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HCF
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sushil hande
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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.'
GMAT assassins aren't born, they're made,
Rich
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.'
GMAT assassins aren't born, they're made,
Rich
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OptimusPrep
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You can write 384 as 128*3 or 2^7*3sushil hande wrote:HCF of 384 & a^5b^2 is 16ab.what is the correct relation between a and b?
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.
Any conditions given for what a and b are? If a and b are prime,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
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!jaetpu wrote:Any conditions given for what a and b are? If a and b are prime,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
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
