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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!!

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? _________________

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

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

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

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!]

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.

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.

Back to hometown after a short trip to New Delhi for my visa appointment. Whoever tells you that the toughest part gets over once you get an admit is...