Hard DS - Number Properties and Exponents - The Beat The GMAT Forum

Hard DS - Number Properties and Exponents

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Tue Apr 13, 2010 8:56 pm

What is the value of x if x is the remainder obtained when 2^(8p+2) + z is divided by 5, and p is a positive integer?

1) z = 6
2) z is even

OA: A

Here is what I did:

Statement 2: Obviously not sufficient.

Statement 1: I plugged in "1" for p to get 2^10 which is 1024. Then I plugged in 6. Which means that the remainder is 0. Now, I don't have time to check the next possible value of p, since that would yield 2^18. However, I noticed a pattern:

2
4
8
16
32
64
128
256
512
1024
etc...

The number is repeating, so I know that 2^(8p+2) will always end on a 4, so +6 = 10. Remainder always 0. Sufficient.

Now, let's say I didn't pick up on the logical pattern next time, is there a more mechanical way to solve this problem?

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Source: Beat The GMAT — Data Sufficiency | Tue Apr 13, 2010 8:56 pm

kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Tue Apr 13, 2010 11:05 pm

student22 has deciphered an imp pattern

my solution would be on these lines

2^(8P+2) = 2^2(4P+1) and it is possible to arrive at an answer

but there may be other methods. Just couldn't stop from posing a teaser.

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thephoenix: Legendary Member; Posts: 1560; Joined: Tue Nov 17, 2009 2:38 amFollowed by: 5 members; Thanked: 137 times

by thephoenix » Tue Apr 13, 2010 11:53 pm

pattern of unit digit is 2,4,8,6,2,4,,8,6........after evry 4th power it repeats
8p+2 when divided by 4 leaves remainder 2 hence the unit digit for 2^(8P+2) and for an int value of P will be same as that of 2^2 which is 4
and 6+4 =10 unit dig is 0 and is div by 5

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dxgamez: Senior | Next Rank: 100 Posts; Posts: 65; Joined: Wed Nov 25, 2009 6:33 pm; Thanked: 3 times

by dxgamez » Wed Apr 14, 2010 12:41 am

@thephoenix

Can you explain further 8p+2 when divided by 4 leaves remainder 2 hence the unit digit for 2^(8P+2) and for an int value of P will be same as that of 2^2 which is 4

I can't seem to get it. Don't get the relation where 8p+2 leaves remainder 2 hence the unit digit .....

Is this the same method that kstv is saying about?

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akhpad: Legendary Member; Posts: 809; Joined: Wed Mar 24, 2010 10:10 pmFollowed by: 4 members; Thanked: 50 times

by akhpad » Wed Apr 14, 2010 7:10 am

What is the value of x if x is the remainder obtained when 2^(8p+2) + z is divided by 5, and p is a positive integer?

1) z = 6
2) z is even

Statement 1:
This is all about just power cycle.
Power cycle of 2 is 2, 4, 8, 6
8p+2 = 4*2*p + 2

Unit digit in 2^(8p+2) is 4

2^(8p+2) + z = Number (having unit digit 4) + 6 = Number having unit digit 0
Divisible by 5

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Wed Apr 14, 2010 10:35 am

Ok, so I guess I went about this problem correctly by noticing that unit digit pattern.

Ktsv, can you please go into more detail on your solution.

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liferocks: Legendary Member; Posts: 576; Joined: Sat Mar 13, 2010 8:31 pmFollowed by: 1 member; Thanked: 97 times

by liferocks » Wed Apr 14, 2010 5:55 pm

I have approached this way

2^8=256 hence the reminder is 1 when devided by 5 -->this I did as I already know the powers of 2 till 12 else i would have take taken the pattern of reminder as 2,4,3,1 in to consideration and concluded that for poser 8 reminder will be 1

we can write the number 2^(8p+2) + z as 4(256^p)+z
when this is divided by 5 reminder will be 4(1^p)+z=4+z
from condition 1 z is 6 hence reminder is 10 i.e 0-->suff
from option 2 we cannot comment abt the reminder value-->insuff

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kstv: Legendary Member; Posts: 610; Joined: Fri Jan 15, 2010 12:33 amFollowed by: 2 members; Thanked: 47 times

by kstv » Wed Apr 14, 2010 9:50 pm

2^(8P+2) = 2^2(4P+1) = 4^(4P+1)
P is a +ve integer
so the expression (4P +1) >=5
ie odd number 5, 9 etc

the power cycle ( got this term from akhp77) of 4 is
4, 4Â², 4^3 etc 4, 16 , 64
notice the odd exponents i.e. 4, 4^3 lead to 4 in the unit place of the no
we know (4P+1) is always odd
so 4^(4P+1) will always have 4 in the unit place
1) z = 6 so 4^(4P+1) + z will have 0 in the unit place
any such number is divisible by 5
and leave no remainder Sufficient
2) z is even can be 2,4, 6 , 8
Insuff

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student22: Master | Next Rank: 500 Posts; Posts: 216; Joined: Mon Dec 21, 2009 10:26 am; Thanked: 16 times

by student22 » Thu Apr 15, 2010 7:50 am

Wow, kstv and liferocks, both very interesting approaches to this problem.

kstv, very good way to prove that the exponent will be odd, and hence always end in 4 for the 4 power cycle.

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