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
finding the last two digits II
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- ajith
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(201*202*203*204*246*247*248*249)=201*202*203*204*246*247*248*249 mod 100bhumika.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
= 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
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why have u taken mod 100??
ajith wrote:(201*202*203*204*246*247*248*249)=201*202*203*204*246*247*248*249 mod 100bhumika.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
= 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
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that is a relatively easy query to answer mod 100 always gives last two digitsbhumika.k.shah wrote:why have u taken mod 100??
Check it out
742 mod 100 = 42
63535 mod 100 = 35
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how have u taken = 98*192*141*46 mod 100
Okay basically i dint get it from d start
Okay basically i dint get it from d start
bhumika.k.shah wrote:why have u taken mod 100??
ajith wrote:(201*202*203*204*246*247*248*249)=201*202*203*204*246*247*248*249 mod 100bhumika.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
= 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?
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Now that is a tough one (told u, the other one is easy)bhumika.k.shah wrote:how have u taken = 98*192*141*46 mod 100
Okay basically i dint get it from d start
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)
Last edited by ajith on Tue Feb 02, 2010 1:02 am, edited 1 time in total.
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That is right, you should add 300 to make it positive.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?
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
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Phewwww! so the explanation did end...FINALLY
What are the chances of such qs appearing on THE GMAT ??? :'(
What are the chances of such qs appearing on THE GMAT ??? :'(
ajith wrote:Now that is a tough one (told u, the other one is easy)bhumika.k.shah wrote:how have u taken = 98*192*141*46 mod 100
Okay basically i dint get it from d start
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)
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exactlyand I honestly do not think GMAT will ask anything like this
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
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Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
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