### Discussion :: Problems on H.C.F and L.C.M - General Questions (Q.No.26)

Sureshkumar said: (Jan 3, 2011) | |

H.C.F example please |

Ram said: (Mar 14, 2011) | |

Plz explain all |

Sakthi said: (Mar 25, 2011) | |

HCF explain please. |

Arun said: (Jun 29, 2011) | |

H.C.F of 1651 and 2032, Explain please. |

Ramesh said: (Jul 2, 2011) | |

I need the calculation process for finding hcf. |

Mini said: (Jul 18, 2011) | |

@Ramesh It is better explained in the following link...check it out http://www.ilovemaths.com/1lcmandhcf.asp |

Patil Bhatu said: (Sep 7, 2011) | |

H.C.F of 1651 and 2032, Explain please. |

Umesh said: (Sep 18, 2011) | |

HCF explain please. |

Santhosh said: (Sep 30, 2011) | |

HCF Explain & give me some examples please. |

Deepak said: (Dec 26, 2011) | |

Division Method: Suppose we have to find the H.C.F. of two given numbers, divide the larger by the smaller one. Now, divide the divisor by the remainder. Repeat the process of dividing the preceding number by the remainder last obtained till zero is obtained as remainder. The last divisor is required H.C.F. ans... Required number = H.C.F. of (1657 - 6) and (2037 - 5) = H.C.F. of 1651 and 2032 let devide the greater no with smaller one means 2030/1651=381 then,1651/381=127 then,381/127=0 so the The greatest number which on dividing 1657 and 2037 leaves remainders 6 and 5 respectively, is:127 |

Ashwarya said: (Dec 29, 2011) | |

Ho we know that 1657 - with 6 and how we know that 2037 is minus with 5 please explain. |

M.V.Krishna/Palvoncha said: (Jan 11, 2012) | |

Here 6 & 5 are remainders of 1657(say A) & 2037(say B) respectively. To make A and B(dividends) perfectly divisible by a divisor, the remainders should be subtracted from dividends. Now if we divide the dividend by a divisor, the remainder will be zero, because we made dividends perfectly divisible. Now the numbers are 1657-6=1651 and 2037-5=2032. HCF of 1651,2032. 1651)2032(1 1651 ------- 381)1651(4 1524 ------ 127)381(3 381 ------ 0 ------- So HCF is 127. Hope you understand. |

Venkat said: (Jan 15, 2012) | |

THANKS krishna your explanation is too easy. |

Kirti said: (Feb 6, 2012) | |

Can we do 1657+6 or 1657-6 both will give same answer. ? |

Anu said: (Aug 16, 2012) | |

No .it is not possible .because 1657-6 gives the remainder 0,but 1657+6 gives the remainder 1. |

Anu said: (Aug 16, 2012) | |

It is not possible . the term 1657+6 is not a perfet multiplier it gives a remainder 1.the term 1657-6 is a perfect multiplier it gives a remainder 0.that's why they doesnot give same answer. |

Loganathan said: (Aug 31, 2012) | |

Thank you mr. M. V. Krishna/Palvoncha. |

Km Manimegan said: (Jul 30, 2013) | |

TO FIND THE HCF OF TWO NUMBERS, DIVIDE THE BIGGER NUMBER BY SMALLER . TAKE 2032 AND 1651, ON DIVIDING. 1651)2032( 1 1651 -------------- 381 ------------- THEN DIVIDE 1651 BY 381. 381)1651(4 1524 -------------- 127 -------------- THEN DIVIDE 381 BY 127. 127)381(3 381 ---------- 0 ---------- THE DIVIDING OF BIGGER NUMBER BY SMALLER NUMBER SHOULD CONTINUE TILL WE GET REMAINDER AS ZERO. THEN THE LAST DIVISOR IS THE H.C.F OF THE TWO GIVEN NUMBERS. |

Thyagarajan said: (Mar 26, 2014) | |

Answer: 13*127 = 1651+6 = 1657. 16*127 = 2032+5 = 2037. |

Prabhakaran said: (Oct 9, 2014) | |

@Thayagarajan. How we know 127 is the correct answer?, that's we need tell clearly. |

Shivam said: (Jun 2, 2016) | |

If we divide 7 by 2 it gives a remainder of 1. Either we add or subtract the remainder 1 from 7 the numbers obtained 8 and 6 which is divisible by 2. How the answer come in this problem, that we have to add or subtract? |

Albert said: (Jul 29, 2016) | |

Please explain the calculation of H.C.F in detail. |

Rajan Gupta said: (Aug 24, 2016) | |

Please explain the method to calculate the HCF with an example. |

K Sreenivasulu said: (Nov 14, 2016) | |

Thank you @M. V. Krishna. |

Sayeeda said: (Mar 29, 2017) | |

Thank you @Deepak. |

Jenifer said: (May 23, 2017) | |

Why should we find HCF? Why not LCM? |

Nidhi said: (Jun 14, 2017) | |

Please explain the solution of this question. |

Janaiah said: (Dec 11, 2017) | |

Thanks @Krishna. |

Ankit said: (Jan 25, 2018) | |

@Jenifer. As you see, the question is asking for greatest number which means greatest factor or divisor. So here we will do HCF/ GCD. |

Rishav said: (Sep 15, 2018) | |

Why HCF and not LCM? Lets see, We have to find the greatest number which on dividing 1657 and 2037 leaves remainder 6 and 5. Forget about 'greatest'. we need a number 'which divides 1657' and '2037' ie, if that number is 'x', we have the expression as 1657/x with remainder 6 and 2037/x with remainder 5. Lets write this in the division form (Dividend = quotient*divisor + remainder) ie, 1657 = q1 * x + 6. 2037 = q2 * x + 5. Therefore, 1651 = q1 * x. 2032 = q2 * x. ie, 'x' (our number) is a common 'FACTOR' of 1651 & 2032 (NOT 'Multiple'). [eg. 24 = 12 *2 or 6*4 or 8*3 ... etc. here, 12 & 2, 6 &4, 8 &3 all are factors of 24 and not multiple]. So, we need common factors of 1651 & 2032 and just select the greatest among them(since the question asks for the greatest). |

Kumar said: (Jul 23, 2019) | |

Super explanation, Thanks all. |

Vivek said: (Jan 16, 2020) | |

How can we find the HCF of Three co-prime numbers? please anyone explain. |

Aayushi said: (Mar 26, 2020) | |

This method of finding HCF of two large no. is called Euclid's division lemma, you can check it further on youtube you will then be able to solve the question. |

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