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Question

The Least Common multiple of \(2^6-1\) and \(2^9-1\) is

a) 2^12+27*2^9-217
b) 2^ 12 +63*2^3-1
c) 2^12+5^29-1
d) 2^12+9*2^8 -1
e) None of these.

Hi Experts,
I would like to know what is the best approach to solve such questions.

Help will be appreciated.

Thanks
H


I feel [B] is the answer. I am not sure my way is the shortest (coz i did have to go thru a lot of calculation), but it is a way nonetheless :)

From the above, one can find that the HCF of the numbers \(2^6-1\) and \(2^9-1\) is 7. By the rule,
LCM*HCF = Product of two numbers
=> LCM = Product of two numbers /7

7 can be expressed as \(2^3 -1\). By checking all the answer choices one can see that only [B] i.e. \(2^ 12 +63*2^3-1\),
also written as, \(2^ 12 +(2^6 - 1)2^3-1 = 2^ 12 +2^9 -2^3 -1\), results the product of numbers i.e. \(2^15 - 2^6 - 2^9 + 1.\)

Hope my answer is accurate! Would certainly appreciate a shorter way on this! :)

Regards,
Arpan

**edited to correct typo
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Bunuel, kindly update the OA in expanded / Mathematical form.
The un-formatted options tend to confuse :)
Thanks
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Bunuel, kindly update the OA in expanded / Mathematical form.
The un-formatted options tend to confuse :)
Thanks
______________
Done.
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The Least Common multiple of 2^6-1 and 2^9-1 is:

The following steps will help to solve the sum very easily-:

First, lets get the units digit values of 2^6-1 and 2^9-1
1.Units digit of 2^6-1 is 3
2.Units digit of 2^9-1 is 1

Consider every option here with respect to taking the units digit of the value.
A. - 5 (Eliminated as the required values are not multiples of 3 and 1)
B. - 3
C. - 0 (Eliminated as the required values are odd numbers, i.e. 3 and 1)
D.- 9

We are left with option B and D

LCM of 3 and 1 is option B., i.e. 3( which is its unit digit)


Official Answer- Option B.


Thanks.

Regards,
Raunak Damle!

Cheers!
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imhimanshu
The Least Common multiple of 2^6-1 and 2^9-1 is:

A. \(2^{12}+27*2^9-217\)

B. \(2^{12} +63*2^3-1\)

C. \(2^{12}+5^{29}-1\)

D. \(2^{12}+9*2^8 -1\)

E. None of these.

The Least Common multiple of 2^6-1 and 2^9-1 is:

\(2^6 -1 = (2^3+1)(2^3-1) \)
\(2^9 -1 = (2^3 -1)(2^6 + 2^3 + 1) \)

\(LCM (2^6 -1,2^9 -1) = (2^3+1)(2^9-1) = 2^{12} +2^9 -2^3 -1 = 2^{12} + 64*2^3 - 2^3 - 1 = 2^{12} +63*2^3 - 1\)

IMO B
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Quote:
Note that option 'none of these' makes it more complicated since you cannot try some more esoteric methods e.g. last digit etc. GMAT doesn't give you this option.

Hi Karishma, this is actually not true. I've seen GMAT questions, on the live test, with this answer choice.
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IMO best approach is do it the boring way if ur quick with calculations.
Both numbers are 63 and 511 respectively
And knowing 63 is divisible by 7 all u need to check is if 511 is also divisible by 7 otherwise the LCM is simply the product
And it is!
Now we find factors of 511=7 and 73
for 63 it is 7*9
So to find LCM it is simply 63*73
Seeing the options ur quick to eliminate all except option B (visually close) and optiin E which is none of these.
Option B fits perfectly hence is the answer.

imhimanshu
The Least Common multiple of 2^6-1 and 2^9-1 is:

A. \(2^{12}+27*2^9-217\)

B. \(2^{12} +63*2^3-1\)

C. \(2^{12}+5^{29}-1\)

D. \(2^{12}+9*2^8 -1\)

E. None of these.
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This alone wont help because none of these is also an option, unless u also figure out the range of values possibly quickly then yes. But under time pressure tunnel visioning units digit with none of these is not advisable IMO.
raunakd11
The Least Common multiple of 2^6-1 and 2^9-1 is:

The following steps will help to solve the sum very easily-:

First, lets get the units digit values of 2^6-1 and 2^9-1
1.Units digit of 2^6-1 is 3
2.Units digit of 2^9-1 is 1

Consider every option here with respect to taking the units digit of the value.
A. - 5 (Eliminated as the required values are not multiples of 3 and 1)
B. - 3
C. - 0 (Eliminated as the required values are odd numbers, i.e. 3 and 1)
D.- 9

We are left with option B and D

LCM of 3 and 1 is option B., i.e. 3( which is its unit digit)


Official Answer- Option B.


Thanks.

Regards,
Raunak Damle!

Cheers!
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