Aptitude - Problems on H.C.F and L.C.M - Discussion
Discussion Forum : Problems on H.C.F and L.C.M - General Questions (Q.No. 1)
1.
Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case.
Answer: Option
Explanation:
Required number = H.C.F. of (91 - 43), (183 - 91) and (183 - 43)
= H.C.F. of 48, 92 and 140 = 4.
Discussion:
212 comments Page 13 of 22.
Sreyas Biju said:
1 decade ago
The smallest number divisible by 12,32,42 and 63 is its LCM which is 2016.
The prime factorization of 2016 is 2x2x2x2x2x3x3x7.
Pick out the pairs in this step. They are 2x2 2x2 3x3.
2 and 7 are not in pairs. But if there is one more 2 and one more 7, both 2 and 7 can also be in pairs.
So the required numbers are one more 2 and one more 7. So it is 2x7= 14. Therefore, 2016x14 = 28224.
So the required square number is 28224.
Thank you!
The prime factorization of 2016 is 2x2x2x2x2x3x3x7.
Pick out the pairs in this step. They are 2x2 2x2 3x3.
2 and 7 are not in pairs. But if there is one more 2 and one more 7, both 2 and 7 can also be in pairs.
So the required numbers are one more 2 and one more 7. So it is 2x7= 14. Therefore, 2016x14 = 28224.
So the required square number is 28224.
Thank you!
Rony said:
1 decade ago
HCF of two numbers is 12, their sum is 72, then the two numbers are?
Moses said:
1 decade ago
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.).
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.).
Kanchan said:
1 decade ago
As it is asking for greatest no among all options the answer for above problem should be 7 as it divides all the no and gives remainder 0.
Lakhan saini said:
1 decade ago
Divided by 4 each number we get remainder 3
Honey said:
1 decade ago
Here 4 is even number. So its multiples will also be even. In the question above all the numbers were odd. If is divide by 4 then it will compulsory leaves an remainder.
Sandeep said:
1 decade ago
Why did we take the differences of the numbers?
Is this the only way to solve it by taking the differences if there is any other method then please reply?
Is this the only way to solve it by taking the differences if there is any other method then please reply?
Neeraj said:
1 decade ago
Why we take HCF? When one should take LCM and HCF.
Dove said:
1 decade ago
I don't understand why you write 2*2 and say the answer is 4.
Dhananjay said:
1 decade ago
@ Sanddep:
Have a look at explanation given by @Prasant. We have to solve equation for 'p'. Hence,we eliminate 'r' by taking differences of numbers.
Have a look at explanation given by @Prasant. We have to solve equation for 'p'. Hence,we eliminate 'r' by taking differences of numbers.
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