bhumika.k.shah wrote:Find the last 2 digits


(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16
5.66

OA C

(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

so last two digits are 76

why have u taken mod 100??

ajith wrote:

bhumika.k.shah wrote:Find the last 2 digits


(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16
5.66

OA C

(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

so last two digits are 76

bhumika.k.shah wrote:why have u taken mod 100??

that is a relatively easy query to answer mod 100 always gives last two digits

Check it out

742 mod 100 = 42

63535 mod 100 = 35

how have u taken = 98*192*141*46 mod 100
Okay basically i dint get it from d start

bhumika.k.shah wrote:why have u taken mod 100??

ajith wrote:

bhumika.k.shah wrote:Find the last 2 digits


(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16
5.66

OA C

(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

so last two digits are 76

i got all the rest but how did you get -224 mod 100 into 76?? I got everything else but this last part? is it you add 300 to change it to a positive number? Also I tried using the 100 mod but on different numbers than what you have done and I didn't get 76. How do you know if you are modifying the correct numbers?

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)
.
.
.

=-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)

diegow77 wrote:i got all the rest but how did you get -224 mod 100 into 76?? I got everything else but this last part? is it you add 300 to change it to a positive number? Also I tried using the 100 mod but on different numbers than what you have done and I didn't get 76. How do you know if you are modifying the correct numbers?

That is right, you should add 300 to make it positive.

If you post the steps you did, may be I can be of help (in case of latter query). You can modify any numbers as long as the rules are not broken and for rules refer the above post

Phewwww! so the explanation did end...FINALLY

What are the chances of such qs appearing on THE GMAT ??? :'(

ajith wrote:

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)
.
.
.

=-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)

and I honestly do not think GMAT will ask anything like this

exactly

THANK GOD!!!

sanju09 wrote:

and I honestly do not think GMAT will ask anything like this

exactly