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HCF

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Source: — Problem Solving |

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by [email protected] » Sat Jul 16, 2016 4:16 pm
Hi sushil hande,

When posting GMAT questions, you should make sure to post the ENTIRE prompt (including the answer choices). In many cases, the answer choices provide a hint as to how you can go about solving the problem, but if we don't have those answers, then we're forced to take the 'math approach.'

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by OptimusPrep » Sun Jul 17, 2016 8:43 pm
sushil hande wrote:HCF of 384 & a^5b^2 is 16ab.what is the correct relation between a and b?
You can write 384 as 128*3 or 2^7*3
The HCF of 2^7*3 and a^5b^2 = 16ab would contain the highest powers of the common multiples.
You can try solving ahead with this.

Let me know if this helps.
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by gauri14 » Sun Dec 30, 2018 10:29 pm
I tried solving this question even with the little hint given but still don't know how to proceed.
If it helps the options given for this question are as follows -
1) a = 2b
2) a + b =3
3) a - b = 3
4) a + b = 5
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by jaetpu » Thu Jan 03, 2019 10:44 pm
gauri14 wrote:I tried solving this question even with the little hint given but still don't know how to proceed.
If it helps the options given for this question are as follows -
1) a = 2b
2) a + b =3
3) a - b = 3
4) a + b = 5
Any conditions given for what a and b are? If a and b are prime,

prime factorization of 384 = 2^7 * 3, 16ab is 2^4*a*b. Thus a and b must be 3 and 2 (in either order). a+b=5
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by gauri14 » Fri Jan 04, 2019 4:45 am
jaetpu wrote:
gauri14 wrote:I tried solving this question even with the little hint given but still don't know how to proceed.
If it helps the options given for this question are as follows -
1) a = 2b
2) a + b =3
3) a - b = 3
4) a + b = 5
Any conditions given for what a and b are? If a and b are prime,

prime factorization of 384 = 2^7 * 3, 16ab is 2^4*a*b. Thus a and b must be 3 and 2 (in either order). a+b=5
No special condition for a and b. The answer for this is 5 like you've shown. Thanks for the help!
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