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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, 14:44

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

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

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

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!]

Answer should be (3).

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

"When the going gets tough, the tough gets going!"

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

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. _________________

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

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28 Apr 2013, 00:54

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 _________________

"When the going gets tough, the tough gets going!"

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

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

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!]

Answer should be (3).

Excellent approach but still taking more than 2 mins _________________

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.

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 _________________

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

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27 Jun 2013, 09:21

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.

Please help explain the logic when we multiply and when we do not.

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

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

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!]

Answer should be (3).

This is a great method. However i am not able to understand why R of 246/50 = -4? Please help to explain. _________________

“Confidence comes not from always being right but from not fearing to be wrong.”

in fact, it is also not that time consuming to multiply all numbers since we do not have to calculate exact and entire digits off the resulting number. All we need is last two digits, so try multiply them all, it is still workable when you cannot think of any shortcuts when in a hurry

gmatclubot

Re: Find the last 2 digits of 65*29*37*63*71*87*62
[#permalink]
21 Sep 2013, 12:08

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