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The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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25 Oct 2011, 01:29
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The LCM and HCF of two numbers are 2376 and 22 respectively. Find the larger of the two numbers if their sum is 682. A) 484 B) 562 C) 352 D) 576 E) 594
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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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24 Nov 2011, 11:28
cleetus wrote: The LCM and HCF o two numbers are 2376 and 22 respectively. Find the larger of the two numbers if their sum is 682. A) 484 B) 562 C) 352 D) 576 E) 594 Methode 2 Now lets solve this using traditional way using variables Since the HCF is 22, Both the 2 numbers should be divisible by the HCF 22. Let the two numbers by 22a and 22b where a and b do not have any common factor. 22ab = 2376 ab = 108 22(a+b) = 682 a+b = 31 Solving for ab=108 and a+b=31, we get a=4 and b=27 or a=27 and b = 4 The larger among the two is 27*22=594 Answer is E.
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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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25 Oct 2011, 02:50
a x b = gcf(a,b) x lcm (a,b) a x b = 2376 x 22 = 52,272 (last number ends in two)
Eliminate b,c,d as 1s digit won't work. Test a and e.
E works.
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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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25 Oct 2011, 03:00
1+ for E, If A and B are two numbers whose LCM and GCF are 2376 and 22 respectively then AB = 2376*22 and A+B is 682 as given. With this much info we can easily calculate A and B using brute force method and the answer is E.



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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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25 Oct 2011, 17:34
pike wrote: a x b = 2376 x 22 = 52,272 (last number ends in two)
Eliminate b,c,d as 1s digit won't work.
Care to elaborate on this part?



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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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25 Oct 2011, 20:53
shanghaizzle wrote: pike wrote: a x b = 2376 x 22 = 52,272 (last number ends in two)
Eliminate b,c,d as 1s digit won't work.
Care to elaborate on this part? When I saw this question, I wrote down the following: a x b = 52272 a + b = 682 This is a clear constraint and its pretty simple to brute force the answer choices, but there is a further constraint, if we just focus on the ones digits. If you think about multiplication we have: Or just focusing on the ones digits: j x k = 2 j+k = 2 There is only one set of numbers, 8 and 4, which will fit that sub criteria. 8*4 = 2, 8+4 = 12. We can see it like this: Hence we know that one of the numbers must end in either 8 or 4, and can eliminate b,c,d. This just jumped into my mind straight away, I didn't spend too much time validating it as I knew brute force would work and I would know with certainty once I found an awer. So it was more a matter of giving me a head start in testing A and E. If it hadn't I would have just kept plugging.



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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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24 Nov 2011, 11:23
cleetus wrote: The LCM and HCF o two numbers are 2376 and 22 respectively. Find the larger of the two numbers if their sum is 682. A) 484 B) 562 C) 352 D) 576 E) 594 There are 2 approaches in solving this. Methode 1.HCF * LCM = The Actual Number. 2376 * 22 = 52272 So the answer which we are looking for has to be a factor of 52272. So among the options shortlist the answers by eliminating those numbers which is not divisible by 52272. and then take the highest number as the answer as the question asks abt the highest number. Here A and E are divisible by 52272 and since E is the highest, its the answer.
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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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24 Nov 2011, 11:45
cleetus wrote: The LCM and HCF o two numbers are 2376 and 22 respectively. Find the larger of the two numbers if their sum is 682. A) 484 B) 562 C) 352 D) 576 E) 594 This is how i solved it. LCM: contains all factors excluding duplicates. HCF: common factors So factors of LCM (2376) are 2,2,2,3,3,3,11. These are the prime factors available in the two numbers. But 22 is the HCF, so the remaining two numbers will have 2,2 and 3,3,3 as the remaining factors of the two numbers as the primes should be unique to each other. Else they would have been counted in the HCF. So we have the two numbers as 22*2*2 = 88 and 22*3*3*3 = 594. The process is pretty quick though Answer is E.



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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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24 Nov 2011, 13:41
Thats pretty easy. Thanks for the explanation +1 to u
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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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18 Dec 2011, 06:21
HI,
I did as follows:
a and bo both have to be multiples of 22, then:
682  22 = 660  not in answer choice 660  22 = 638  not in answer choice 638  22 = 616  not in answer choice 616  22 = 594  YES in answer choice
Answer E



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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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21 Apr 2016, 10:24
Sudhanshuacharya wrote: cleetus wrote: The LCM and HCF o two numbers are 2376 and 22 respectively. Find the larger of the two numbers if their sum is 682. A) 484 B) 562 C) 352 D) 576 E) 594 This is how i solved it. LCM: contains all factors excluding duplicates. HCF: common factors So factors of LCM (2376) are 2,2,2,3,3,3,11. These are the prime factors available in the two numbers. But 22 is the HCF, so the remaining two numbers will have 2,2 and 3,3,3 as the remaining factors of the two numbers as the primes should be unique to each other. Else they would have been counted in the HCF. So we have the two numbers as 22*2*2 = 88 and 22*3*3*3 = 594. The process is pretty quick though Answer is E. Hi, How did you determine that the remaining numbers i.e 2^2 and 3^3 are unique to each number? Could it not be 22*2*3 and 22*2*3*3 or some other combination? Please help.



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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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09 Aug 2017, 10:03
OreoShake wrote: Sudhanshuacharya wrote: cleetus wrote: The LCM and HCF o two numbers are 2376 and 22 respectively. Find the larger of the two numbers if their sum is 682. A) 484 B) 562 C) 352 D) 576 E) 594 This is how i solved it. LCM: contains all factors excluding duplicates. HCF: common factors So factors of LCM (2376) are 2,2,2,3,3,3,11. These are the prime factors available in the two numbers. But 22 is the HCF, so the remaining two numbers will have 2,2 and 3,3,3 as the remaining factors of the two numbers as the primes should be unique to each other. Else they would have been counted in the HCF. So we have the two numbers as 22*2*2 = 88 and 22*3*3*3 = 594. The process is pretty quick though Answer is E. Hi, How did you determine that the remaining numbers i.e 2^2 and 3^3 are unique to each number? Could it not be 22*2*3 and 22*2*3*3 or some other combination? Please help. Because HCF is 22. Otherwise HCF would have been larger.



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Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the [#permalink]
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09 Aug 2017, 13:00
cleetus wrote: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the larger of the two numbers if their sum is 682.
A) 484 B) 562 C) 352 D) 576 E) 594 if 682/22=31, then larger number must be a multiple of 22≥16*22 that divides into 2376 only 594 fits E




Re: The LCM and HCF of two numbers are 2376 and 22 respectively. Find the
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