i think b alone is not sufficient coz if m=1,
then the expression will be 34n + 2 +1 i.e.
34n + 3
now take n = 1 then 34n+3 = 37. hence remainder 7
take n = 2 , 34n+3 = 71. hence remainder 1.
hence insufficient..
IMO C.
ds
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Source: Beat The GMAT — Data Sufficiency |
I know what is wrong with this. The question says [3^(4n+2)] + m, and here is why the answer is B:shibal wrote:ketkoag i think so too, but oa in gmat prep was b
A. INSUFF -> You don't know the value of m
B. SUFF. Check out the values of x = 3^(4n+2) below:
n= 1 x= 3^6 = 9^3 = 729
2 x= 3^10 = 9^5 = 729 * 81
3 x= 3^14 = 9^7 = 9 * (729)^2
See the pattern? Regardless of the value of n, the value of the last digit of x will always be 9 and, when you add 1, you are dividing by a number that is divisible by 10.
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sreak1089
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If you look closely, 3 ^ 1 always ends in 3
3 ^ 2 always ends in 9
3 ^ 3 always ends in 7
3 ^ 4 always ends in 1
3 ^ 5 always ends in 3
3 ^ 6 always ends in 9
3 ^ 7 always ends in 7
3 ^ 8 always ends in 1
and so on this pattern repeats
3 ^ 2, 3 ^ 6, 3 ^ 10, and so on can be represented as 3 ^ (4n + 2)
which will always end in 9, since m = 1, which means 3 ^ (4n + 2) + m
ends in 0, hence the remainder will always be zero.
Also, 3 ^1, 3 ^ 5, 3 ^ 9, so on is represented as 3 ^ (4n + 1) which always ends in 3.
3 ^ 3, 3 ^ 7, 3 ^ 11, so on is represented as 3 ^ (4n + 3) which always end in 7.
3 ^ 4, 3 ^ 8, 3 ^ 12, so on is represented as 3 ^ (4n) always ends in 1.
3 ^ 2 always ends in 9
3 ^ 3 always ends in 7
3 ^ 4 always ends in 1
3 ^ 5 always ends in 3
3 ^ 6 always ends in 9
3 ^ 7 always ends in 7
3 ^ 8 always ends in 1
and so on this pattern repeats
3 ^ 2, 3 ^ 6, 3 ^ 10, and so on can be represented as 3 ^ (4n + 2)
which will always end in 9, since m = 1, which means 3 ^ (4n + 2) + m
ends in 0, hence the remainder will always be zero.
Also, 3 ^1, 3 ^ 5, 3 ^ 9, so on is represented as 3 ^ (4n + 1) which always ends in 3.
3 ^ 3, 3 ^ 7, 3 ^ 11, so on is represented as 3 ^ (4n + 3) which always end in 7.
3 ^ 4, 3 ^ 8, 3 ^ 12, so on is represented as 3 ^ (4n) always ends in 1.
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zaarathelab
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It's imp to remember the cyclicity of different numbers
3 has a cyclicity of 4. This means that when 3 is raised to an exponential power, the last digit of the final value will repeat itself after every 4 powers.
Ex -
3^1 =3 (last digit 3)
3^2 = 9 (last digit 9)
3^3 = 27 (last digit 7)
3^4 = 81 (last digit 1)
3^5 = 243 (last digit 3 - - - repeat pattern)
Now here, Statement 1 is clearly insufficient as m can take on different value, correspondingly the remainder will change
Statement 2 -
Sufficient - cyclicity of 3 which will repeat itself after 4 powers. plug in different values of n and see
3 has a cyclicity of 4. This means that when 3 is raised to an exponential power, the last digit of the final value will repeat itself after every 4 powers.
Ex -
3^1 =3 (last digit 3)
3^2 = 9 (last digit 9)
3^3 = 27 (last digit 7)
3^4 = 81 (last digit 1)
3^5 = 243 (last digit 3 - - - repeat pattern)
Now here, Statement 1 is clearly insufficient as m can take on different value, correspondingly the remainder will change
Statement 2 -
Sufficient - cyclicity of 3 which will repeat itself after 4 powers. plug in different values of n and see
