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# Let M = 10! and let N = M3. If a is the greatest integer val

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Let M = 10! and let N = M3. If a is the greatest integer val [#permalink]  31 Jul 2020, 10:08
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87% (01:43) correct 12% (02:14) wrong based on 8 sessions
Let $$M = 10!$$ and let $$N = M^3$$. If a is the greatest integer value of x such that $$2^x$$ is a factor of N, and b is the greatest integer value of y such that $$3^y$$ is a factor of N, then what is the value of a + b?

A. 18

B. 24

C. 33

D. 36

E. 54
[Reveal] Spoiler: OA

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Re: Let M = 10! and let N = M3. If a is the greatest integer val [#permalink]  01 Aug 2020, 22:55
M = 10! will have 2^8 and 3^4 along with other factors.
N = M^3, N will have 2^24 and 3^12 along with other factors.
Since a and b are greatest integer. a = 24 and b= 12.
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Re: Let M = 10! and let N = M3. If a is the greatest integer val [#permalink]  02 Aug 2020, 00:30
Didnt understand this
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Re: Let M = 10! and let N = M3. If a is the greatest integer val [#permalink]  02 Aug 2020, 06:40
Prime factorization of 10! will give us 2 to the power 8 and 3 to the power 4.

Hence, a cube of 10! will have 2 to the power of 24 and 3 to the power of 12.

so, a=24 and b=12.

a+b = 36.

Option D.
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Re: Let M = 10! and let N = M3. If a is the greatest integer val [#permalink]  03 Aug 2020, 01:49
M=10!=10*9*8*7*6*5*4*3*2*1

N=M^3= (10*9*8*7*6*5*4*3*2*1)^3 = 10^3 * 9^3 * 8^3 * 7^3 * 6^3 * 5^3 * 4^3 * 3^3 * 2^3 * 1^3

The greatest value of x such that 2^x is a factor of N is the number of times 2 appears as a factor in N
Therefore from N we get:

10^3=(2*5)^3= 2^3 * 5^3 -->3 two's
8^3= (2^3)^3 = 2^9 -->9 two's
6^3= (2*3)^3 = 2^3 * 3^3 -->3 two's
4^3= (2^2)^3 = 2^6 -->6 two's
2^3 -->3 two's

Total: 24 two's
a=24

We repeat the same process in order to look for b the max number that 3 appears as a factor in N

9^3= (3^2)^3= 3^6 -->6 three's
6^3= (2*3)^3 = 2^3 * 3^3 -->3 three's
3^3 -->3 three's

Total: 12 three's
b=12

a+b=24+12=36

Re: Let M = 10! and let N = M3. If a is the greatest integer val   [#permalink] 03 Aug 2020, 01:49
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