Divisibility/ Number Props
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Source: Beat The GMAT — Problem Solving |
Took me maybe a little over 2 minutes I think...gotta start timing myself. Anyway, if x is a positive integer in the equation, just use 1. Now, you can simplify this to 7^15 + 3. You have to find the units digit of 7^15. If you do 7^1 through 7^5 written down, you can see a pattern emerge. This pattern tells you that the units digit of 7^15 is 3. Adding 3 to this units digit gives you a 6 as the units digit of 7^15 + 3. Dividing that number by five would give you a remainder of one (since 6/5 = 1 as quotient with 1 as remainder). Answer is B.
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sixpointer
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euro wrote:If x is a positive integer, what is the remainder when [ 7^(12x+3) ] + 3 is divided by 5?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4
[spoiler]OA (B)[/spoiler]
7^(12x+3) % 5 = 7^12x*7^3 % 5 => 2^12x*2^3 % 7
Now 4%5 = -1(remainder)
So, 2^4 will give remainder as 1 with 5
Take any value of x u will always get 1 as remainder
8 divided by 5 leaves remainder 3
So 3+3 divided by 5 will leave reaminder as 2 .
P.S if u are not able to understand . Please Pm me I will explain
Hey Six, you lost me with that one.
I'm going to assume that the (%) sign I'm seeing is meant to represent division?
(7^12x * 7^3) / 5 looks like it was factored correctly, but I'm not quite sure how you converted that into (2^12x * 2^3) / 7. I'm guessing you obtained this by logic, as it doesn't seem to follow any exponent rule I'm familiar with (correct me if I'm wrong)? Just curious how this was derived because this quickly arrived at the correct answer of remainder = 1.
I came to remainder 1 using the same method as Arcane. Except I just focused on multiplying the units digit by 7 each time because that's all we'll need to know to determine the repeating pattern. Multiplying the units digit by 7 each time we see that:
7^1 = 7
7^2 = 49
7 ^3 = 9*7 = 3
7^4 = 3*7 = 1
7^5 = 1* 7 = 7
From here we have determined the pattern and see that 7^15 ends in 3.
I'm going to assume that the (%) sign I'm seeing is meant to represent division?
(7^12x * 7^3) / 5 looks like it was factored correctly, but I'm not quite sure how you converted that into (2^12x * 2^3) / 7. I'm guessing you obtained this by logic, as it doesn't seem to follow any exponent rule I'm familiar with (correct me if I'm wrong)? Just curious how this was derived because this quickly arrived at the correct answer of remainder = 1.
I came to remainder 1 using the same method as Arcane. Except I just focused on multiplying the units digit by 7 each time because that's all we'll need to know to determine the repeating pattern. Multiplying the units digit by 7 each time we see that:
7^1 = 7
7^2 = 49
7 ^3 = 9*7 = 3
7^4 = 3*7 = 1
7^5 = 1* 7 = 7
From here we have determined the pattern and see that 7^15 ends in 3.
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Rahul@gurome
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Hi!rros0770 wrote:Hey Six, you lost me with that one.
I'm going to assume that the (%) sign I'm seeing is meant to represent division?
(7^12x * 7^3) / 5 looks like it was factored correctly, but I'm not quite sure how you converted that into (2^12x * 2^3) / 7. I'm guessing you obtained this by logic, as it doesn't seem to follow any exponent rule I'm familiar with (correct me if I'm wrong)? Just curious how this was derived because this quickly arrived at the correct answer of remainder = 1.
I came to remainder 1 using the same method as Arcane. Except I just focused on multiplying the units digit by 7 each time because that's all we'll need to know to determine the repeating pattern...
What sixpointer wrote in his reply was an application of Modulo Arithmetic. The % (percentage) sign is actually the modulus sign.
(a mod n) = a%n = b, means when a is divided by n, the remainder is b.
Some basic theorems of modulo arithmetic are:
1. (a + b) mod n = [(a mod n) + (b mod n)] mod n
2. (ab) mod n = [(a mod n)*(b mod n)] mod n
3. a^b mod n = [(a mod n)^b] mod n
These formulas can be proved very easily. In fact, (1) is very much trivial.
Now using these formulas here,
[7^(12x + 3) + 3] mod 5
= [7^(12x + 3) mod 5 + 3 mod 5] mod 5
= [[(7 mod 5)^(12x + 3)] mod 5 + 3 ] mod 5
= [[2^(12x + 3)] mod 5 + 3] mod 5
= [[8*(2^12x)] mod 5 + 3 ] mod 5
= [[(8 mod 5)*(2^12x)] mod 5 + 3 ] mod 5
= [[3*(2^12x) mod 5] mod 5 + 3] mod 5
= [[3*(4096^x) mod 5] mod 5 + 3] mod 5
= [[3*(4096 mod 5)^x mod 5] mod 5 + 3] mod 5
= [[3*1] mod 5] mod 5 + 3 ] mod 5
= [3+3] mod 5
= 1
Though the alternate solution provided to this question is standard and easy, this method will help you to solve this kind of problems real fast. (Provided you have mastered it!
)Hope this helps.
Last edited by Rahul@gurome on Thu Oct 28, 2010 12:13 pm, edited 1 time in total.
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waltz2salsa
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For this question i believe the easier method would be:
Last digit of 7^(12x+3) is 3 for any x ( 7 has a cyclicity of 4)
hence, Last digit of (7^(12x+3) + 3) is 6
So our question is now remainder when 6 is divided by 5 and the correct answer is 1.
Regards,
Shashwat
Last digit of 7^(12x+3) is 3 for any x ( 7 has a cyclicity of 4)
hence, Last digit of (7^(12x+3) + 3) is 6
So our question is now remainder when 6 is divided by 5 and the correct answer is 1.
Regards,
Shashwat
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pesfunk
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Perfect, Easiest and Un-debatable !!
@Rahul - We don't need FORMULA to solve this question....just make life simple...
@Rahul - We don't need FORMULA to solve this question....just make life simple...
waltz2salsa wrote:For this question i believe the easier method would be:
Last digit of 7^(12x+3) is 3 for any x ( 7 has a cyclicity of 4)
hence, Last digit of (7^(12x+3) + 3) is 6
So our question is now remainder when 6 is divided by 5 and the correct answer is 1.
Regards,
Shashwat
