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

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Joined: 12 Aug 2015
Posts: 2610
Schools: Boston U '20 (M)
GRE 1: Q169 V154
What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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14 Oct 2016, 07:21
4
00:00

Difficulty:

75% (hard)

Question Stats:

58% (03:20) correct 42% (03:32) wrong based on 70 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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Director
Joined: 05 Mar 2015
Posts: 992
What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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14 Oct 2016, 10:57
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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14 Oct 2016, 11:51
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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Thanks and Regards

Abhishek....

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Joined: 05 Mar 2015
Posts: 992
Re: What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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14 Oct 2016, 12:00
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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14 Oct 2016, 12:10
rohit8865 wrote:
if i m not wrong please reconsider the highlighted part

thanks

Thanx bro for highlighting the error...

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..

PS: Thanks for pointing out my error, I got it , I was excited seeing stonecold's post after long time...
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Thanks and Regards

Abhishek....

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Posts: 3625
Re: What is the remainder when A*B*C+D+E*F*G is divided by 10  [#permalink]

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15 Oct 2016, 02:02
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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08 Jan 2019, 06:37
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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Veritas Prep GMAT Instructor

Re: What is the remainder when A*B*C+D+E*F*G is divided by 10   [#permalink] 08 Jan 2019, 06:37
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