bhumika.k.shah wrote:how have u taken = 98*192*141*46 mod 100
Okay basically i dint get it from d start
Now that is a tough one (told u, the other one is easy)
mod is a function which gives us the remainder a mod b gives us the remainder when a is divided by b
45 mod 10 =5
I hope you understood why we are interested in mod 100 (yeah it gives the last two digits) ----------(1)
Now there is a theorem which says a*b mod c = a mod c * b mod c -----------(2)
I will give you an example 19*5 mod 4 = 19 mod 4* 5 mod 4 = 3*1 =3 (19*5 = 95 which indeed leaves a remainder of 3 when divided by 4)
Now these are the ground rules - if do not understand all of the above - Please do not try the below one.
(201*202*203*204*246*247*248*249)=201*202*203*204*246*247*248*249 mod 100 (yup we are interested in last two digits)
= 1*2*3*4*46*47*48*49 mod 100 (refer (2))
= 98*192*141*46 mod 100 (combined 49 and 2, 48 and 4, 47 and 3 and left 46 alone)
= -2*-8*41*46 mod 100 (refer (2))
[Now there is one rule I dint mention if you encounter -ve values in mod operations; to get the remainder you just have to add the denominator till the number is positive
a remainder of -2 in this case suggests a remainder of -2+100 = 98] ----------(3)
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=-224 mod 100 (refer (3))
= 76
Now it takes only a bit of figuring out after this (and I honestly do not think GMAT will ask anything like this)
Always borrow money from a pessimist, he doesn't expect to be paid back.