Latest TCS Aptitude Question SOLUTION: what is the last non-zero digit of 96!?

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what is the last non-zero digit of 96!?

Asked In TCS garima kadam (11 years ago)

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ashish  (11 years ago)

Rightmost non-zero digit of n!
R(n!)=last digit of [2^a. R(a!).R(b!)]

where n=5a+b
so 96=5*19+1
i.e. a=19 and b=1

last digit of [2^19. R(19!).R(1!)]-------(1)
now again break 19 in the form of 5a+b
19=5*3+4
last digit of [2^3. R(3!).R(4!)]
i.e. 2
and put the value in (1)
ans : 6

Is this solution Helpfull? Yes (22) | No (4)  

SUMAN KUMAR  (11 years ago)

1*2*3*4*5*6*7*8*9*10
last non zero digit=8
11*12*13*.........*20
last non zero digit=8
21*.....*30=8(last non zero digit=8)
31*.....*40=8(last non zero digit=8)
41*.....*50=8(last non zero digit=8)
51*.....*60=8(last non zero digit=8)
61*.....*70=8(last non zero digit=8)
71*.....*80=8(last non zero digit=8)
81*.....*90=8(last non zero digit=8)
91*92*..*96=8(last non zero digit=2)
so=8^9*2
last digit=6

Is this solution Helpfull? Yes (17) | No (2)  

mukesh  (11 years ago)

there is a formula for getting last non-zero digit of n!
R(n!)=last digit of[2^a*R(a!)*R(b!)]
where n=5*a+b.
here n=96=5*19+1.
so, a=19&b=1
R(96!)=last digit of[2^19*R(19!)*R(1!)]
=last digit of[(2^4)^4*2^3*2*1]
=last digit of[6*8*2*1]
=last digit of[36]
=6.



last nonzero digit of 19!
19=5*3+4.
a=3,b=4
R(19!)=last digit of[2^3*R(3!)*R(4!)]
=8*6*4
=2.

Is this solution Helpfull? Yes (7) | No (1)  

saurabh agarwal  (11 years ago)

1*2*3*4*5*6*7*8*9*10
last non zero digit=8
11*12*13*.........*20
last non zero digit=8
21*.....*30=8(last non zero digit=8)
31*.....*40=8(last non zero digit=8)
41*.....*50=8(last non zero digit=8)
51*.....*60=8(last non zero digit=8)
61*.....*70=8(last non zero digit=8)
71*.....*80=8(last non zero digit=8)
81*.....*90=8(last non zero digit=8)
91*92*..*96=2(last non zero digit=2)
so 8^9*2=6
last digit=6

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Ajay Poshak  (11 years ago)

@SUMAN KUMAR
by having 91*92*93.....*96 = we have last non zero digit as 8, then how you considered it as 2.

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Arindam Mukherjee  (11 years ago)

last nonzero digit of 96!=last digit of[Right most digit of 2^19*right most digit of 19!* right most digit of 1!]
now right most nonzero digit of 19!=[right most digit of 2^3*right most digit of 3!*right most digit of 4!]=right most of[8*6*4]=2

So,last nonzero digit of 96! is last digit of 8*2*1=16..i.e is 6 is the answer

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Awesome  (11 years ago)

how u getting the values of A & B ?

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