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

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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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26 Feb 2013, 16:40
Very interesting, but this is not a real GMAT question... isn't it?
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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09 Mar 2013, 13:44
Awesome job, sriharimurthy. I just want to add that this particular calculation could have been performed in a slightly different way for more clarity, i.e. by applying the general formula twice: "R of 13*29*37*63*71*87*31(/10)" = R of 3*9*7*3*1*7*1(/10)" = R of 81*49(/10) = 9

Thank you a lot for the very nice formula!
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Re: Find the last two digits [#permalink]

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27 Apr 2013, 11:33
sriharimurthy wrote:
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!]

Can someone explain the highlighted part. Why R of (x-n)/n = R of (6-25)/25 = +6?
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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27 Apr 2013, 22:30
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.

For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows.

5*-1*-3*3*1*-3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90.

Q2.

same approach. 1*2*3*4*-4*-3-*-2-*-1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36.

To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36
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Re: Find the last two digits [#permalink]

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27 Apr 2013, 22:48
sdas wrote:

Can someone explain the highlighted part. Why R of (x-n)/n = R of (6-25)/25 = +6?

I am not sure from where you are reading this and what this means but I think you are confused with the concept of negative remainders. It is discussed here: http://www.veritasprep.com/blog/2011/05 ... emainders/

When the remainder of x/n is 6, there is no confusion. But when the remainder of x/n comes out to be -6, we need to adjust the remainder (make it positive) by making it n - 6 (why has been discussed in the post)

Also, when talking about divisibility and remainders, you don't usually take negative integers into account. Divisibility is a positive integer concept. You can divided 6 balls among 3 kids but not -6 balls among 3 kids.

So (6-25) = -19 divided by 25 doesn't really make sense.

If instead it were (25n - 19) divided by 25, then the remainder will be -19 since 25n is divisible by 25. A remainder of -19 is same as a remainder of 6.
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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27 Apr 2013, 23:54
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
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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29 Apr 2013, 05:52
Thanks Karishma, for clearing my doubt.
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Re: Find the last two digits [#permalink]

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25 Jun 2013, 07:44
sriharimurthy wrote:
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!]

Excellent approach but still taking more than 2 mins
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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25 Jun 2013, 08:49
riyazv2 wrote:
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.

For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows.

5*-1*-3*3*1*-3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90.

Q2.

same approach. 1*2*3*4*-4*-3-*-2-*-1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36.

To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36

How would you find the last two digit of 33*33*33*33 using the method mentioned here.
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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25 Jun 2013, 21:34
koolgmat wrote:
riyazv2 wrote:
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.

For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows.

5*-1*-3*3*1*-3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90.

Q2.

same approach. 1*2*3*4*-4*-3-*-2-*-1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36.

To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36

How would you find the last two digit of 33*33*33*33 using the method mentioned here.

33*33*33*33 /100

33*33*33*33 /5 -----------------> Divide the denominator by 20.

-2*-2*-2*-2/5= 16/5 = 1

Now lets multiply 1 with 20 (since we divided it with 20) = 20
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]

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27 Jun 2013, 08:21
prateekbhatt wrote:
koolgmat wrote:
riyazv2 wrote:
Q1.

For these type of questions, always try to relate the numbers to a common divider. Here it will be 10 and hence we can minimize the products are as follows.

5*-1*-3*3*1*-3*2 = 10*27 = 270 = the final digit will be 0. But to get the tenth digit, please add the remaining two = (2+7). In the end, the answer will be 90.

Q2.

same approach. 1*2*3*4*-4*-3-*-2-*-1 = 24*24 = 576, so tenth digit will be (5+7) & again (1+2) = 3. Finally the last two digit of the expresiion the bracket will be 36.

To find the last two digits after applying the square function is to follow the same procedure. Take the last digit, which is 6. Hence, 6*6 = 36

How would you find the last two digit of 33*33*33*33 using the method mentioned here.

33*33*33*33 /100

33*33*33*33 /5 -----------------> Divide the denominator by 20.

-2*-2*-2*-2/5= 16/5 = 1

Now lets multiply 1 with 20 (since we divided it with 20) = 20

If i apply this logic to find last two digit of 77*77*77*77 then

77*77*77*77/100

77*77*77*77/5 -----------------> Dividing the denominator by 20.

2*2*2*2 /5 = 16/5 = 1

We multiply 1 with 20 ( since we divided it by 20) = 20
This gives answer as 21 and not 41.

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Re: Find the last two digits [#permalink]

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20 Sep 2013, 19:25
sriharimurthy wrote:
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!]

This is a great method. However i am not able to understand why R of 246/50 = -4? Please help to explain.
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Re: Find the last two digits [#permalink]

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21 Sep 2013, 05:25
ankur1901 wrote:
This is a great method. However i am not able to understand why R of 246/50 = -4? Please help to explain.

No, -4 is the Negative Remainder and not actual remainder. We can get actual remainder by adding divisor in the negative remainder i.e. -4+50 = 46

Refer my this post, if you still have any query about remainder theorem :- find-the-last-2-digits-of-86325-60.html#p1218383
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Re: Find the last two digits   [#permalink] 21 Sep 2013, 05:25

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