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

Question

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

Sudhanshuacharya · Accepted Answer

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.

cleetus · Answer

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.

pike · Answer

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.

Posted from my mobile device

arjunbt · Answer

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.

shanghaizzle · Answer

Care to elaborate on this part?

pike · Answer

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:
 https://i.imgur.com/iJCV6.png

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:
 https://i.imgur.com/W5cYD.png

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.

cleetus · Answer

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.

cleetus · Answer

Thats pretty easy. Thanks for the explanation
+1 to u

Saurajm · Answer

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

OreoShake · Answer

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.

vs224 · Answer

Because HCF is 22. Otherwise HCF would have been larger.

gracie · Answer

if 682/22=31,
then larger number must be a multiple of 22≥16*22 that divides into 2376
only 594 fits
E

bumpbot · Answer

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

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

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