If k is an non-negative integer and

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BTGmoderatorRO: Moderator; Posts: 772; Joined: Wed Aug 30, 2017 6:29 pm; Followed by:6 members

If k is an non-negative integer and

by BTGmoderatorRO » Sat Feb 17, 2018 12:16 pm

If k is an non-negative integer and $$^{15^k}$$ is a divisor of 759,325 then $$^{3^k}$$ âˆ’ $$^{k^3}$$ =

A. 0
B. 1
C. 37
D. 118
E. 513

OA is B
How do I approach this? very urgent, please. An Expert should come to my aid.

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Source: Beat The GMAT — Problem Solving | Sat Feb 17, 2018 12:16 pm

deloitte247: Legendary Member; Posts: 2214; Joined: Fri Mar 02, 2018 2:22 pm; Followed by:5 members

by deloitte247 » Sun Apr 22, 2018 9:37 am

759, 325 is not divisible by 3, hence not divisible by 15 either.
The only way $$15^k$$ where k is a non- negative integer and will divide
759, 325, is if k= 0
substituting 0 for k
$$3^k-k^3=3^0-0^3$$
$$=1-0=1$$
hence the correct answer is B.

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Scott@TargetTestPrep: GMAT Instructor; Posts: 8086; Joined: Sat Apr 25, 2015 10:56 am; Location: Los Angeles, CA; Thanked: 43 times; Followed by:29 members

by Scott@TargetTestPrep » Tue Apr 24, 2018 10:47 am

BTGmoderatorRO wrote:If k is an non-negative integer and $$^{15^k}$$ is a divisor of 759,325 then $$^{3^k}$$ âˆ’ $$^{k^3}$$ =

A. 0
B. 1
C. 37
D. 118
E. 513

Since 15 = 3 x 5, we see that both 3 and 5 must evenly divide into 759,325. Since 759,325 ends in "5," we know it is evenly divisible by 5.

Recall the shortcut to determine if a number is evenly divisible by 3: Add the digits of the number; if that sum is evenly divisible by 3, then the number itself is divisible by 3. For 759,325,the sum of the digits is 7 + 5 + 9 + 3 + 2 + 5 = 29; we see that 3 can't evenly divide into 759,325. Thus, k can't be a positive integer, so k = 0.

Thus, 3^0 - 0^3 = 1 - 0 = 1.

Answer: B

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