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

There is the difference andthe cut remainder due to following reason ...
@Prashant

{We can represent any integer number in the form of: pq+r.
Where p is the dividend, q is the quotient, r is a reminder.

so:
43 = pq1 + r;
91 = pq2 + r;
183 = pq3 + r;

Take r same in the above three equations as given in the question.
p is the value that we want to find out. which should be greatest.

On solving three equations we get:

p(q2-q1)= (91-43)=48;
p(q3-q2)= (183-91)=92;
p(q3-q1)= (183-43)=140.

Here, you have to take the difference due to cancel the remainder and cancellation is applied
{
p(q2-q1)= (91-43)=48; (to get the p[divisor*which is common among all of them eq 1,2,3*** ]and two quotient is given.
p(q3-q2)= (183-91)=92;
p(q3-q1)= (183-43)=140; }.

Let neglect two quotient[q2-q1..etc]//// 'p' is that value which had to be common so you have to calculate given HCF of eqns.

48 = 4 * 12
92 = 4 * 23
140 = 4 * 35.
Here consider p is 4 [which is hcf &&common..which is 'p' remember] ..and 12 is quotient [q2-q1].

I hope you will understand.

Assume that the required answer is "D" and the remainder that it leaves when dividing 43, 91 and 183.

This means that,

43 = d*q1 + r,

91 = d*q2 + r,

183 = d*q3 + r,

Where q1, q2 and q3 are the respective quotients.

Subtracting the first equation from second, we get:

91 - 43 = (d*q2 + r) - (d*q1 + r).

Hence, 48 = d*(q2 - q1) = d*(an integer).

Similarly, subtracting second equation from first, we get:

183 - 91 = (d*q3 + r) - (d*q2 + r).

Hence, 92 = d*(q3 - q2) = d*(another integer).

Now, 48 = d*(an integer) and 92 = d*(another integer) mean that both 48 and 92 are divisible by d. (In other words, "d" is a common divisor of 48 and 92. )

Now common divisors of 48 and 92 are 1, 2 and 4 only. Out of which, highest is 4. (Alternatively, you can directly find the HCF (or GCD) of 48 and 92). Hence 4 is the answer.

Note:

There is no need to calculate the HCF of 3 numbers, namely (91 - 43), (183 - 91) and (183 - 43). It is sufficient to take any two of these three numbers. The HCF will still be 4.

Hi @Ajeeth, very simple for the H.C.F.

For the numbers, you have given (455, 522 & 533).

H.C.F=1, Because the factor of these numbers are 455 = 1 x 5 x 7 x 13,
522 = 1 x 2 x 261 &
533 = 1 x 533.

So except 1, no digit is common in each number. Hence the common digit 1 is the H.C.F of above numbers.
This is the common rule of finding H.C.F. That those digit/digits which will be common in all the numbers.

Eg. You have to find the H.C.F. of 24, 60 & 81. Now factors of each number will be 24 = 2 x 2 x 2 x 3, 60 = 2 x 2 x 3 x 5, & 81 = 3 x 3 x 3 x 3.

Here only the digit '3' is common in all, hence 3 will be H.C.F of these numbers. Also note that if more than 1 digit is common in all the numbers then multiplication of those digits will be the H.C.F (suppose 2 & 5 both are common in all the numbers then the multiplication of both the digit will be 10) hence 10 will be the H.C.F of those numbers.

Thanks.

@All.

Here you need the highest common factor of three numbers those leave the same remainder

Let's do with the formula.

(p)q1 + r = 43.
(p)q2 + r = 91.
(p)q3 + r = 183.
we have used the same 'r' because all three leave the same remainder.
we need to find the 'p' value, which is the highest common factor.
to find the 'p' value, we need to remove 'r' first, ---------> (important point).
it seems that we can do it by making a difference between two equations.

Let's do it.

=> (pq2 +r) - (pq1 +r)
=> pq2+r-pq1-r
=> pq2-pq1 ----> now the reminder is taken out.
=> p(q2-q1) -----> need to find p value ----> (91-43)----> 1

Other numbers after diff will be like;

=> p(q3-q2) --------->(183-91)---------------> 2.
=> p(q3-q1) -------->(183-43) ---------------> 3.

Now we can find HCF of these three numbers to find p-value. Because 'p' is common in all numbers.

Hope you got it.

1.The HCF of two or more numbers is the largest common factor of the given numbers. It is the smallest number which divides the two or more given numbers.

The LCM of two or more numbers is the smallest number that is a common multiple of two or more numbers. It is the smallest number which is divisible by two or more given numbers.

Consider two numbers 120 and 144.
Prime factorization of 120 = 2 x 2 x 2 x 3 x 5
Prime factorization of 144 = 2 x 2 x 2 x 2 x 3 x 3
HCF of 120 and 144 = 2 x 2 x 2 x 3 = 24. 24 is the smallest number which divides both 120 and 144.

LCM of 120 and 144 = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720. 720 is the smallest number which is divisible by both 120 and 144. NOTE IT IS THE INTERSECTION OF THE PRIME FACTORS WHEN MULTIPLIED THAT GIVES THE HCF. AND AUNIVERSAL THAT GIVES THE LCM.

HIGHEST COMMON FACTOR (H.C.F).
LOWEST COMMON MULTIPLE (L.C.M.).

An indirect way to understand.

Let's take the numbers are 28, 35, 56 and the divisor is 7.

When 7 divide these remainder are 0, 0, 0.

Now see 35 - 28 = 7, (Divisible by divisor). Also 56 - 28 = 28 and 56 - 35 = 21 both are divided by divisor. This is like property check for any number.

So we can use this in our problem. Two or more no. when giving the same remainder with common divisor then the difference between the original numbers are also divided by divisor.

Further using in question. 91, 43, 183. See we don't have divisor or factor.

See d or factor divide the no 91, 43, 183 and the differences.

91 - 43, 183 - 43, 183 - 91 say 48, 140, 92.

Since any common divisor or heresy HCF should divide the numbers and differences so we find the HCF of 48, 140, 92.

Hope this will clear why we take difference.

1.The HCF of two or more numbers is the largest common factor of the given numbers. It is the smallest number which divides the two or more given numbers.

The LCM of two or more numbers is the smallest number that is a common multiple of two or more numbers. It is the smallest number which is divisible by two or more given numbers.

Consider two numbers 120 and 144.
Prime factorization of 120 = 2 x 2 x 2 x 3 x 5
Prime factorization of 144 = 2 x 2 x 2 x 2 x 3 x 3
HCF of 120 and 144 = 2 x 2 x 2 x 3 = 24. 24 is the smallest number which divides both 120 and 144.

LCM of 120 and 144 = 2 x 2 x 2 x 2 x 3 x 3 x 5 = 720. 720 is the smallest number which is divisible by both 120 and 144.

2.HIGHEST COMMON FACTOR (H.C.F).
LOWEST COMMON MULTIPLE (L.C.M.).

Here is the reason why you take HCF of differences:

43 = p * q1 + r;
91 = p * q2 + r;
183 = p * q3 + r;

Here p is the dividend that we need to maximize. Let's solve these linear equations. First, lets move r to one side, since the remainder is constant.

43 - pq1 = r ----> Eq1
91 - pq2 = r ----> Eq2
193 - pq3 = r ----> Eq3

Solving the three equations:

p(q2 - q1) = 91 - 43 = 48.
p(q3 - q2) = 183 - 91 = 92.
p(q3 - q1) = 183 - 43 = 140.

We can rewrite this as:
pK1 = 48.
pK2 = 92.
pK3 = 140.
or,
48 = pK1 + 0.
92 = pK2 + 0.
140 = pK3 + 0.

Here K1, K2, K3 are new quotients. Our goal is still to maximize the value of p. So of course we will take HCF of 48, 92 and 140 to get the maximum p.

The answer is 4 and how it is 4 is below,

We can represent any integer number in the form of: D*q + r.
Where D is divisor, q is quotient, r is the remainder.

So each number can be written accordingly:
43 = D*q1 + r1;
91 = D*q2 + r2;
183 = D*q3 + r3;

r1, r2 & r3 will be same in above three equations according to the question.
D is the value that we want to find out. which should be greatest.

On solving three equations we get:

D*(q2-q1)= (91-43)=48
D*(q3-q2)= (183-91)=92
D*(q3-q1)= (183-43)=140
It is obvious that q3>q2>q1.

For the greatest value of D that divide each equation we take the HCF of 48,92,140

THEREFORE ANSWER IS 4.

Dear friends, I am trying to satisfy this question, that why we have to take differences of those numbers.

We all know that the result of H.C.F of any two or more numbers is only their difference between them, such as H.C.F. OF 7& 14 = 7.

Now we calculate the H.C.F. of 7, 14 &21.
We see that the differences between them are actually (14 - 7 = 7, 21 - 7 = 7 & 21 - 7=14.

But the difference of 21 - 7 = 14, whose factor is 2 & 7.

Hence, only 7 will be the H.C.F. OF 7, 14 & 21.
On the above basis, the difference between them is 48, 92 & 140.

Therefore, the H.C.F.of 48, 92 & 140 will be 4.