Aptitude - Problems on H.C.F and L.C.M - Discussion

I think it is 23*14 the largest one.

23*13 =299.

23*14=322 so this is the large.

23*14=322 largest.

We can represent any integer number in the form of: pq+r.
Where p is dividend, q is quotient, r is reminder.
So: A = pq1 + r;
B = pq2 + r;

HERE P=23 (HCF GIVEN).
LCM FACTOR GIVEN ,SO q1 =13,q2 =14 , REMAINDER (r=0).
So, A = 23*13 = 299 , B= 23*14= 322.
ANSWER = 322.

Can anyone please make it clear with the basic concept, please?

Here, 'other two factors of their lcm should be taken into consideration carefully.

Suppose there are two numbers, 15 and 25.Its hcf is 5 and lcm is 75.now write its factors.. 3*5*5, among these factors one of three is hcf, so when we multiply by its hcf with remaining other two factors i.e 5*3 and 5*5 gives original number.

Yes, you are correct, Thanks @Dheeraj.

I am not understanding it. Please help me to get it.

Let's understand what lcm and HCF are;

HCF of 2 no is the greatest number can divide both;
so 36 = 2x18,
46=2x23.

Here HCF is 2 and remainings are related in the relation of coprime only as 18 and 23 are coprime factors of these both numbers.

Therefore;

number A = HCF x coprime a.
number B= HCFx coprime b(Formula 1).

Now move to question
we know that number will be HCF x co-prime number, therefore, numbers will be A=23 x a
B=23 x b(where a & b are co-primes)

Now look at the formula;

HCF x LCM= A x B(here A and B are the numbers).

We can write A= HCF x coprime a.
&
B=HCF x coprime b(by formula 1).

Put this in formula
HCF x LCM =(HCF x A)(HCF x B).

Now if we take hcf in right hand side.
LCM =HCFx A x HCFx B/HCF.
therefore LCM=HCF x ab (where a & b are co primes).

and we get formula here
LCM= HCF x coprimes ab (formula 2)

we have in question

LCM 's factor 13 &14 which are coprimes.
means lcm =13 x 14 x LCM (as formula 2 tells us).
therefore by starting formula;
HCF x a =A -> 23 x 13=A=299.

HCF x b=B -> 23 x 14=B=322.

Thanks @Chiya.