Latest SSC Aptitude Question SOLUTION: Find last two digits of the following expression (201*202*203*204*246*247*248*249)^2

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Find last two digits of the following expression (201*202*203*204*246*247*248*249)^2

Asked In SSC MAN (10 years ago)

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SURAJ KUMAR  (8 years ago)

(01*02*03*04*46*47*48*49)^2
(01*04*09*16*16*09*04*01) on sq.
(01*04*09*16)^2
(576)^2
Last 2 digit is 76

Is this solution Helpfull? Yes (3) | No   

Sourav Dutta  (10 years ago)

To find th last two digit the number must be divided by 100 and the remainder will be the ans .
now (201*202*203*204*....248*249)divided by 100
remainder=(1*2*3*....48*49)=49!
from online calculator u can easily calculate the last two digit of 491 i.e 24

now the no is in whole square form so it will be (24)^2=576
so the last two digit is 76

Is this solution Helpfull? Yes (2) | No (8)  

divya singh  (10 years ago)

Answer will be 00

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

milan jena  (8 years ago)

(201*202*203*204*246*247*248*249)=201*202*203*204*246*247*248*249 mod 100
= 1*2*3*4*46*47*48*49 mod 100
= 98*192*141*46 mod 100
= -2*-8*41*46 mod 100
=16*41*46 mod 100
=92*8*41 mod 100
= 92*328 mod 100
= -8*28 mod 100
=-224 mod 100
=76 mod 100

Now if (201*202*203*204*246*247*248*249) is 76 mod 100
(201*202*203*204*246*247*248*249)^2 = 76^2 mod 100
=5776 mod 100
= 76

Is this solution Helpfull? Yes | No (1)  

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