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Bunuel
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38!(39.40...-1)=38!(3z-1)
only need to find power of 3 in 38!

[38/3]+[38/9]+[38/27]=12+4+1=17
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Bunuel
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­What is the highest power of 3 available in the expression 58! - 38!?

A. 16
B. 17
C. 18
D. 19
E. 20
­58! - 38! =

= 38!(58 * 57 * 56 * ... * 39 - 1)

58 * 57 * 56 * ... * 39 is a multiple of 3, thus 58 * 57 * 56 * ... * 39 - 1 is 1 less than a multiple of 3, so not a multiple of 3 and thus will not contribute any 3's to the entire product.

The highest integer power of 3 in 38! is 38/3 + 38/9 + 38/27 = 12 + 4 + 1 = 17 (remember we take only the quotients of the divisions, that is 38/3 would give 12, not 12 2/3).

Answer: A­.

­
Hope it helps.­

Can anyone explain the following line in detail?

58 * 57 * 56 * ... * 39 is a multiple of 3, thus 58 * 57 * 56 * ... * 39 - 1 is 1 less than a multiple of 3, so not a multiple of 3 and thus will not contribute any 3's to the entire product.
­
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Zeus_
Bunuel
Bunuel
­What is the highest power of 3 available in the expression 58! - 38!?

A. 16
B. 17
C. 18
D. 19
E. 20
­58! - 38! =

= 38!(58 * 57 * 56 * ... * 39 - 1)

58 * 57 * 56 * ... * 39 is a multiple of 3, thus 58 * 57 * 56 * ... * 39 - 1 is 1 less than a multiple of 3, so not a multiple of 3 and thus will not contribute any 3's to the entire product.

The highest integer power of 3 in 38! is 38/3 + 38/9 + 38/27 = 12 + 4 + 1 = 17 (remember we take only the quotients of the divisions, that is 38/3 would give 12, not 12 2/3).

Answer: A­.

­
Hope it helps.­

Can anyone explain the following line in detail?

58 * 57 * 56 * ... * 39 is a multiple of 3, thus 58 * 57 * 56 * ... * 39 - 1 is 1 less than a multiple of 3, so not a multiple of 3 and thus will not contribute any 3's to the entire product.
­
­
3n, where n is an integer, is a multiple of 3 and thus always divisible by 3.

3n - 1 is one less than a multiple of 3, so NOT a multiple of 3, and thus is not divisible by 3.­
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