Latest TCS Aptitude Question SOLUTION: if x^y denotes x raised to the power y, find last two digits of (1507^3381) + (1457^3757)

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if x^y denotes x raised to the power y, find last two digits of (1507^3381) + (1457^3757)

Asked In TCS Deepak Kumar (8 years ago)

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ashish  (8 years ago)

last digit=7
7^1=07
7^2=49
7^3=343
7^4=2401
7^5=16807
7^6=117649
repeating cycle of last two digits of 7=4
1507^3381=07 ; 1457^3757=07
last two digit=07+07=14

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sandip kumar subuddhi  (8 years ago)

lat digit of power of 7 are as follows:
7,9,3,1,7,9,3,1,7,9,3,1,.........(because,7^1=7,7^2=49,....)
((1507^4)^845)*1507 + ((1457^4)^939)*1457 = 64(last two digit)

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Suryaharsh Seethamsetti  (8 years ago)

((1507^4)^845)*1507 + ((1457^4)^939)*1457 =(...01^845)*1507+(...01^939)*1457=...07+...57=64

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Mohd Ashif Ali  (8 years ago)

= 1507^3381 + 1457^3757 can also be written as

= (1507)^(4*845)X1507 + (1457)^(4*939)X1457

now if we calculate 1507^4 , and 1457^4 the last two digit for both of them is (01), and also last two digit for (01)^n is always (01) therefore,

= (01)^(845)X1507 + (01)^(939)X1457
= (01)X1507 + (01)X1457
= 1507 + 1457
= 2964

Hence last two digit will be "64"..

Is this solution Helpfull? Yes (7) | No   

Bisu  (8 years ago)

For unit place 7^1=7,,7^3757=last digit 7.so 7+7=14
FOr 10th digit,0*1=0,,5*5=25.so 5+0=5
so 64

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sonali  (8 years ago)

(7^3381+57^3757)/100
3381 and 3757 can be written in the form of 4n+1
so we have 7+57=64
last two digits are 64.

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akansha  (8 years ago)

4. as we know that 7 raise to the power any number will always end at 7,9,3,1. and the number 3381 can be written as 4n+1 form. hence it will end at 7. similarly 3757 can also be written as 4n+1 and also ends at 7. hence adding we will get 4 as unit digit,.

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vatsal agrawal  (8 years ago)

sum is 49+93=142

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Avani Tiwari  (8 years ago)

7^1=07
7^2=49
7^3=343
7^4=2401
7^5=16807
7^6=117649
cycle repeats => 7^2=49(last 2 digits of 3381 since 1507=1505/5 so 2 powers remain)

Similarly 81^1=81(last 2 digits of each)
81^2=61
81^3=41
81^4=21
81^5=01
81^6=81
this pattern repeats
1457=1455/5 and hence 2 powers remain therefore last 2 digits= 61

so,40+61=104
so the last two digits=04

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ethan hunt  (8 years ago)

7 powers are 7,9,3,1 3381/4 and 3757/4
7+7=14

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lavanya  (5 years ago)

1507^(3381)+1457^3757
consider last digit:-
= 7^(4*845+1)+7^(4*939+1) [cyclicity of 7 is 4)
:7^1+7^1=14
therefore,last two digits=14

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