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

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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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.

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.

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.

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

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

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.

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

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