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ash_maverick
- Senior | Next Rank: 100 Posts
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Last digit in the expansion of 43^43=7
Last digit in the expansion of 33^33=3
Last digit of 43^43+33^33=0
Hence remainder=0
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exponent problem
Source: Beat The GMAT — Problem Solving |
i'm curious if it can be solved like below.
The problem is to ask
43^43 + 33^33 mod 10
Modular Exponentiation
X^a (mod n).
43^43 mod 10
First,
43^1 mod 10 = 3
43^2 mod 10 = 3^2 mod 10 = 9
43^4 mod 10 = 9^2 mod 10 = 1
43^8 mod 10 = 1^2 mod 10 = 1
....
43^32 mod 10 = 1 mod 10 = 1
43 = 32 + 8 + 2 + 1
43^43 mod 10
= 43^32 mod 10 * 43^8 mod 10 * 43^2 mod 10 + 43^1 mod 10
= (1 * 1 * 9 * 3 ) mod 10
= 27 mod 10
= 7
Same procedure is done on 33
33^1 mod 10 = 3
33^2 mod 10 = 3^2 mod 10 = 9
33^4 mod 10 = 9^2 mod 10 = 1
.....
33^32 mod 10 = 1 mod 10 = 1
33 = 32 + 1
33^33 mod 10
=33^32 mod 10 * 33^1 mod 10
= (1 * 3 ) mod 10
= 3 mod 10
= 3
Therefore, their sum must be 3 + 7 which gives the unit digit 0.
The remainder is obviously 0.
The problem is to ask
43^43 + 33^33 mod 10
Modular Exponentiation
X^a (mod n).
43^43 mod 10
First,
43^1 mod 10 = 3
43^2 mod 10 = 3^2 mod 10 = 9
43^4 mod 10 = 9^2 mod 10 = 1
43^8 mod 10 = 1^2 mod 10 = 1
....
43^32 mod 10 = 1 mod 10 = 1
43 = 32 + 8 + 2 + 1
43^43 mod 10
= 43^32 mod 10 * 43^8 mod 10 * 43^2 mod 10 + 43^1 mod 10
= (1 * 1 * 9 * 3 ) mod 10
= 27 mod 10
= 7
Same procedure is done on 33
33^1 mod 10 = 3
33^2 mod 10 = 3^2 mod 10 = 9
33^4 mod 10 = 9^2 mod 10 = 1
.....
33^32 mod 10 = 1 mod 10 = 1
33 = 32 + 1
33^33 mod 10
=33^32 mod 10 * 33^1 mod 10
= (1 * 3 ) mod 10
= 3 mod 10
= 3
Therefore, their sum must be 3 + 7 which gives the unit digit 0.
The remainder is obviously 0.
how did you get the last digit? 8)chandra_adesh wrote:Last digit in the expansion of 43^43=7
Last digit in the expansion of 33^33=3
Last digit of 43^43+33^33=0
Hence remainder=0
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Stacey Koprince
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Whenever they ask you a "units digit" question with really high exponents, there will be a pattern and you will only have to follow the units digit through the problem (b/c it is all multiplication).
43^43:
3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.
43^43:
3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.
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Stacey Koprince
GMAT Instructor
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Manhattan GMAT
Contributor to Beat The GMAT!
Learn more about me
Stacey Koprince
GMAT Instructor
Director of Online Community
Manhattan GMAT
Contributor to Beat The GMAT!
Learn more about me
thats my suicide note 
none but this problem is responsible for it. :twisted: haha i know such big numbers are justbubbles. i dealt with units 5 6 1 they are of course easy. but this with 3 as a unit must belong to ''tough'' category.
Getting started @BTG?
Beginner's Guide to GMAT | Beating GMAT & beyond
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Beginner's Guide to GMAT | Beating GMAT & beyond
Please do not PM me, (not active anymore) contact Eric.
WOW, beautiful trick. I totally missed that one 8) ThanksStacey Koprince wrote:Whenever they ask you a "units digit" question with really high exponents, there will be a pattern and you will only have to follow the units digit through the problem (b/c it is all multiplication).
43^43:
3^1 = 3^2 = 3^3 = 23^4 = 83^5 = etc. so the units digit pattern is 3-9-7-1. The pattern repeats every 4th term. So 3^4, 3^8, 3^12, etc, will all have the units digit 1. 3^40 will be 1, 3^41 will be 3, 3^42 will be 9, 3^43 will be 7. Same pattern as above. 3^32 will be 1, 3^33 will be 3. units digit 7 + units digit 3 will equal units digit 0. Anything that ends in 0 will have a remainder of 0 when divided by 10.
