Exponents-

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[email protected]: Master | Next Rank: 500 Posts; Posts: 429; Joined: Wed Sep 19, 2012 11:38 pmFollowed by: 4 members; Thanked: 6 times

by [email protected] » Tue Oct 29, 2013 9:04 pm

If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 12

Ans E

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Source: Beat The GMAT — Problem Solving | Tue Oct 29, 2013 9:04 pm

theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Tue Oct 29, 2013 9:23 pm

(2^3)(3^4)(5^7) = b*(a^3)

b = (2^3)(3^4)(5^7)/a^3

Value of A = 1, 2, 3, 5, 25(5x5), 6(2x3), 10(2x5), 50(2x5x5), 15(3x5), 45(3x5x5), 30(2x3x5), 150(2x3x5x5)

Correspondingly the value of "b" will change.

So, 12

Answer [spoiler]{E}[/spoiler]

R A H U L

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Tue Oct 29, 2013 9:58 pm

theCodeToGMAT wrote:(2^3)(3^4)(5^7) = b*(a^3)

b = (2^3)(3^4)(5^7)/a^3

Value of A = 1, 2, 3, 5, 25(5x5), 6(2x3), 10(2x5), 50(2x5x5), 15(3x5), 45(3x5x5), 30(2x3x5), 150(2x3x5x5)

Correspondingly the value of "b" will change.

So, 12

Answer [spoiler]{E}[/spoiler]

Rahul,
Can you post any other method ? (Since we may miss any one of the possibility during exam time)
or Instead writing down all possible options is there a way to find 12 possibilities from the numbers 2,3,5.

Thanks in advance

Regards,
Uva.

Known is a drop Unknown is an Ocean

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Tue Oct 29, 2013 10:13 pm

Hi All,
If the above question has been restated with a as Prime number then can we use the below method ?
(2^3)(3^4)(5^7) = b*(a^3)

In LHS you can see there are (3+1)*(4+1)*(7+1) = 4*5*8 factors

In RHS you can see (3+1)*b =4*b factors

So b can take 40 factors

so b can take 40 different possible values

Please help me whether I am wrong or right.

Thanks in advance.

Regards,
Uva.

Known is a drop Unknown is an Ocean

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theCodeToGMAT: Legendary Member; Posts: 1556; Joined: Tue Aug 14, 2012 11:18 pmFollowed by: 34 members; Thanked: 448 times; GMAT Score:650

by theCodeToGMAT » Tue Oct 29, 2013 10:22 pm

~~mistakenly posted~~

R A H U L

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Uva@90: Master | Next Rank: 500 Posts; Posts: 490; Joined: Thu Jul 04, 2013 7:30 amFollowed by: 5 members; Location: Chennai, India; Thanked: 83 times

by Uva@90 » Tue Oct 29, 2013 10:33 pm

theCodeToGMAT wrote:
Uva@90 wrote:Hi All,
If the above question has been restated with a as Prime number then can we use the below method ?
(2^3)(3^4)(5^7) = b*(a^3)

In LHS you can see there are (3+1)*(4+1)*(7+1) = 4*5*8 factors

In RHS you can see (3+1)*b =4*b factors

So b can take 40 factors

so b can take 40 different possible values

Please help me whether I am wrong or right.

Thanks in advance.

Regards,
Uva.
See UVa, you cannot assume "a" & "b" as Prime Number..

Rahul
I am asking , if in question if they had mentioned Explicitly as Prime number then is my approach correct ?

Known is a drop Unknown is an Ocean

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ganeshrkamath: Master | Next Rank: 500 Posts; Posts: 283; Joined: Sun Jun 23, 2013 11:56 pmFollowed by: 26 members; Location: Bangalore, India; Thanked: 97 times; GMAT Score:750

by ganeshrkamath » Wed Oct 30, 2013 3:03 am

[email protected] wrote:If a and b are positive integers, and (2^3)(3^4)(5^7) = (a^3)*b, how many different possible values of b are there?
(A) 2
(B) 3
(C) 4
(D) 6
(E) 12

Ans E

(2^3)(3^4)(5^7) = (a^3)*b

Here, since all the factors have powers greater than or equal to 3, 'a' can be any one number or a combination of the 3.

Total possibilities:
None of the given factors (1^3): 1
One at a time: 3
Two at a time: 3C2 + 1 = 3 + 1 = 4
Three at a time: 3C3 + 2C1 = 1 + 3 = 3
Four at a time: 1

Total = 1 + 3 + 4 + 3 + 1 = 12

Choose E

Cheers

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GMAT Score: 750 V40 Q51 AWA 5 IR 8
https://www.beatthegmat.com/first-attemp ... tml#688494

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Matt@VeritasPrep: GMAT Instructor; Posts: 2630; Joined: Wed Sep 12, 2012 3:32 pmFollowed by: 119 members; Location: East Bay all the way; Thanked: 625 times; GMAT Score:780

by Matt@VeritasPrep » Wed Oct 30, 2013 10:33 pm

Another way of doing this is just grouping the terms into threes.

(2 * 2 * 2) (3 * 3 * 3) (5 * 5 * 5) (5 * 5 * 5) * 3 * 5

We have four different "cubes" here, so we could choose 0 to 4 of them - but be careful, as two of them are identical! (If they weren't, the problem would have a nice shortcut.)

Now we just need to decide how many of these "cubes" we want in our a. Each of these are small enough that you're probably best just counting out each case. (For "2 of them", for instance, I would chose 2+3, 2+5, 3+5, or 5+5.)

0 of them: 1 way
1 of them: 3 ways
2 of them: 4 ways
3 of them: 3 ways
4 of them: 1 way

And we're done!

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