### Discussion :: Problems on H.C.F and L.C.M - General Questions (Q.No.1)

Nagu said: (Jul 15, 2010) | |

Why they took difference between those nos. Please explain me. |

Anu said: (Jul 27, 2010) | |

Explain the logic of this solution. |

Sourabh Das said: (Jul 28, 2010) | |

I also want an explanation about the logic of the solution. |

Raj said: (Aug 2, 2010) | |

Why do we do the difference of number to each other? |

Anirudh Rai said: (Aug 6, 2010) | |

Is this the only way to this type of problem? If yes then would you please explain it. |

Arya said: (Sep 19, 2010) | |

If three nos as in this case is 43, 91 and 183 are given then to find HCF the shortest solution is take the diff: 1) 91 - 43 = 48 (4*3*7) 2) 183 - 91 = 92 (4*23) 3) 183 - 43 = 140 (4*7*5) Thus the HCF is 4. |

Prashant said: (Sep 22, 2010) | |

We can represent any integer number in the form of: pq+r. Where p is dividend, q is quotient, r is reminder. so: 43 = pq1 + r; 91 = pq2 + r; 183 = pq3 + r; Take r same in above three equations as given in 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; For the greatest value of p that divide each equation we take the HCF of 48,92,140 THEREFORE ANSWER IS 4. |

Sakthivel said: (Oct 28, 2010) | |

How to arrive at this equation p(q2-q1)= (91-43) from 43 = pq1 + r; 91 = pq2 + r; 183 = pq3 + r; |

Hary said: (Oct 29, 2010) | |

Solving the equations: 91=pq2+r 43=pq1+r _________________ 48=pq2-pq1 thus 48=p(q2-q1)=91-43. |

Shwetha said: (Dec 15, 2010) | |

Simply divide each number by 4 the remainder is same for all so the answer is 4. |

Dhanasekar said: (Jan 8, 2011) | |

Simply try divide to divide each by the options, by this which is exactly divided then that is the answer. |

Nam said: (Jan 12, 2011) | |

Acc to the formulae H.C.F. of [(H.C.F. of any two) and (the third number)] gives the H.C.F. of three given number. So as per the first formulae the hcf of 183 & 43 is comming 10 n the third no. is 91 so how will 4 be the ans |

Hind said: (Jan 12, 2011) | |

I go with same method as shweta and dhanasekar has said in the forum. |

Navneet said: (Jan 12, 2011) | |

Divide each with each option. |

Pankaj Maharaj said: (Jan 22, 2011) | |

Yesss! you sud simply divide and find the ans in quicker way. |

Durga said: (Jan 23, 2011) | |

Getting answer is not important but knowing how to approach that answer is important. Thanks for Prashant. |

Titus said: (Jan 29, 2011) | |

That is true. Understanding is more important. |

Sweety said: (Feb 6, 2011) | |

I don't understand why we need to find the difference. Please explain the logic behind it. |

Komal said: (Feb 9, 2011) | |

The difference is used to get the common factors between two nos. it will eliminate the uncommon factors. And at-last we have only to pick up the highest common factor... you can try with an example for 30 and 10. factors of (30) = 5,2,3 factors of (10) = 5,2 factors of (30-10) = 5,2 |

Habib said: (Feb 22, 2011) | |

The difference is taken to analyze the pattern. Amongst the numbers ! Ie each difference tells us that these numbers are multiples of 4 and have the same remainder because all numbers can be divided by 4. Hence you get 4. |

Jayalalitha said: (Feb 28, 2011) | |

WHY WE HAVE TO TAKE DIFFERENCE OF THOSE TWO NUMBERS |

Prathyusha said: (Mar 6, 2011) | |

SMART WAY OF SOLVING BY @SHWETHA |

Yuva Kumar said: (Mar 12, 2011) | |

Dear Sir/madam please expain I could not understand. |

R. Ahmad said: (Mar 13, 2011) | |

Why it is used two times 183 nos Explain? |

Fil said: (Mar 21, 2011) | |

If you mult 4*3*7 its coming 84 how 4 will be the greatest no please explain me. |

Krishna said: (Apr 30, 2011) | |

4*3*7 is wrong right one is 4*12 k |

Vidhya said: (May 9, 2011) | |

Why we have to take the difference of those two numbers? |

Sita said: (Jun 9, 2011) | |

I'm not clear with any of the above explanation. Can anyone tell me clearly. |

Mini said: (Jul 12, 2011) | |

Why they took the differences of given 3 numbers and then calculated the H.C.F. Why didn't they simply calculate the H.C.F OF GIVEN 3 NUMBERS. |

Mysterious said: (Jul 17, 2011) | |

@Mini and all friends If it was just asked to find the HCF of 43,91,183 then simply the Division method or Factorization method had been applied. But here it is stated that "so as to leave the same remainder in each case", that's why the difference of the numbers is calculated. |

Sandeep.Ch said: (Aug 13, 2011) | |

Please tell me which is the perfect explanation. |

Abhishek said: (Aug 28, 2011) | |

@Prashant: thanks for best explanation. |

Ajit said: (Sep 1, 2011) | |

Thanks prashant for best explanation. |

Nadhirsha said: (Sep 8, 2011) | |

Well done Prashant. |

Samir said: (Sep 12, 2011) | |

Which is the highest five digit number which is equally divisible by 60, 80, and 90 and how to arrive this answer ? |

Hapreet said: (Sep 14, 2011) | |

Why we have to less these numbers? |

Ranjeet said: (Oct 7, 2011) | |

Simply because there is no same factor of these no and they don't divide by each other when we divide 91 by 43 there remainder is 5 that is not the divisible of 43. So we take this method to explanation. |

Uzair said: (Oct 11, 2011) | |

U guys are not understanding the meaning of H.C.F actually...i'll give an example and an explanation! Highest common factor Here are the list of prime factors of 24 and 36: 24 = 2 x 2 x 2 x 3 36 = 2 x 2 x 3 x 3 If we write down the numbers that are the same in both lists, they will give us the highest common factor of 24 and 36: HCF of 24 and 36 is 2 x 2 x 3 = 12 This is the meaning of H.C.F...so simple! :D |

Ashish Saini said: (Oct 12, 2011) | |

Thanks my dear parshant. |

Sowjanya said: (Oct 28, 2011) | |

Thanks prashanh, good explanation of sum, but I don't know some people are not at all trying to understand the topic first, then it is very easy my dear friends. |

Dhanasekar said: (Nov 8, 2011) | |

Can any one explain how to identify. Whether the problem is based on HCF or least common multiple. Is there any key words? |

Raje said: (Nov 19, 2011) | |

Simply way is divide 43, 81 and 183 by 4 we get the same remainder 3 for all. |

Lakshmisha said: (Nov 24, 2011) | |

@Prashant, thanks for explanation |

Pooja Madrewar said: (Dec 3, 2011) | |

Explain the logic of this solution |

Rajendra said: (Dec 25, 2011) | |

Thanks prashant your explaination is really good... |

Anand said: (Dec 30, 2011) | |

No need to difference between them. Simply take one by one option and divide given number in question (43, 91 and 183) and equate remainder. No need to difference between them. Okay friends. So simple ! |

Manindra said: (Feb 8, 2012) | |

We take difference as it is very simple for calculation. |

Hemalatha said: (Feb 15, 2012) | |

I'm not clear with any one of the above explanation, why we take the difference of those numbers. |

Lakshmi said: (Feb 17, 2012) | |

I am not clear please explain the above explanation |

Ravi Varma 501 said: (Mar 26, 2012) | |

Swetha and prashanth explanation is good. And abve some are all in kids formulas. . . So dont look others friends. |

Surbhi Swaraj said: (Apr 17, 2012) | |

Is there any trick to solve this problem? |

Heena said: (May 11, 2012) | |

Here take the difference between the numbers then start dividing. 48/2, 92/2, 140/2 Then 48/2=2*2*2*2*3 92/2=2*2*23 140/2=2*2*35 So HCF is [nothing but the highest common factor]. So HCFis 2*2=4. |

Lovely said: (May 14, 2012) | |

Hey prashant your explanation is too good. Thanks. Is there more short method to do solve this. |

Aadhilakshmi said: (May 22, 2012) | |

Thanks for the clear explanation prasanth. |

Parthiban said: (May 22, 2012) | |

It is very simple consider 43,91,183 are the 3 numbers first , find the diff between them 91-43=48; 183-91=92; 183-43=140; let divide the 48,92,140,by 2 we have 48=2*2*2*2*3 92=2*2*23 140=2*2*149 here common factor is 2*2 so answer is 4 |

Mayank said: (Jun 5, 2012) | |

4*22-gives rem 3. 4*10-gives rem3. 4*45-also gives rem 3. All three numbers have equal rem 3 satisfies condition given in number. Hence ans is 4. |

Cathy said: (Jun 15, 2012) | |

How do you know we are to find the h.C.F ? I seem to be getting so confused. |

Karthick said: (Jun 16, 2012) | |

@Cathy. The keyword is greatest or largest- if you find these words, we are required to find HCF. If in case you find least or lowest in the problem, we need to find LCM. |

Ankur Jindal said: (Jun 17, 2012) | |

If three nos as in this case is 43, 91 and 183 are given then to find HCF the shortest solution is take the diff: 1) 91 - 43 = 48 (4*3*7). 2) 183 - 91 = 92 (4*23). 3) 183 - 43 = 140 (4*7*5). Thus the HCF is 4. |

Merry said: (Aug 28, 2012) | |

When pratibhan said the common factor is 2 ok this is write but why she multiple two no of 2 whyn't three no of 2 means the result is 8. |

Rdbaby said: (Oct 1, 2012) | |

@ Merry. Pratibhan meant, the common thing coming in prime factor list of all 3 nos is 2*2, i.e 4. |

Sruthi said: (Oct 31, 2012) | |

Why we have to take the difference of the numbers please explain? |

Hema said: (Nov 14, 2012) | |

Why we have to take the difference of the numbers can you please explain? |

Furqan said: (Dec 28, 2012) | |

What is different between HCF & LCM? And what is the meaning HCF & LCM? |

Prasand. said: (Jan 24, 2013) | |

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.). |

Lavan said: (Apr 24, 2013) | |

Why we have to take difference of two number? |

T Shyam Sunder said: (May 7, 2013) | |

Why you had taken difference between this numbers? |

Saurabh said: (Jun 17, 2013) | |

What happens if there are only two numbers, instead of three? |

Nikita said: (Jun 25, 2013) | |

HCF of any two number is the number which divide each of them exactly. And any no which divides each of two number exactly also divides their difference, their sum and sum and difference of any multiple of that numbers. Ex- if 5 is them common factor of 25 and 15. It is also a factor of (25-15) , (25+15) and (25a+15b). Therefore here we have taken the difference to find the HCF. Because when we will subtract them remainder will get cancelled. |

Raj said: (Jul 13, 2013) | |

Actually hcf means highest common factor. Which means"highest number which divide the numbers leaving reminder as 0". Look at the question they asked to find a hcf which leaves the sme reminder in each case. By outer observation we can say 4 is hcf because 43/4=3, 91/4=3, and 143/4=3. Its a simple method of analysing hcf. |

Sendilnathan said: (Jul 31, 2013) | |

How to find highest number which divides some numbers and divide the same reminder? |

Shital said: (Aug 5, 2013) | |

@Parthiban. If there is large numbers then it is reliable method to divide and then finding highest common factor? |

Shital said: (Aug 5, 2013) | |

Mysterious please tell me how will be that example converted if divide method and factorization method has given in example? |

Chaitanya said: (Sep 15, 2013) | |

How to calculate HCF ? |

Lohit said: (Oct 2, 2013) | |

Explain logic of the solution ? |

Amarnath said: (Oct 11, 2013) | |

Hey friends I will say one example of HCF. Eg..72and 126...then, Using Division method First find H.C.F. of 72 and 126.
H.C.F. of 72 and 126 = 18. Similarly calculate H.C.F. of 18 and 270 as 18. Hence H.C.F. of the given three numbers = 18. |

Niveditaa Singh said: (Nov 12, 2013) | |

Is this only method to solve this? If it can be solve by another way then please explain. |

Jaggu said: (Dec 31, 2013) | |

43%4 = 3. 91%4 = 3. 183%4 = 3. That's why answer is 4. |

Sharmi said: (Jan 7, 2014) | |

Why you have to divide all the number with 4. What is the reason, please explain? |

Afsa said: (Jan 8, 2014) | |

Some one help! What is the logic behind H. C. F? |

Mani Pandian said: (Jan 8, 2014) | |

Here are the list of prime factors of 24 and 36: 24 = 2 x 2 x 2 x 3. 36 = 2 x 2 x 3 x 3. If we write down the numbers that are the same in both lists, they will give us the highest common factor of 24 and 36: HCF of 24 and 36 is 2 x 2 x 3 = 12. This is the meaning of H.C.F. |

Rajvir Singh said: (Jan 17, 2014) | |

Why do need to take difference three times? (91-43), (183-91) was enough. and then take h.c.f. of 48 and 92 please someone give another example why do we need to do (183-43)? |

Tahir said: (Feb 19, 2014) | |

Just simple according to the question there is no compilation is 43/4 the reminder is 3 then 91/4 the remainder is 3 and now 183/4 the reminder is once again three. F look at question the reminder is same ie 3. |

Sarath said: (Mar 12, 2014) | |

I think the smarter way will be to divide the numbers with options and check. But your speed has to be so high. |

Kumari said: (Apr 26, 2014) | |

Dear friends, Why they took difference between them and is any other easy as well as short method ? please communicate this answer as soon as possible. Thanking you. |

Prem said: (May 15, 2014) | |

I am unable to understand this problem. Can anyone explain all steps involved? |

Madhukumar E said: (Jun 5, 2014) | |

2(48 2(24 2(12 3(6 (2 So now 48 = 2*2*2*3*2. Use the same method for 92 and 140. Then they give = 2*2*23 and, = 2*2*5*7. Now, take the common factors. Here we go, 2*2*2*3*2. 2*2*23. 2*2*5*7. 2*2 are common factors. So, H.C.F. of 48, 92 and 140 = 4. |

Aditya said: (Jun 16, 2014) | |

Find the least number divisible by 12, 32, 42, and 63 and it must be a perfect square? Please give me the solution. |

Sreyas Biju said: (Jul 11, 2014) | |

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! |

Rony said: (Jul 19, 2014) | |

HCF of two numbers is 12, their sum is 72, then the two numbers are? |

Moses said: (Jul 27, 2014) | |

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.). |

Kanchan said: (Jul 29, 2014) | |

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. |

Govind Paswan said: (Sep 16, 2014) | |

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. Explanation: 43, 91, 183. 183 = (183-91) = 92. 91 = (183-43) = 140. 43 = (91-43) = 48. 92, 140, 48. 92 = 2*2*23. 140 = 2*2*35. 48 = 2*2*2*3*2. Ans 4. |

Lakhan Saini said: (Oct 30, 2014) | |

Divided by 4 each number we get remainder 3 |

Honey said: (Nov 26, 2014) | |

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: (Dec 1, 2014) | |

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? |

Neeraj said: (Dec 26, 2014) | |

Why we take HCF? When one should take LCM and HCF. |

Dove said: (Jan 22, 2015) | |

I don't understand why you write 2*2 and say the answer is 4. |

Dhananjay said: (Jan 24, 2015) | |

@ 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. |

Vamshi said: (Feb 27, 2015) | |

48, 92 and 140 = 4 how now am confusing, please explain. |

Mandeep said: (Mar 11, 2015) | |

@Prashant. P is not a dividend its divisor. |

Vignesh Kumar said: (Mar 15, 2015) | |

Simple way to calculate HCF: 2|48 92 140 ----------- 2|24 46 70 ----------- |12 23 35 ------------ Stop if you can't divide by common divisor, so answer = 2*2 = 4. |

Abhishek Ray said: (Apr 8, 2015) | |

Kindly revert back with logically, how this problem was processed? |

Saumya Tyagi said: (Apr 20, 2015) | |

Well I also cannot understand the correct logic of solving this question please explain. |

Jayanta Atkari said: (Apr 25, 2015) | |

(91 - 43 = 48) = 4*12. (183 - 91 = 92) = 4*23. (183 - 43 = 140) = 4*35. Thus HCF is = 4. |

Asif said: (Apr 30, 2015) | |

Please if you know this answer you tell with clarification. |

Saroja said: (May 6, 2015) | |

43 = HA+R. 91 = HB+R. 183 = HC+R. Here we are expressing the numbers as HCF*Prime number. 43-91 = H(A-B) = 48. 183-91 = H(C-B) = 92. 183-43 = H(C-A) = 140. So when we take HCF of 48, 92, 140 we get "H". *When we are expressing two numbers in terms of their HCF the format is HCF*co-primes. So we have expressed. 43 = HA, 91 = HB, 183 = HC. |

Daniel said: (May 15, 2015) | |

13 and 7 don't leave remainder in 91 while they leave in other and 9 clearly leave different remainder in 43 and 91. So it is 4 you can think logically or use hit and trial method or simply use formulas given by others. |

Suganthy said: (Jun 11, 2015) | |

Thanks for all who have explained the concept. |

Vinitha said: (Jun 15, 2015) | |

Hey friends. H.C.F. of (91-43=48), (183-91=92), (183-43=140). H.C.F. of 48, 92, 140 = 4. 48 = 4*12, 92 = 4*23, 140 = 4*35. So we divided by 4 in all 48, 92, 140. Thus H.C.F. is 4. |

Phaaani said: (Jul 21, 2015) | |

What could be the value of x if HCF of 12, 24, x is as same as GCD of 18, 36 and x? |

Rajesh said: (Jul 30, 2015) | |

1.91-43 = 48, 183-91 = 92, 183-43 = 140. 2.92-48 = 44, 140-92 = 48, 140-48 = 92. For every time we eliminating uncommon Factors. |

Vandana said: (Aug 14, 2015) | |

But there is that that same remainder in each case. |

Naresh said: (Aug 18, 2015) | |

Please explain again simply way. |

Somanath said: (Aug 18, 2015) | |

Find the highest number which divides 25 and 29 equally leaving 5 in each case. |

M Sai Teja said: (Aug 25, 2015) | |

I think 43 and 91 are prime numbers. Prime numbers don't have factors, to find H.C.F we need factors, for that we are subtracting numbers to get non-prime numbers. |

Iegowg said: (Aug 28, 2015) | |

Just divide each option with all figures it is that simple. |

Anukriti Srivastava said: (Sep 19, 2015) | |

Is there any quick and efficient method to find HCF and LCM of big numbers such as HCF of 38094, 43762, 84632 for competitive exam if yes then please help me out waiting. |

Manjunath said: (Sep 21, 2015) | |

I want the explanation of the logic. |

Mamang said: (Oct 7, 2015) | |

Please tell me the most simple formula. |

Vimal Kumar said: (Oct 11, 2015) | |

If the sum of two numbers are 348 and their HCF is 48 then what is the difference of the numbers? |

Trishana said: (Nov 15, 2015) | |

Here in this question greatest number is 7 and it gives the same remainder that is 1 when divided by all. So how is the answer 4? |

Sowjanya said: (Jan 6, 2016) | |

In @Prasanth answer, p is the divisor not dividend and the analysis of sum is correct. |

Dileep Gowda said: (Jan 8, 2016) | |

Yes your right @Sowjanya. But @Prasanth is almost right I think you need more information. |

Saurabh said: (Jan 15, 2016) | |

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. |

Chandu said: (Jan 18, 2016) | |

1st divide 48 with 4 then remainder is 0. 2nd divide 92 with 4 then remainder is 0. 3rd divide 140 with 4 then remainder is 0. Hence so 4 is a perfect number. |

Chirag said: (Jan 31, 2016) | |

Easy calculation, 7 and 13 are directly eliminated. 4 and 9 are the required. So divide it by the 9, no remainder will be same. So 4 is the option. |

Umar said: (Mar 3, 2016) | |

How to solve the same question if remainder is not same in each case? |

Daniel said: (Mar 11, 2016) | |

I can't understand the question. |

Devinder said: (Mar 29, 2016) | |

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. |

Pradeep said: (May 5, 2016) | |

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. |

Mz Gucchi said: (Jun 17, 2016) | |

Hi, how to solve the Lcm? Please help me. |

Kiran Kumar said: (Jul 17, 2016) | |

Better to go by basic methods and simple operations to arrive the answer. So taking the options and dividing each no is the best solution. Don't complicate things by doing too much of research. Just solve it in a smart and easy way and arrive the answer. Wherever the concept is actually required use it only in those cases and not in all the cases. Because the smart and easy approach is more important than the steps. |

Senthmaizhan said: (Aug 12, 2016) | |

All this explanation are not clear to me. Please explain it in a simple way. |

Shyam Yadav said: (Aug 19, 2016) | |

What are the numbers to find the difference? |

Mr.Newton said: (Aug 24, 2016) | |

Everything is fine, but I am getting doubt with HCF. In three numbers common we get 2. If we divided 2 * 2 * 2 get 8. Why are you telling 4? |

Ajeeth said: (Sep 16, 2016) | |

How to take HCF I don't know how to take HCF please anybody explains me. How to find the HCF for any unknowns for ex: 533, 522, 455 for this number? |

Pradeep said: (Sep 20, 2016) | |

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. |

Brianna said: (Oct 6, 2016) | |

I don't understand I'm only eleven, please someone help me to get this. |

Khushi said: (Oct 16, 2016) | |

I'm confused by this one, Please explain with division method. |

Vimalsankar said: (Nov 8, 2016) | |

43/4 the remainder is 3. 91/4 the remainder is 3. 183/4 the remainder is 3. So, finally we get the answer 4. |

Tamzid Khan said: (Nov 14, 2016) | |

Is there anyone who will clearly explain why one has to take the difference of those numbers? I mean what is the key note in the question seeing which I will solve the problem in this way? Please explain someone. |

Phani said: (Nov 17, 2016) | |

Is this the common HCF problem if so when I try to solve it using Euclid's method I got the answer '1'. Am I right? |

Ramya said: (Dec 31, 2016) | |

I'm not understanding the logic. Please someone explain me. |

Tanvi said: (Jan 4, 2017) | |

I am not understanding please explain me. |

Manoj Rmk said: (Jan 6, 2017) | |

Find the greatest number that will divide 43, 91 and 183 so as to leave the same remainder in each case. Explanation: 43, 91, 183. 183 = (183-91) = 92. 91 = (183-43) = 140. 43 = (91-43) = 48. 92, 140, 48. 92 = 2 * 2 * 23. 140 = 2 * 2 * 35. 48 = 2 * 2 * 2 * 3 * 2. Ans 4. |

Bisht Ashish said: (Feb 10, 2017) | |

@Prashant. In your explanation, 'p' must be a Divisor, not the Dividend. |

Arsh said: (Mar 15, 2017) | |

Here, the remainder is unknown that is why we have only chance to find their difference. |

Sowmya said: (Mar 20, 2017) | |

Well explained, thanks @Prashant. |

Mani said: (Mar 23, 2017) | |

Why can't we think like this, In given four option one is HCF of 43, 91, 183. And leave the same reminder as per the question. Let me consider 4 is HCF then divide the numbers by four. It shows the same reminder hence 4 is the answer. |

Stuti said: (Mar 25, 2017) | |

I understood the sum now. Given explanations was good. Thankyou all. |

Rituraj said: (Apr 12, 2017) | |

Well done @Saurabh. |

Lokesh said: (May 27, 2017) | |

Easy to understand. Thanks to all the given explanation. |

Alex said: (Jun 17, 2017) | |

Good explanation @Prashant. |

Logesh said: (Sep 1, 2017) | |

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. |

Mohan said: (Oct 15, 2017) | |

Hi. In question, they gave a greatest number, in the sense. Ex- if you take 4 as a greater num for the given question. Means this is the greater num which divides all the dividend if you take 5 it won't divide all' the dividends. If they ask greater num -then take smaller num in the choices and divide all the dividends by that num if you get remainder same then that is the answer. |

Rajat said: (Nov 16, 2017) | |

Why we taking the difference between these, first explain this. |

Gowtham said: (Nov 20, 2017) | |

Thanks @Komal. I could finally get it. |

Sayyidah Shah said: (Dec 7, 2017) | |

Best solution, Thanks @Nikita. |

Shankar said: (Dec 27, 2017) | |

Let x be the greatest possible number such that it leaves the same remainder when it divides 183, 91 or 43. Since the remainder is the same in each case, the difference of the terms must be exactly divisible by x. Also, x must the greatest possible number that exactly divides the difference between the terms. Required number, x = HCF of (183 " 91, 91 " 43, 183 " 43) = HCF of (92, 48, 140) = 4. |

Mahima said: (Feb 10, 2018) | |

Nice explanation @Parthiban. Thanks. |

Mrunmay said: (Feb 18, 2018) | |

Why numbers are being subtracted? I still have a doubt. Please explain me. |

Jyothi said: (Feb 28, 2018) | |

Not getting it, can anyone please explain it? |

Ankush said: (Mar 24, 2018) | |

@Mrunmay. Solution they take difference because to make remainder zero. And than that divisor will divine these number Completely. |

G Vinay said: (Apr 13, 2018) | |

How? Please explain the answer. |

Ahmad Zaidi said: (Apr 15, 2018) | |

@Samir. First find the LCM of 60, 80, 90, which is 720. Then divide it by 99999 which is highest 5 digit number, and less the reminder which is 639 and answer is 99360. Which is equally divisible by 60, 80, 90. Thanks. |

Chary said: (May 30, 2018) | |

Thanks for giving nice explanation @Prasanth. |

Sixface said: (Jun 27, 2018) | |

It's very simple guys => 43/4=3, 92/4=3, 183/4=3. The remainder is 3 for all divide so 4 is the correct answer. |

Shanmukh Adari said: (Jul 12, 2018) | |

Your explanation is fantastic, Thank you so much @Prashant. |

Mahima Chaurasia said: (Sep 13, 2018) | |

In the question it is asked the remainder should be same so, Answer is option A ie 4. 43/4 remainder is 03. 91/4 remainder is 03. 183/4 remainder is 03. |

Pabitra Bhusan Saha said: (Sep 15, 2018) | |

Your explanation is very easy to understand and this is the best process, Thanks @Saroja and @Sourav. |

Shreekanth said: (Sep 18, 2018) | |

Keep on dividing all the three numbers by 2 so that reminder should be the same ie 2[43 91 183] [ 21 45 91] remainder will be 1 for all the three. Divide again by 2. 2[21 45 91] 10 22 45 remainders will be same ie 1. So after this, if you try to divide all three again by 2 then you get a different value for the reminder. And hence; 2 * 2 = 4. |

Magnus Udoka said: (Dec 1, 2018) | |

Your explanation was superb, Thanks @Prashant. |

Davoas said: (Dec 30, 2018) | |

What if we have a fixed remainder. For example take 90, 114 and 230 with remainder 6 what will be the divisor? Anyone explain me. |

Kuldeep Sharma said: (Mar 1, 2019) | |

let the greatest number is H. Now; H*Quotient(Q1)+Reminder(R)= 43 -------> (1) H*Quotient(Q2)+Reminder(R)= 91 -------> (2) H*Quotient(Q3)+Reminder(R)= 183 -------> (3) Now, (1)-(2), (3)-(2), (3)-(1) H*(Q2-Q1)= 48, H*(Q3-Q2)= 92, H*(Q3-Q1)= 140. So 48, 92 and 140 are completely divisible by H, that will be the HCF of these numbers. So, the HCF of 48, 92 and 140 is 4. |

Pabitra Saha said: (Jun 27, 2019) | |

Very good explanation, Thanks @Kuldeep Sharma. |

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