply with 1 (4181)
6*1=6,take last digit from this is 2 ...........(2)
now (1)(2)
so last two digit is 61
last two digits of (1941^3843)+(1961^4181) is 61+21=82 (ans)

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ADRIJA CHATTERJEE  (11 years ago)

82..
last 2 digit of (1941^3843)is last 2 digit of (((1941^3840)*41)*41)*41)i.e, 21. in this way last 2 digit of (1961^4181) is 61. so 61+21=82.

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KAUSTAV CHAUDHURI  (10 years ago)

When an integer with an ODD UNITS DIGIT OTHER THAN 5 is raised to A POWER THAT IS A MULTIPLE OF 20, the last two digits of the result are 01.

Thus, the last two digits of 1941³⁸⁴⁰ and 1961⁴¹⁸⁰ are 01.
To determine the last two digits of 1941³⁸⁴³ and 1961⁴¹⁸¹, keep multiplying the last two digits by 41.

Last two digits of 1941³⁸⁴³:
1941³⁸⁴⁰ --> 01.
1941³⁸⁴¹ --> 01*41 = 41.
1941³⁸⁴² --> 41*41 = 1681.
1941³⁸⁴³ --> 81*41 = 3321.
Thus, the last two digits of 1941³⁸⁴³ are 21.

Last two digits of 1961⁴¹⁸¹:
1961⁴¹⁸⁰ --> 01.
1961⁴¹⁸¹ --> 01*61 = 61.
Thus, the last two digits of 1961⁴¹⁸¹ are 61.

Thus:
(last two digits of 1941³⁸⁴³) + (last two digits of 1961⁴¹⁸¹) = 21+61 = 82.

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shivani tyagi  (11 years ago)

82 is d correct ans.

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Vikas  (11 years ago)

plz elaborate....

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manish jaiswal  (11 years ago)

anirudh explain

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anirudh anand  (11 years ago)

ans is 42?

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