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. 4)
4.
Let N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. Then sum of the digits in N is:
Answer: Option
Explanation:
N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305)
= H.C.F. of 3360, 2240 and 5600 = 1120.
Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4
Discussion:
158 comments Page 12 of 16.
Chandra said:
1 decade ago
[(6905-1305)-(4665-1305)] / 2 = [2240] / 2 = 1120.
@Ravi can you explain how you did this step?
@Ravi can you explain how you did this step?
RAVI JAIN said:
1 decade ago
EASY ONE NO. ARE 6905, 4665, 1305
SO [(6905-1305)-(4665-1305)] / 2 = [2240] / 2 = 1120.
NOW 1+1+2+0 = 4.
SO [(6905-1305)-(4665-1305)] / 2 = [2240] / 2 = 1120.
NOW 1+1+2+0 = 4.
Suleman said:
1 decade ago
People try dividing each number by 8 and you will get the same remainder.
Gopalakrishnan said:
1 decade ago
@All.
Is very simple to find the answer!
In question clearly given N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. So we dividing the each number by giving option as 4 5 6 8 and we get same remainder as option A) 4.
Is very simple to find the answer!
In question clearly given N be the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. So we dividing the each number by giving option as 4 5 6 8 and we get same remainder as option A) 4.
Adarsha Gowda Mandya said:
1 decade ago
Hello @Sarswathi, why are you taking 10 to multiply number after the subtraction. Could you please tell me?
Manish said:
1 decade ago
People try dividing each number by 8 and you will get the same remainder.
Sahid said:
1 decade ago
8 will divide each number and leave reminder 1, and 8 is greatest among all of the option.
Helena said:
1 decade ago
The clue is in the question. A digit by definition is any number between 0 & 9.
N = 1120.
The digits of 1120 are the digits, 1, 1, 2, 0.
There are four of them.
Hope that helps.
N = 1120.
The digits of 1120 are the digits, 1, 1, 2, 0.
There are four of them.
Hope that helps.
Gargi said:
1 decade ago
5 is not correct answer. As question is for sum of the greatest devider. And greatest number which devides all the given numbers is : 1120 so, digit sum of the same is 4.
Correct answer is: 4.
Correct answer is: 4.
Rangan said:
1 decade ago
@Ramana : The Greatest common divisor (or the HCF) is not the solutiom needed here, it is rather the SUM OF THE DIGITS of THE HCF.
The Remainder being the same for each case, it is clear that subtracting the lesser no. from the greater ones renders the remainder zero. Also the HCF of te no.s left thereafter just wait to be known, because it perfectly divides the no.s and thus gets the same remainder in each case.
The Remainder being the same for each case, it is clear that subtracting the lesser no. from the greater ones renders the remainder zero. Also the HCF of te no.s left thereafter just wait to be known, because it perfectly divides the no.s and thus gets the same remainder in each case.
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