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# Find the last 2 digits of 65*29*37*63*71*87*62

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Manager
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Kudos [?]: 98 [1] , given: 46

Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]  04 Nov 2009, 00:14
1
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18
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Question Stats:

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Find the last 2 digits
Q1.
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90

Q2.
(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16
Manager
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Kudos [?]: 800 [14] , given: 18

Re: Find the last two digits [#permalink]  15 Nov 2009, 13:28
14
KUDOS
1
This post was
BOOKMARKED
Hussain15 wrote:
Find the last 2 digits
Q1.
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90

Hi, there is a very quick way to solve these questions:

Since we want to find the last two digits, we have to find the remainder when divided by 100. (in decimal format)

Therefore,

$$R of (65*29*37*63*71*87*62)/100 = R of (13*29*37*63*71*87*31)/10$$
Note: Divided 65 by 5 and 62 by 2.

Now, $$R of (13*29*37*63*71*87*31)/10 = R of [3*(-1)*(-3)*3*1*(-3)*1]/10 = R of -81/10 = -1$$

Since remainder is coming negative, we have to add it to 10.

Thus remainder is in fact 9. In decimal format, it is expressed as 0.9 or (9/10)

Thus the remainder when 65*29*37*63*71*87*62 is divided by 100 in decimal format is 0.9.

The last two digits will therefore be 0.9*100 = 90.

_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
Followers: 78

Kudos [?]: 800 [14] , given: 18

Re: Find the last two digits [#permalink]  15 Nov 2009, 13:37
14
KUDOS
2
This post was
BOOKMARKED
Quote:
Q2.
(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16

Similarly for this question,

$$R of (201*202*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/100$$

$$= R of (201*101*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)/50$$

Note: I have left denominator as 50 since it will be easier in calculations.

$$= R of [(1*1*3*4*(-4)*(-3)*(-2)*(-1)]*[(1*2*3*4*(-4)*(-3)*(-2)*(-1)]/50$$

$$= R of (12*24*24*24)/50 = R of (6*24*24*24)/25 = R of [6*(-1)*(-1)*(-1)]/25 = -6$$

Since remainder is coming negative, we add 25 to it.

Thus Remainder is 19. In decimal format, it is 19/25 or 0.76

Thus last two digits will be 0.76*100 = 76

[Note: Rather than calculating the decimal value first, it will be faster to combine the last two steps as follows: (19/25)*100 = 19*4 = 76. This is how I did it and it saved me valuable seconds!]

_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

Last edited by sriharimurthy on 15 Nov 2009, 14:07, edited 1 time in total.
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Re: Numbers 4 [#permalink]  04 Nov 2009, 01:34
7
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1
This post was
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65*29*37*63*71*87*62 = (60+5)*(30-1)*(40-3)*(60+3)*(70+1)*(90-3)*(60+2)

multiplying 5*1*3*3*1*3*2 = 90

last two digits are 90
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Kudos [?]: 98 [3] , given: 46

Re: Numbers 4 [#permalink]  06 Nov 2009, 04:19
3
KUDOS
Vyacheslav wrote:
Ans to 2nd question should be 56.

I am sure its not a GMAT question, even after using techniques it takes more than 2-3 min..
Here we go:
(201*202*203*204*246*247*248*249)^2

we can make pairs within the numbers inside the brackets such as
(200+1)(200+49) which gives us last 2 digits = 49
(200+2)(200+48) which gives us last 2 digits = 96
(200+3)(200+47) which gives us last 2 digits = 141
(200+4)(200+46) which gives us last 2 digits = 184

No again the last two dogits can be paired as
49*141 = (100-41)(100+41) or (100^2 - 41^2) which gives us last 2 digits = 19 and
96*184 = (140-44)(140+44) or (100^2 - 44^2) which gives us last 2 digits = 64

Now, 19*64 gives us last 2 digits of 16 and square of 16 will be 56. Phewwww!!

Pls tell me I am right, else i am not attempting the first one!!

49=100-51 but not 41. Or it doesn't matter?

Could you also explain how to calculate that (100^2 - 44^2) has 64 as last two digits?

concept to be used for such sums is REMAINDER THEOREM

to get last 2 digits divide by 100

(65*29*37*63*71*87*62)/100=

13*29*37*63*71*87*62)/20= ....dividing by 5 both numerator n deno

Remainder Thm---> -7*9*-3*3*11*7*2/20 ( ie 13/20 gives us remainder -7 or 13;29/20 gives us rem 9.......
= -63*-99*14/20
= 63*99*14/20
Remainder Thm--->3*-1*-6/20
= 18/20
that gives us remainder 18.....but 1st step we had divided by 5 therfore multiply by 5 now
ie remainder = 18*5=90
Manager
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Re: Find the last two digits [#permalink]  15 Nov 2009, 13:44
2
KUDOS
In case you are not able to follow my steps completely, view my post on tips and tricks to deal with remainders.

Take your time with them and make sure you understand all the points properly (especially points 5 and 6). Then try to follow the steps I've done.

Once you understand those concepts thoroughly you should be able to solve these questions really fast.

In fact, it took me less than 2 minutes to solve each of the above problems.

All the best!

Cheers.
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

Manager
Joined: 29 Oct 2009
Posts: 211
GMAT 1: 750 Q50 V42
Followers: 78

Kudos [?]: 800 [2] , given: 18

Re: Numbers 4 [#permalink]  15 Nov 2009, 14:14
2
KUDOS
Bunuel wrote:

Excellent! I was just posting the solutions for these two questions with similar remainder approach but no need for them now.

+1 for each.

Thanks Bunuel... Makes it all the more special coming from you!
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

Manager
Joined: 13 Dec 2009
Posts: 130
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Kudos [?]: 218 [2] , given: 10

Re: Numbers 4 [#permalink]  21 Apr 2010, 07:30
2
KUDOS
Hussain15 wrote:
Great explanation. Kudos!!

I just couldn't understand the $$-ve$$ remainders, I was aware of only positive values like $$Rof 29/10=9$$.

Can you elaborate that how do you get $$Rof 29/10=-1$$ ?

you are correct, there is no concept of $$-ve$$ remainder as such.
this is just a way how to get remainder specially if divisor is big and numbers are just short of the divisor. in above example it is easy to work with
the remainder concept i.e. the concept you have mentioned but have a look on the following problem:

$$\frac{73*75*74}{76}$$ if we need to find the remainder of this expression then normal remainder concept will not work, because for each number the remainder is same. in such cases we need to find some other way to work with and that's where $$-ve$$ remainder concept came into picture.

ROF of above expression will be $$\frac{-3*-1*-2}{76} = Rof \frac{-6}{76}$$
$$-ve$$ values we get just counting number required to make remainder zero. e.g. -3 came since we need 3 more to make it 76. similarly others.
since $$-ve$$ can not be remainder so the actual remainder of the expression will be $$-6+76 = 70$$

hope this will help.
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Re: Numbers 4 [#permalink]  04 Nov 2009, 03:09
1
KUDOS
Ans to 2nd question should be 56.

I am sure its not a GMAT question, even after using techniques it takes more than 2-3 min..
Here we go:
(201*202*203*204*246*247*248*249)^2

we can make pairs within the numbers inside the brackets such as
(200+1)(200+49) which gives us last 2 digits = 49
(200+2)(200+48) which gives us last 2 digits = 96
(200+3)(200+47) which gives us last 2 digits = 141
(200+4)(200+46) which gives us last 2 digits = 184

No again the last two dogits can be paired as
49*141 = (100-41)(100+41) or (100^2 - 41^2) which gives us last 2 digits = 19 and
96*184 = (140-44)(140+44) or (100^2 - 44^2) which gives us last 2 digits = 64

Now, 19*64 gives us last 2 digits of 16 and square of 16 will be 56. Phewwww!!

Pls tell me I am right, else i am not attempting the first one!!
Manager
Joined: 29 Oct 2009
Posts: 211
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Kudos [?]: 800 [1] , given: 18

Re: Numbers 4 [#permalink]  15 Nov 2009, 13:48
1
KUDOS
Hi guys,

I have solved these questions here (Two similar topics are merged-Moderator). It took me less than 2 minutes to solve each of these questions. Kindly have a look at my method and try to understand it. It will really help you solve these problems really fast even if they come on the GMAT.

Cheers.
_________________

Click below to check out some great tips and tricks to help you deal with problems on Remainders!
compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html#p651942

1) Translating the English to Math : word-problems-made-easy-87346.html

Manager
Joined: 13 Dec 2009
Posts: 130
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Kudos [?]: 218 [1] , given: 10

Re: Find the last two digits [#permalink]  20 Apr 2010, 20:40
1
KUDOS
Hussain15 wrote:
sriharimurthy wrote:

Now, $$R of (13*29*37*63*71*87*31)/10 = R of [3*(-1)*(-3)*3*1*(-3)*1]/10 = R of -81/10 = -1$$

I have read the post & read the tips & tricks on remainders too but could not understand how srihari is getting $$-1, -3 & -3$$ in above equation?

Can anyone help me out?

$$-1, -3 & -3$$ came after dividing $$29,37 & 87$$ with $$10$$, as -ve remainder. this is approach to solve such problem. when -ve value came 10 is added because remainder is always +ve.

if this approach is troubling you, you can still solve as follows by doing division normal way:

$$R of (13*29*37*63*71*87*31)/10 = R of [3*9*7*3*1*7*1]/10$$

or $$R of [3*9*7*3*1*7*1]/10 = R of [27*21*7]/10$$

or $$R of [7*1*7]/10 = R of [49]/10$$

=> remainder is 9

http://gmatclub.com/forum/tips-how-to-get-remainders-92928.html

hope this will help
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Re: Find the last two digits [#permalink]  21 Apr 2010, 07:42
1
KUDOS
Hussain15 wrote:
Also, can anyone explain the red bold portion highlighted above?? I have read the post of tips & tricks on remainders but couldn't understand this sentence.

let us take a simple example, if want to know the unit digit of 9*8 what i will do is just multiply and see the unit digit
one other way around to find the unit digit is just divide the expression with 10 and see the reminder, and that will nothing but the unit digit of the expression.

so remainder of the expression $$\frac{9*8}{10} = 2$$ and that is what the unit digit of the expression..

based on similar logic we can get last two digits of expression by dividing 100..
you can extend this login up to any number e.g. to get last 3 digits of the any expression, just divide it by 1000 and count the remainder.

hope this will help.

Last edited by einstein10 on 21 Apr 2010, 17:09, edited 1 time in total.
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Re: Numbers 4 [#permalink]  21 Apr 2010, 07:43
1
KUDOS
Hussain15 wrote:
Waooo!!

Great explanation!!! Got it now!!

One more for you!!!

hey thanks. happy to see it worked for you.. that's more than kudos
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]  09 Dec 2012, 01:29
1
KUDOS
virtualanimosity wrote:
Find the last 2 digits
Q1.
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90

Q2.
(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16

It seems there's a shortcut for the second problem:
1.) Forget about 201 and 246 and simplify it to 1*2*3*4*4*3*2*1, since 246 is the same as (250-4) giving 4, not 6.
1*2*3*4*4*3*2*1=2^2*3^2*4^2=574

2.)Now, take 74 out of 574 and calculate a square of 74.
It doesn't matter what it is but it ends with 76

Last edited by felixjkz on 10 Dec 2012, 00:14, edited 1 time in total.
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]  29 Apr 2013, 02:47
1
KUDOS
Expert's post
sdas wrote:
Hi Karishma, many thanks for your response. I am a fan of your blogs, do see my Quant Consolidated notes, Carcass's efforts of your blogs are best in there...

My confusion is how do we get to remainder 6 and not -19?
6/25 if this has remainder 6 then why should 24/25 have remainder of -1 (is the formula not same (x-n)/n
The negative part I understand completely, your blogs are too good

When the divisor is 25, a remainder of 6 is the same thing as a remainder of -19. Since GMAT doesn't give you negative remainders, you will not have -19 in the options. So you mush choose 6.
24/25 has remainder 24 which is also same as remainder of -1. Both are correct but again, GMAT will only give you 24 in the options.
Think of it: when you have 24 balls and you must distribute them equally among 25 kids, you can say that you have 24 balls remaining and you gave each kid 0 (the quotient) ball. Or you can say that you have -1 ball remaining (i.e. you gave one extra ball from your side) and each kid got 1 (the quotient) ball.
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]  29 Apr 2013, 10:55
1
KUDOS
Expert's post
virtualanimosity wrote:
Find the last 2 digits
Q1.
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90

Q2.
(201*202*203*204*246*247*248*249)^2
1.36
2.56
3.76
4.16

Q1.
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90

Finding the last two digits of an expression

Thru the Application of REMAINDER THEOREM

Remainder of the above expression when divided by 100 will give the answer of this question

$$\frac{65*29*37*63*71*87*62}{100}$$

$$\frac{13*29*37*63*71*87*62}{20}$$ ------------------> on dividing by 5

$$\frac{13*9*17*3*11*7*2}{20}$$ ------------------> taking remainder when each number divided by 20

$$\frac{117*51*77*2}{20}$$

$$\frac{17*11*17*2}{20}$$ ------------------> taking remainder when each number divided by 20

$$\frac{289*22}{20}$$

$$\frac{9*2}{20}$$ ------------------> taking remainder when each number divided by 20

$$\frac{18}{20}$$

So 18 is the remainder
However since initially we divided the numerator and denominator by 5, now we need to multiply 18 by 5

So last two digits = 90

Thru the Application of NEGATIVE REMAINDERS

$$\frac{13*29*37*63*71*87*62}{20}$$

$$\frac{(-7)*9*(-3)*3*(-9)*7*2}{20}$$ ----------> taking remainder when each number divided by 20

$$\frac{21*27*(-9)*14}{20}$$

$$\frac{1*7*(-9)*(-6)}{20}$$ ----------> taking remainder when each number divided by 20

$$\frac{7*54}{20}$$

$$\frac{(-13)*(-6)}{20}$$ ----------> taking remainder when each number divided by 20

$$\frac{78}{20}$$

$$\frac{18}{20}$$ ----------> taking remainder when each number divided by 20

18*5 = 90

NOTE :- If we are asked to find last 3 digits of the expression, we will obtain individual remainders when divided by 1000

Although Indian CAT is fond of such problems, i have never seen those in my any GMAT practice material

Regards,

Narenn
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Re: Numbers 4 [#permalink]  04 Nov 2009, 02:07
What is this stuff. I thought answer is "A".

Can some one explain in detail?
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Re: Numbers 4 [#permalink]  04 Nov 2009, 04:20
Hussain15 ..... Please explain how you reached to ans A
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Re: Numbers 4 [#permalink]  04 Nov 2009, 12:30
It looks like that it is not GMAT question...
Not reasonable to do in less than 2 minutes.
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Re: Numbers 4 [#permalink]  04 Nov 2009, 14:59
Ans to 2nd question should be 56.

I am sure its not a GMAT question, even after using techniques it takes more than 2-3 min..
Here we go:
(201*202*203*204*246*247*248*249)^2

we can make pairs within the numbers inside the brackets such as
(200+1)(200+49) which gives us last 2 digits = 49
(200+2)(200+48) which gives us last 2 digits = 96
(200+3)(200+47) which gives us last 2 digits = 141
(200+4)(200+46) which gives us last 2 digits = 184

No again the last two dogits can be paired as
49*141 = (100-41)(100+41) or (100^2 - 41^2) which gives us last 2 digits = 19 and
96*184 = (140-44)(140+44) or (100^2 - 44^2) which gives us last 2 digits = 64

Now, 19*64 gives us last 2 digits of 16 and square of 16 will be 56. Phewwww!!

Pls tell me I am right, else i am not attempting the first one!!

49=100-51 but not 41. Or it doesn't matter?

Could you also explain how to calculate that (100^2 - 44^2) has 64 as last two digits?
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Re: Numbers 4   [#permalink] 04 Nov 2009, 14:59

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