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What is the remainder when A*B*C+D+E*F*G is divided by 10

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What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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New post 14 Oct 2016, 06:21
3
00:00
A
B
C
D
E

Difficulty:

  75% (hard)

Question Stats:

60% (02:55) correct 40% (03:19) wrong based on 60 sessions

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If =>
A= (128)^55
B=99^23
C=112^5987
D=1287^2235
E=53^129
F=29^83
G=58^111

What is the remainder when A*B*C+D+E*F*G is divided by 10


A) one
b) two
c) three
d) four
e) five

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What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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New post 14 Oct 2016, 09:57
stonecold wrote:
If =>
A= (128)^55
B=99^23
C=112^5987
D=1287^2235
E=53^129
F=29^83
G=58^111

What is the remainder when A*B*C+D+E*F*G is divided by 10


A) one
b) two
c) three
d) four
e) five


Unit digits of
A=2
B=1
C=8
D=3
E=3
F=1
G=2
unit digit of A*B*C+D+E*F*G=xxxxxx25/10=xxxxxx5/2
last digit xxxxx5 /2 then remainder is thus 1

Ans A
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Re: What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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New post 14 Oct 2016, 10:51
stonecold wrote:
If =>
A= (128)^55
B=99^23
C=112^5987
D=1287^2235
E=53^129
F=29^83
G=58^111

What is the remainder when (A*B*C)+D+(E*F*G) is divided by 10


A) one
b) two
c) three
d) four
e) five


A number is divisible by 10 only if the last digit is 0....

Further the remainder of a number divided by 10 will be the units digit ( If the last digit is not 0 ), check this property using any number....


Find the units digit of the numbers -

\(A= (128)^{55}\) will have units digit as 2 ; since cyclicity of 8 is 4

\(B=99^{23}\) will have units digit as 1 ; since cyclicity of 9 is 2

\(C=112^{5987}\) will have units digit as 8 ; since cyclicity of 2 is 4

\(D=1287^{2235}\) will have units digit as 3 ; since cyclicity of 7 is 4

\(E=53^{129}\) will have units digit as 3 ; since cyclicity of 3 is 4

\(F=29^{83}\) will have units digit as 9 ; since cyclicity of 9 is 2

\(G=58^{111}\) will have units digit as 2 ; since cyclicity of 8 is 4


Now, (A*B*C) will have units digit as ( 2 *1 *8 ) = 6
And , (E*F*G) will have units digit as ( 3*9*2) = 4

So, (A*B*C)+D+(E*F*G) = 6 + 3 + 4 => 13

So, IMHO remainder will be 3...


PS: Good to see you Mr stonecold , please post the OA ....
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Re: What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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New post 14 Oct 2016, 11:00
1
Abhishek009 wrote:
stonecold wrote:
If =>
A= (128)^55
B=99^23
C=112^5987
D=1287^2235
E=53^129
F=29^83
G=58^111

What is the remainder when (A*B*C)+D+(E*F*G) is divided by 10


A) one
b) two
c) three
d) four
e) five


A number is divisible by 10 only if the last digit is 0....

Further the remainder of a number divided by 10 will be the units digit ( If the last digit is not 0 ), check this property using any number....


Find the units digit of the numbers -

\(A= (128)^{55}\) will have units digit as 2 ; since cyclicity of 8 is 4

\(B=99^{23}\) will have units digit as 1 ; since cyclicity of 9 is 2

\(C=112^{5987}\) will have units digit as 8 ; since cyclicity of 2 is 4

\(D=1287^{2235}\) will have units digit as 3 ; since cyclicity of 7 is 4

\(E=53^{129}\) will have units digit as 3 ; since cyclicity of 3 is 4

\(F=29^{83}\) will have units digit as 9 ; since cyclicity of 9 is 2

\(G=58^{111}\) will have units digit as 2 ; since cyclicity of 8 is 4


Now, (A*B*C) will have units digit as ( 2 *1 *8 ) = 6
And , (E*F*G) will have units digit as ( 3*9*2) = 4

So, (A*B*C)+D+(E*F*G) = 6 + 3 + 4 => 13

So, IMHO remainder will be 3...


PS: Good to see you Mr stonecold , please post the OA ....


Hi Abhishek

if i m not wrong please reconsider the highlighted part

thanks
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What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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New post 14 Oct 2016, 11:10
rohit8865 wrote:
if i m not wrong please reconsider the highlighted part

thanks


Thanx bro for highlighting the error... :oops:

9^1 = 9
9^2 = 1
9^3 = 9

So, every odd power of 9 will have units digit as 9 and every even power of 9 will have even power as 1

So, \(B = 99^{23}\) will have units digit as 9
And, \(F =29^{83}\) will have units digit as 9

Now, Now, (A*B*C) will have units digit as ( 2 *9 *8 ) = 4
And , (E*F*G) will have units digit as ( 3*9*2) = 4


(A*B*C) + D + (E*F*G) = ( 4 + 3 + 4 ) = 11

So, 11/10 will leave a remainder 1..

Answer will be (A) 1

PS: Thanks for pointing out my error, I got it , I was excited seeing stonecold's post after long time... :oops: :shock:
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Re: What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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New post 15 Oct 2016, 01:02
1
stonecold wrote:
If =>
A= (128)^55
B=99^23
C=112^5987
D=1287^2235
E=53^129
F=29^83
G=58^111

What is the remainder when A*B*C+D+E*F*G is divided by 10


A) one
b) two
c) three
d) four
e) five


Nice Question.

Good to see you back stonecold :)
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Re: What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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New post 08 Jan 2019, 05:37
stonecold wrote:
If =>
A= (128)^55
B=99^23
C=112^5987
D=1287^2235
E=53^129
F=29^83
G=58^111

What is the remainder when A*B*C+D+E*F*G is divided by 10


A) one
b) two
c) three
d) four
e) five


Cyclicity helps you solve this problem. Check these posts for more details:

https://www.veritasprep.com/blog/2015/1 ... -the-gmat/
https://www.veritasprep.com/blog/2015/1 ... at-part-2/
https://www.veritasprep.com/blog/2015/1 ... questions/
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Re: What is the remainder when A*B*C+D+E*F*G is divided by 10 &nbs [#permalink] 08 Jan 2019, 05:37
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