Question

what is the power of 6 in 60!

Answer 1

\[\Large\left\lfloor\dfrac{60}6\right\rfloor+\left\lfloor\dfrac{60}{36}\right\rfloor\]

Answer 2

Can you explain...
why you wrote like that.....

Answer 3

Wait sorry it's not correct

Answer 4

\[\left\lfloor\frac{60}2\right\rfloor+\left\lfloor\frac{60}4\right\rfloor+\left\lfloor\frac{60}8\right\rfloor+\left\lfloor\frac{60}{16}\right\rfloor+\left\lfloor\frac{60}{32}\right\rfloor+\left\lfloor\frac{60}3\right\rfloor+\left\lfloor\frac{60}9\right\rfloor+\left\lfloor\frac{60}{27}\right\rfloor\]

Answer 5

check oly powers of 3, thats enough kc

Answer 6

oh sorry i did it wrong again, @ganeshie8 please do it, i've lost my qualification.

Answer 7

lol no, just dont wry about powers of 2, as there will be plenty of them compared to powers of 3

Answer 8

i know how to do it now, but i've lost my qualification, so lol

Answer 9

\[\left\lfloor\frac{60}3\right\rfloor+\left\lfloor\frac{60}9\right\rfloor+\left\lfloor\frac{60}{27}\right\rfloor\]
If the number is divisible by 3, there is one power of 3. If it's further divisible by 9, it has one more power of 3. If it's still divisible by 27, it'll have even one more power of 3.

Answer 10

Explanation???

Answer 11

Recall that \(n!=n\cdot(n-1)\cdot(n-2)\dots1\) and \(6=3\cdot2\)

the power of \(6\) can then be counted by the number of times pairs of \(2,3\) appear in \(60!\)

Answer 12

since there are more multiples of \(2\) than \(3\), we know the number of \(3\)s is our "limiting" factor here

Answer 13

to count the number of \(3\)s in \(60!\), recognize that every multiple of \(3\) up to \(60\) appears in \(60!=60\cdot59\cdot58\dots3\cdot2\cdot1\). to count the number of such multiples we find:$$60/3=20$$

Answer 14

now recognize that multiples of \(3^2=9\), however, contribute not just one \(3\) as the above calculation suggests but in fact also provide an *additional* \(3\), hence we must also count the number of multiples of \(3^2=9\) up to \(60\)
$$60/3^2=6.666\dots$$the whole number of times \(3^2\) goes into \(60\) is just the floor of this number, i.e. \(\lfloor60/3^2\rfloor=6\)

Answer 15

in the same sense, multiples of \(3^3, 3^4, \dots\) contribute additional \(3\)s as well; we must count the number of mulitples of \(3^3,3^4,\dots\) in the same way. we know when to stop when there are *no* non-zero multiples of \(3^n\) below \(60\), which corresponds to \(3^4=81>60\).

we only need to count once more, then: \(\lfloor 60/3^3\rfloor =2\)

Answer 16

so in total we have \(\lfloor60/3\rfloor+\lfloor60/3^2\rfloor+\lfloor60/3^3\rfloor=20+6+2=28\)

Answer 17

compare this to the prime factorization of \(60!\):
http://www.wolframalpha.com/input/?i=prime+facotrization+of+60%21

$$2^{56}×3^{28}×5^{14}×7^9×11^5×13^4×17^3×19^3×23^2×29^2×31×37×41×43×47×53×59$$

notice we have \(3^{28}\) and a surplus of \(2\)s to there are \(28\) factors of \(6\) in our number

this same trick allows you to count the number of digits at the end of \(60!\) only instead you'd count the number of multiples of \(5\) (since \(10=5\cdot2\))