Latest TCS Aptitude Question SOLUTION: Find the last 3 digit of 2988^687

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(#M40024587) TCS Logical Reasoning question Keep an EYE

Find the last 3 digit of 2988^687

Asked In TCS sarika nayak (10 years ago)

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kalyani dangeti  (10 years ago)

(2988%1000)^687
988^687%1000
(988^684)*(988^3)
(988%1000)^3
(976144%1000)*988
(144*988)%1000
142272%1000
272
ans=272

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RAKESH  (10 years ago)

dividing a no. by 1000,remainder is the last three digit of number
2988^687/1000
=(3*1000-12)^687/1000
=(-12)^687/1000
={(-12)^3}^229/1000
=(-1728)^229/1000
={(-1728)^2}^114*(-1728)/1000
=(2985984)^114*(-1728)/1000
=(984)^114*(-1728)/1000
=(-16)^114*(-1728)/1000
={(-16)^3}^38*(-1728)/1000
=(-4096)^38*(-1728)/1000
=(-96)^38*(-1728)/1000
=(9216)^19*(-728)/1000
=(216)^19*272/1000
=(46656)^9*216*272/1000
=(656^9)*58752/1000
=(656^3)^3*752/1000 (consider last 3 or 4 digit of product)
=(416)^3*752/1000
=296*752/1000
=592/1000
=rem(592)

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satheesh k  (10 years ago)

988^2=976144%1000=144
988^4=144^2%1000=736
988^8=736^2%1000=696
988^16=696^2%1000=416
988^32=416^2%1000=056
988^64=056^2%1000=136
988^128=136^2%1000=496
988^256=496^2%1000=016
988^512=016^2%1000=256

988^687=988^512.988^128.988^32.988^8.988^4.988^2.988

988^687=256.496.56.696.736.144.988

988^687=592

i think 592 is answer it's correct or not ?

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sandy  (10 years ago)

(..8)^(2)=...4 ie. a num ending with 8 raised to the power of 2 gives a num ending with 4 (since 8^2 is 64).similarly 8^3=512..ends with 2.
8^4=4096 ends with 6..
8^5=32768..again this series repeats...the cycle is 4...
hence 687%4=3
8 6 2..the ans is 2

Is this solution Helpfull? Yes (3) | No (11)  

Guru Pranav  (10 years ago)

142272%1000
272
ans=272

Is this solution Helpfull? Yes (2) | No (1)  

shivam shivhare  (10 years ago)

koi k pass simple explanation nahi hai kyaa...?

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sandy  (10 years ago)

this is for last digit alone

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Divya  (10 years ago)

272 is the answer

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ks  (10 years ago)

592 is the correct answer

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DHIRENDRA   (10 years ago)

last to digit is 88
88^687=== 687%4==3 is remainder
88^3===472 answer

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ayesha zulkha  (10 years ago)

687/4=remainder 3,
for remainder 3 base unit=(base unit)^3
i.e. 8^3=512

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dipanjan biswas  (9 years ago)

the formula for this problem is
(x+1)^n = 1+nx+n(n-1)*x^2 therefore the ans will be

(2987+1)^687= 1 + 069 + 641*169
= 070 + 329
= 399 answer

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