Latest TCS Aptitude Question SOLUTION: 5036^214 last two digit ans is 96. how to solve it

answer is 16,,,cyclicity of 5
36,96,56,16,76

last digit will be 16
36^214
6^428
(2*3)^428
2^428 * 3^428
((2)^10)^48 * 2^8 * 61
(24)^48 *56*61
=>76*56=56
56*61
=>16
so ans is 16

96 will be ans when ques will be 5306^214

last two digit will be 26 (not 96).
To find digit in the 10th place we have to multiply,10th digit in the base*unit digit in the power

firstly,divide the power 214 by 4 i.e, 214/4 , then the remainder will be 2... now put (5036)^2 = 96 (last two digits)

last digit = 6 last power =4
(6)^4=1296
last 2 digit =96

6^4=6..then 3*4=12
ans is 26

5036^214=2^428*1259^214
->(2^10)^42*2^8*(...81)^107 (by squaring 1259 we get ...81)
we know,2^10^even power is always 76
and 81^107 will be 1^unit digit of power=1 and 8*unit digit of power i.e 7=56 or 6
->76*56*61
=16(answer)
so last two digit is 16

AMIT how you got this step (1024)^85×64×61
24×64×61=---96

5036^214
Unit Digit of 6^214 is 6
As 6^anything last digit is always 6

Tens Digits will be 10th unit of number^unit digit of power
Tens Digit = 3 ^ 4 = 81 therefore 1

There fore answer is 16

(2^4×1259)^214
2^856×((1259)^2)^(214÷2)
(2^10)^85×2^6×(-----81)^107
(1024)^85×64×61
24×64×61=---96

ans is 16..
it is solved based on using power cycle method.every 5 cycle it repeats the series..after that the mod has been taken

take 36 which repeats 36 at 36^6 so consider upto 36^5 so dividing 214/5 we get remainder as 4 so its 16 is the answer

last two digit will be 96
because units digit power 4 and tens digit * 1 which gives 396... and last two digits is 96.. (4&1 are units &tens digit in power)

ANS IS 16
5036^214=(2)^428*1259^214
(2)^420*2^8*(1259^2)^107
76*56*61=16 LAST DIGIT

prakhar jain doing wrong calculaton ... he factorized 2^428= (2^10)^48 * 2^8...

Last two digits of numbers ending in 1

Let's start with an example.

Find the last two digits of 41^{2789}

Multiply the tens digit of the number (4 here) with the last digit of the exponent (9 here) to get the tens digit. The unit digit will always equal to one.

In no time at all you can calculate the answer to be 61 (4 × 9 = 36). Therefore, 6 will be the tens digit and one will be the unit digit


Here are some more examples:

What are the last two digits of 3178631786?

Last two digits will be 81 (3 × 6 gives 8 as tens digit and 1 as unit digit)

Find the last two digits of 71567477156747

Last two digits will be 91 (7 × 7 gives 9 as tens digit and 1 as unit digit)

Now try to get the answer to this question within 10 seconds:

Find the last two digits of 5145651456 × 6156761567

The last two digits of 5145651456 will be 01 and the last two digits of 6156761567 will be 21. Therefore, the last two digits of 5145651456 × 6156761567 will be the last two digits of 01 ×21 = 21

Last two digits of numbers ending in 3, 7 or 9

Find the last two digits of 1926619266.

1926619266 = (192)133(192)133. Now, 192192 ends in 61 (192192 = 361) therefore, we need to find the last two digits of (61)133(61)133.

Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. The last two digits are 81 (6 × 3 = 18, so the tens digit will be 8 and last digit will be 1)

Find the last two digits of 3328833288.

3328833288 = (334)72(334)72. Now 334334 ends in 21 (334334 = 332332 × 332332 = 1089 × 1089 = xxxxx21) therefore, we need to find the last two digits of 21722172. By the previous method, the last two digits of 21722172 = 41 (tens digit = 2 × 2 = 4, unit digit = 1)

So here's the rule for finding the last two digits of numbers ending in 3, 7 and 9:

Convert the number till the number gives 1 as the last digit and then find the last two digits according to the previous method


Now try the method with a number ending in 7:

Find the last two digits of 8747487474.

8747487474 = 8747287472 ×872872 = (874)118(874)118 × 872872 = (69×69)118(69×69)118 × 69 (The last two digits of 872872 are 69) =6111861118 × 69 = 81 × 69 = 89




Last two digits of numbers ending in 2, 4, 6 or 8

There is only one even two-digit number which always ends in itself (last two digits) i.e. 76 raised to any power gives the last two digits as 76. Therefore, our purpose is to get 76 as last two digits for even numbers. We know that 242242 ends in 76 and 210210 ends in 24. Also, 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore, 24342434 will end in 76 and 24532453 will end in 24.

Let's apply this funda:

Find the last two digits of 25432543.

25432543 = (210)54(210)54 × 2323 = (24)54(24)54 (24 raised to an even power) × 2323 = 76 × 8 = 08

(NOTE: Here if you need to multiply 76 with 2n2n, then you can straightaway write the last two digits of2n2n because when 76 is multiplied with 2n2n the last two digits remain the same as the last two digits of 2n2n. Therefore, the last two digits of 76 × 2727 will be the last two digits of 2727 = 28)

Same method we can use for any number which is of the form 2^{n}. Here is an example:

Find the last two digits of 6423664236.

6423664236 = (26)236(26)236 =2141621416 = (210)141(210)141 × 2626 = 2414124141 (24 raised to odd power) × 64 = 24 × 64 = 36


Here are some more examples:

Find the last two digits of 6258662586.

6258662586 = (2×31)586(2×31)586 = 25862586×35863586 = (210)58(210)58×2626 × 3158631586 = 76 × 64 × 81 = 84

Find the last two digits of 5438054380.

5438054380 = (2×33)380(2×33)380 = 23802380×3114031140 = (210)38(210)38 × (34)285(34)285 = 76 × 8128581285= 76 × 01 = 76.

Find the last two digits of 5628356283.

5628356283 = (23×7)283(23×7)283 = 28492849×72837283 = (210)84(210)84 × 2929 × (74)70(74)70 × 7373 = 76 × 12 × (01)70(01)70 × 43 = 16

Find the last two digits of 7837978379.

7837978379 = (2×39)379(2×39)379 = 23792379 × 3937939379 = (210)37(210)37 × 2929 × (392)189(392)189 × 39 = 24 × 12 × 81 × 39 = 92