For what values of n will n! end in zero, when multiplied out
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Answered by
1
the answer is
whenever n is greater than or equal to 5, the number will always end in zero.
The reason is here,
lets say we have a number n, such that,
n ≥ 5.
Then,
n = (n)(n-1)(n-2)(n-3).....5×4×3×2×1
so,
n = (n)(n-1)(n-2)(n-3).....(5×2)×12
n = (n)(n-1)(n-2)(n-3).....12×[10]
That's why, it works whenever n ≥ 5.
HOPE YOU GOT HELP FROM MY ANSWER.
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Farewell!!
whenever n is greater than or equal to 5, the number will always end in zero.
The reason is here,
lets say we have a number n, such that,
n ≥ 5.
Then,
n = (n)(n-1)(n-2)(n-3).....5×4×3×2×1
so,
n = (n)(n-1)(n-2)(n-3).....(5×2)×12
n = (n)(n-1)(n-2)(n-3).....12×[10]
That's why, it works whenever n ≥ 5.
HOPE YOU GOT HELP FROM MY ANSWER.
Mark my answer as brainliest if you got help.
Farewell!!
Answered by
0
5! = 120
For all values of n> 5, answer = 120*6, 120*6*7, 120*6*7*8 and so on.
So for all values of n>=5, n! will end in 0
For all values of n> 5, answer = 120*6, 120*6*7, 120*6*7*8 and so on.
So for all values of n>=5, n! will end in 0
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