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14 Oct 2016, 07:21
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
49% (03:03) correct
51%
(03:15)
wrong
based on 174
sessions
History
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Result
Not Attempted Yet
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= (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
14 Oct 2016, 10:57
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stonecold
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
14 Oct 2016, 11:51
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stonecold
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 ....
