|
Author |
Message |
|
Director
Joined: 16 May 2007
Posts: 552
Followers: 3
Kudos [?]:
10
[0], given: 0
|
Exponents Problem (m04q20) [#permalink]
 19 Jun 2007, 20:15
Question Stats:
91% (01:30) correct
8% (00:29) wrong based on 2 sessions
What is the last digit of the following number 2^{22} * 3^{15} * 5^{16} * 7^1 ? (A) 6 (B) 5 (C) 2 (D) 1 (E) 0 Source: GMAT Club Tests - hardest GMAT questions Can someone please explain how to solve these kind of problems Thanks
|
|
|
|
|
|
|
|
|
Senior Manager
Joined: 04 Jun 2007
Posts: 357
Followers: 1
Kudos [?]:
9
[0], given: 0
|
We need to find the last digit of each term first.
2^1=2; 2^2=4; 2^3=8; 2^4=16; 2^5=32.
So the last digit of 2^x is repeated every 4th term.
so, the last digit of 2^22 will be the last digit of 2^2=4.
Similarly, for 3^x the last digit is repeated every 4th term.
3^1=3; 3^2=9; 3^3=27; 3^4=81; 3^5=243
so, the last digit of 3^15 will be last digit of 3^3=7
The last digit of 5^x is always 5 and that of 7^1 is 7.
So the last digit of the product will be the last digit of 4*3*5*7=0.
|
|
|
|
|
|
Director
Joined: 26 Feb 2006
Posts: 919
Followers: 3
Kudos [?]:
28
[2] , given: 0
|
2
This post received
KUDOS
sumande wrote: We need to find the last digit of each term first.
2^1=2; 2^2=4; 2^3=8; 2^4=16; 2^5=32.
So the last digit of 2^x is repeated every 4th term.
so, the last digit of 2^22 will be the last digit of 2^2=4.
Similarly, for 3^x the last digit is repeated every 4th term.
3^1=3; 3^2=9; 3^3=27; 3^4=81; 3^5=243
so, the last digit of 3^15 will be last digit of 3^3=7
The last digit of 5^x is always 5 and that of 7^1 is 7.
So the last digit of the product will be the last digit of 4*3*5*7=0.
the process is perfect and result too but a small mistake: 4 x 7 x 5 x 7 = 0.
|
|
|
|
|
|
Senior Manager
Joined: 04 Jun 2007
Posts: 357
Followers: 1
Kudos [?]:
9
[0], given: 0
|
Thanks for pointing that out !!
But if there is a 2 and a 5, the other numbers don't matter 
|
|
|
|
|
|
Manager
Joined: 25 Apr 2007
Posts: 147
Followers: 2
Kudos [?]:
14
[1] , given: 0
|
1
This post received
KUDOS
To be honest, I found this solution after reading yours. But the methodology may be helpful for solving a similar problem :
2^22*3^15*5^16*7^1 = 2*5*(2^21*3^15*5^15*7)
2^22*3^15*5^16*7^1 = 10 * integer
10 * integer ends with 0
|
|
|
|
|
|
Manager
Joined: 17 May 2007
Posts: 181
Followers: 1
Kudos [?]:
6
[1] , given: 0
|
1
This post received
KUDOS
2^22)*(3^15)*(5^16)*(7^1)
7^1 = 7 then
5^16 will result 5 in units digit
7 * 5 reults 5 in units digit
Now 5 * 2^anything - will result in 0 as the units digit
Now 0* 3^anything - will give 0 in the units digit.
Answer: Last digist is 0
|
|
|
|
|
|
Current Student
Joined: 03 Oct 2006
Posts: 86
Followers: 1
Kudos [?]:
0
[0], given: 0
|
I found this another way as well. However, I think the method used in the second post is the best one to follow:
(2^15)(3^15)(5^16)(7^1)
=(6^15)(5^16)(7^1)
=(6^15)(5^15)(5^1)(7^1)
=(30^15)(35^1)
30^x = the last digit will always = 0
35 * 30 = the last digit will always = 0
|
|
|
|
|
|
Intern
Joined: 28 Jul 2010
Posts: 10
Followers: 0
Kudos [?]:
1
[0], given: 1
|
Re: Exponents Problem (m04q20) [#permalink]
12 Oct 2010, 05:14
We have 2 and 5 in the product, so last digit always be 0 irrespective of any other number.
|
|
|
|
|
|
Manager
Joined: 07 Aug 2010
Posts: 90
Followers: 1
Kudos [?]:
13
[0], given: 9
|
Re: Exponents Problem (m04q20) [#permalink]
12 Oct 2010, 22:54
trahul4 wrote: What is the last digit of the following number 2^{22} * 3^{15} * 5^{16} * 7^1 ? (A) 6 (B) 5 (C) 2 (D) 1 (E) 0 Source: GMAT Club Tests - hardest GMAT questions Can someone please explain how to solve these kind of problems Thanks E 2() * 3() * 5() * 7() = multiple of 10 ==> 0
_________________
Click that thing  - Give kudos if u like this
|
|
|
|
|
|
Intern
Joined: 10 Jun 2010
Posts: 8
Concentration: Marketing, General Management
GPA: 3.8
WE: Web Development (Investment Banking)
Followers: 0
Kudos [?]:
3
[3] , given: 12
|
Re: Exponents Problem (m04q20) [#permalink]
13 Oct 2010, 09:03
3
This post received
KUDOS
The first thing i saw in this question was a presence of a 5. That means the answer had to have either a 0 or a 5. So now I have automatically eliminated 3 answers. Second, all powers of 3 result in an odd number. (Check it up: 3^1=3, 3^4=81). Third, all powers of 2 are even. So its basically even (because of 2) mulitplied by odd (because of 3) even*odd = even even number*(any power of 5) = ends with a zero eg. 2*5=10, 2*25=50, etc. Hence E
|
|
|
|
|
|
Intern
Joined: 17 Oct 2007
Posts: 46
Followers: 0
Kudos [?]:
2
[0], given: 60
|
Re: Exponents Problem (m04q20) [#permalink]
13 Oct 2010, 17:54
Look no further if you see 2 and 5 as one of the factors.
|
|
|
|
|
|
Manager
Joined: 29 Sep 2008
Posts: 157
Followers: 2
Kudos [?]:
8
[0], given: 1
|
Re: Exponents Problem (m04q20) [#permalink]
14 Oct 2010, 07:16
take the last digit of all of them and t hen multiply and find
2^0=1;2^1=2;2^2=4;2^3=8;2^4=6;2^5=2;so it repeats after 4 times
3^0=1;3^1=3;3^2=9;3^3=7;3^4=1;3^5=3;so it repeats after 4 times
5^0=1;5^1=2;5^2=5;so it repeats after 1 times
7^1=7
2^22=22/4=2 is remainder.means it should repeat after 2 times.i.e 8
3^15=15/4=3 is remainder.means it should repeat after 3 times.i.e 7
5 and 7
so total=8*7*5*7=last digit is 0
|
|
|
|
|
|
Manager
Joined: 08 Sep 2010
Posts: 242
Location: India
WE 1: 6 Year, Telecom(GSM)
Followers: 3
Kudos [?]:
39
[0], given: 21
|
Re: Exponents Problem (m04q20) [#permalink]
14 Oct 2010, 20:43
2^x will repeat the last digit 2 after each 4th power.So
2^22=(2^4)^5 * 2^2 = yyy2*4=yyy8 I am just considering the last digit.
now 3^4=y1
3^15= (3^4)^3 * 3^3 = yy1*27 =yyy7
5^16=yyy5
7^1=7
Multiplying all last digit=8*7*5*7=yy0 so last digit is 0.Ans must be E
Consider KUDOS if u like the answer.
|
|
|
|
|
|
Intern
Joined: 13 Oct 2011
Posts: 4
Followers: 0
Kudos [?]:
0
[0], given: 0
|
Re: Exponents Problem (m04q20) [#permalink]
14 Oct 2011, 05:29
i also got the ans..which will be 4*7*5*7 which will make the unit digit 0
|
|
|
|
|
|
Manager
Joined: 14 Aug 2009
Posts: 56
Followers: 1
Kudos [?]:
12
[0], given: 29
|
Re: Exponents Problem (m04q20) [#permalink]
14 Oct 2011, 12:44
This was easier than I thought. I did no math. This really wasn't an exponent question but a number properties question: Since 2 is a factor the answer cannot be odd so B and D are out Since 5 is a factor and all multiples of 5 are end in 5 or 0 then the answer is E. 0
_________________
My will shall shape the future. Whether I fail or succeed shall be no man's doing but my own. I am the force; I can clear any obstacle before me or I can be lost in the maze. My choice; my responsibility; win or lose, only I hold the key to my destiny - Elaine Maxwell
|
|
|
|
|
|
Manager
Joined: 01 Nov 2010
Posts: 204
Location: India
Concentration: Technology, Marketing
GMAT Date: 08-27-2012
GPA: 3.8
WE: Marketing (Manufacturing)
Followers: 5
Kudos [?]:
10
[0], given: 26
|
Re: Exponents Problem (m04q20) [#permalink]
15 Oct 2011, 04:14
zero.. any 5 when multiplied by 2 ll result a zero as its unit digit.
_________________
kudos me if you like my post.
Attitude determine everything.
all the best and God bless you.
|
|
|
|
|
|
Intern
Joined: 09 Mar 2010
Posts: 33
Location: India
Schools: NYU, Yale, Duke
Followers: 0
Kudos [?]:
1
[0], given: 16
|
Re: Exponents Problem (m04q20) [#permalink]
17 Oct 2011, 03:40
trahul4 wrote: What is the last digit of the following number 2^{22} * 3^{15} * 5^{16} * 7^1 ? (A) 6 (B) 5 (C) 2 (D) 1 (E) 0 Source: GMAT Club Tests - hardest GMAT questions Can someone please explain how to solve these kind of problems Thanks
for 2 : 2*1 = 2 , 2*2 = 4 , 2^3 = 8 , 2^4 = 16
hence it follows : 2,4,8,6 -- 4 is the count
similarly for 3 : 3,9,7,1 -- 4
for 5 : 5 -- count is 1
for 7 : 7,9,3,1 -- count is 4
Hence
2^22 = 2^(22/4) = 2^2 = 4
3^15 = 3^(15/4) = 3^1 = 3
5^16 = 5^1 = 5
7^1 = 7
4*3*5*7 = 420 .. untis digit 0
|
|
|
|
|
|
Manager
Joined: 16 Sep 2010
Posts: 230
Location: United States
Concentration: Finance, Real Estate
GMAT 1: 740 Q48 V42
Followers: 5
Kudos [?]:
42
[0], given: 2
|
Re: Exponents Problem (m04q20) [#permalink]
17 Nov 2011, 23:01
2- 2,4,8,6
3 - 3,9,7,1
5- always 5
So we know that 2^22 will end in a 4 (goes through five rotations and then two more places), 3 will end on a 7 (3 rotations then 3 more places), 5 is 5, and 7 is 7.
So 4x3x7x5x7 = 0
Answer E.
|
|
|
|
|
|
Intern
Joined: 04 Oct 2012
Posts: 12
Concentration: General Management, Human Resources
GMAT Date: 02-02-2013
GPA: 3.55
Followers: 0
Kudos [?]:
4
[1] , given: 7
|
Re: Exponents Problem (m04q20) [#permalink]
16 Oct 2012, 07:01
1
This post received
KUDOS
2*7*5*7 --- i like this thought process i would have never thought of it this way! i would have seen the 2 and 5 and thought oh it's 0 but now when i'm faced with a question like this and there is no 2 or 5 i know what to do!!!!
|
|
|
|
|
|
|
Re: Exponents Problem (m04q20)
[#permalink]
16 Oct 2012, 07:01
|
|
|
|
|
|
|
|
|
|
|
close
