Latest other Aptitude Question SOLUTION: (10!+111)/143 what is the remainder

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(#M40037500) OTHER Numerical Ability question Keep an EYE

(10!+111)/143 what is the remainder

Asked In other himanshu rai (9 years ago)

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Saraswathy  (9 years ago)

143 => 11*13

10! mod 11 + 111 mod 11 = R1
R[(p-1)!/p] = p-1 [Provided p is a prime no.]... So R[10!/11] = 10...
10 + 1 mos 11 => 11 mod 11 = 0
R1 = 0

10! mod 13 + 111 mod 13 = R2
6 + 7 mod 13
13 mod 13 = 0
R2 = 0

Rem = 11x+R1 = 13y+R2
Rem = 11x+0 = 13y+0
x = 13, y = 11
143 mod 143 = 0

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Hardip Bhalani  (9 years ago)

10!=10*9*8*7*6*5*4*3*2*1
Now,when (a+b) is devided by n then remainder will be (a/n)+(b/n)
& when (a*b) is devided by n then remainder will be (a/n)*(b/n)
So,111/143 will give remainder 111.
Now,10!=(3*5*10)*(1*2*4*6*7)*(8*9)
A * B * C
when A is devided by 143 will give remainder 7
when B is devided by 143 will give remainder 50
when C is devided by 143 will give remainder 72
A*B=350 when devided by 143 gives 64 as remainder.
64*C=64*72=16*4*9*8=(144)*(32) when devided by 143 gives remainder 1*32=32
So,finally 32+111=143 when devided by 143 gives remainer 0

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