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

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Re: Numbers 4 [#permalink]  02 Dec 2010, 10:55
Expert's post
Of course last 2 digits of the first question can be easily obtained:
65*29*37*63*71*87*62
1. 70
2. 30
3. 10
4. 90

We have a 5 (in 65) and a 2 ( in 62)
So 65*29*37*63*71*87*62 = 5*13 * 29*37*63*71*87*2 * 31
10 *13 * 29*37*63*71*87*2 * 31
The last digit will be 0. The second last digit will be the last digit of 13*29*37*63*71*87*31 which we get as 9.
So last two digits will be 90.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews SVP Joined: 16 Jul 2009 Posts: 1627 Schools: CBS WE 1: 4 years (Consulting) Followers: 39 Kudos [?]: 653 [0], given: 2 Re: Numbers 4 [#permalink] 13 Dec 2010, 12:24 pradhan wrote: 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 can you explain that technique? _________________ The sky is the limit 800 is the limit GMAT Club Premium Membership - big benefits and savings Intern Joined: 01 Mar 2011 Posts: 8 Followers: 0 Kudos [?]: 3 [0], given: 1 Re: Find the last two digits [#permalink] 07 May 2011, 08:19 sriharimurthy wrote: 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. Thus answer is (4). In your operation you are doing division to make the problem simpler specifically, Quote: 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. You cannot divide to simplify when you are setting out to find a remainder. For example what is the remainder of $$14/10$$? 4 what is the remainder of $$7/5$$? 2 Notice that both 14/10 and 7/5 are both equivalent. But the remainders are different VP Status: There is always something new !! Affiliations: PMI,QAI Global,eXampleCG Joined: 08 May 2009 Posts: 1354 Followers: 15 Kudos [?]: 191 [0], given: 10 Re: Numbers 4 [#permalink] 07 May 2011, 12:11 terrific way to solve this. _________________ Visit -- http://www.sustainable-sphere.com/ Promote Green Business,Sustainable Living and Green Earth !! Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 6216 Location: Pune, India Followers: 1674 Kudos [?]: 9586 [0], given: 196 Re: Find the last two digits [#permalink] 07 May 2011, 18:25 Expert's post gprasant wrote: You cannot divide to simplify when you are setting out to find a remainder. For example what is the remainder of $$14/10$$? 4 what is the remainder of $$7/5$$? 2 Notice that both 14/10 and 7/5 are both equivalent. But the remainders are different Yes. You cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, remember, if you do want to cancel off to make life easier for you, you can do it, provided you remember to multiply it back again. So say, I want to find the remainder when 14 is divided by 10 (remainder 4) and cancel off the common 2 to get 7 divided by 5 giving me a remainder of 2, I can multiply back the 2 I canceled with the remainder to get a remainder of 4. That's a valid technique. e.g. What is the remainder when 85 is divided by 20? It is 5. or I might rephrase it as what is the remainder when 17 is divided by 4 (I cancel off 5 from the numerator and the denominator). The remainder in this case is 1. I multiply the 5 back to 1. I get the remainder as 5! _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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Re: Find the last two digits [#permalink]  07 May 2011, 18:31
VeritasPrepKarishma wrote:
gprasant wrote:

You cannot divide to simplify when you are setting out to find a remainder. For example what is the remainder of $$14/10$$? 4
what is the remainder of $$7/5$$? 2
Notice that both 14/10 and 7/5 are both equivalent. But the remainders are different

Yes. You cannot just cancel off the common terms in the numerator and denominator to get the remainder. But, remember, if you do want to cancel off to make life easier for you, you can do it, provided you remember to multiply it back again.
So say, I want to find the remainder when 14 is divided by 10 (remainder 4) and cancel off the common 2 to get 7 divided by 5 giving me a remainder of 2, I can multiply back the 2 I canceled with the remainder to get a remainder of 4. That's a valid technique.

e.g. What is the remainder when 85 is divided by 20? It is 5.
or I might rephrase it as what is the remainder when 17 is divided by 4 (I cancel off 5 from the numerator and the denominator). The remainder in this case is 1. I multiply the 5 back to 1. I get the remainder as 5!

thanks for the tip karishma. This will definitely save a lot of time during certain PS questions
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Re: Numbers 4 [#permalink]  04 Jul 2011, 08:38
really good discussion.
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Re: Numbers 4 [#permalink]  21 Aug 2012, 06:34
Bunuel wrote:
sriharimurthy wrote:
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.

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

+1 for each.

Hi

This link is no longer active, please show us how to solve these kind of questions
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]  09 Dec 2012, 01:29
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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: Numbers 4 [#permalink]  10 Dec 2012, 00:04
Thanks for this post. I am sharing it on my blog.
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Re: Find the last two digits [#permalink]  24 Dec 2012, 16:36
sriharimurthy wrote:
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.

please explain how did you get [3*(-1)*(-3)*3*1*(-3)*1]/10 from (13*29*37*63*71*87*31)/10
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Re: Numbers 4 [#permalink]  08 Jan 2013, 08:04
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

??? wrong.... 5*1*3*3*1*3*2 does NOT equal 90...
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Re: Numbers 4 [#permalink]  08 Jan 2013, 08:14
Expert's post
mbaiseasy wrote:
Thanks for this post. I am sharing it on my blog.

what is the url of your blog ???
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Re: Find the last two digits [#permalink]  09 Feb 2013, 05:11
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!]

In this part of the solution you subtracted only three 24's, but not 6. Why?
Following your formula it should be $$= R of (12*24*24*24)/50 = R of (6*24*24*24)/25 = R of [(-19)*(-1)*(-1)*(-1)]/25 = 19/25 ----> 76/100$$
Is there any difference if you subtract some numbers and some leave untouched? Do you have to subtract all of them, or you can do it randomly?
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Re: Find the last two digits [#permalink]  13 Feb 2013, 06:30
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 explanation!
I think that this way is a lot quicker...
$$(201*202*203*204*246*247*248*249)^2$$
$$(201*202*203*204*246*247*248*249)*(201*202*203*204*246*247*248*249)$$
$$(1*2*3*4)*((-4)*(-3)*(-2)*(-1)) = 24^2 = 576$$
So the last two numbers are 76.
Solution for 30 seconds!
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Re: Find the last two digits [#permalink]  26 Feb 2013, 09:32
sriharimurthy wrote:
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.

When i calculate is get
$$R of (13*29*37*63*71*87*31)/10 =R of [3*(9)*(7)*3*1*(7)*1]/10 = R of 3969/10 = R = 9$$

how did you get [3*(-1)*(-3)*3*1*(-3)*1]/10 but I am getting [3*(9)*(7)*3*1*(7)*1]/10? What am i doing wrong?

reference: compilation-of-tips-and-tricks-to-deal-with-remainders-86714.html
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]  26 Feb 2013, 10:15
didn't understand the concept
R of (13*29*37*63*71*87*31)/10 =R of [3*(9)*(7)*3*1*(7)*1]/10 = R of 3969/10 = R = 9
why 13/10 Remainder is 3 or 63/10 remainder is 3, but when 29/0 the remainder us -1 and not 9 or 37/10 remainder is -3 and not 7??
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Re: Find the last 2 digits of 65*29*37*63*71*87*62 [#permalink]  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]  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]  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 two digits   [#permalink] 27 Apr 2013, 11:33

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