14/02/2018 10:35 am
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Post your questions on HCF LCM under this sub topic
This topic was modified 7 months ago by TathaGat
26 questions & discussions are there under this sub-topic
0
14/02/2018 12:10 pm
Find the greatest no., which will divide 215,167 and 135 so as to leave the same remainder in each case.
a) 64
b) 32
c) 24
d)16
LydiaMadini 02/06/2024 4:30 pm
The numbers are 215, 167 and 135.
The difference between the numbers (so that the equal remainder gets cancelled out) is - 48, 80, 32.
Ans = HCF (48, 80, 32) = 16.
0
14/02/2018 12:20 pm
The smallest square no. which is exactly divisible by 2,,3,4,-9,6,18,36&60, is
a)900
b)1600
c)3600
d) none of these
This post was modified 7 months ago by TathaGat
Anu 02/06/2024 4:36 pm
The smallest square no. which is exactly divisible by 2,,3,4,-9,6,18,36&60, is
The required number the smallest multiple of t he LCM of the numbers that forms a perfect square.
LCM of the given numbers – 180 = 22x32x5
So the smallest square no. which is exactly divisible by the given numbers is 900.
0
14/02/2018 4:33 pm
a,b are factors of 21600.How many pairs of (a,b)are there such that hcf(a,b) = 45 ?
LydiaMadini 02/06/2024 5:09 pm
The pair will be (45a, 45b) where a and b will be co-prime to each other. Now 21600 = 25*33*52. To find a and b, we first take the factor of 45 from 21600, which leaves 25*3*5. Now we need to find the number of co-prime pairs (a, b) that we can make out of 25*3*5. Let's write down the powers of the prime factors in order to find the co-prime factors:
(2, 22, 23, 24, 25), 3, 5
Therefore, the number of co-prime pairs is found by various combinations of these prime factors:
(Prime factor, Prime factor) (2, 3), (22, 3), (23, 3), (24, 5), (25, 5), (3, 5) ----- 11 in number
(Two prime factors, prime factor) (2 * 3, 5), (22 * 3, 5) (24 * 5, 3), (25*5, 3), (3*5, 2)...(3 *5, 25) --- 15 in number
(1, prime factor) (1, 2), (1, 22) .....(1, 25), (1, 3), (1, 5) --- 7 in number
(1, two prime factors) (1, 2*3), (1, 22*3)..... (1, 24 * 5), (1, 25 * 5), (1, 3* 5) --- 11 in number
(1, three prime factors)- (1, 2 *3 * 5), (1, 22 * 3 * 5)...... (1, 25 * 3 * 5) --- 5 in number.
Therefore, total number of co-prime pairs (a, b) = 49.
0
14/02/2018 4:36 pm
If the sum of two natural numbers and their LCM is 89, then how many such pairs of numbers are possible?
Manish Jindal 19/05/2018 11:31 pm
This post was modified 7 months ago by TathaGat
Let the two numbers be x and y
such that HCF(x,y) = h
=> x = ha, y = hb and LCM(x,y) = hab
=> h(a + b + ab) = 89
=> h = 1
and a + b + ab = 89
=> a = (89 - b)/(1 + b) = 90/(1 + b) - 1.
Number of positive divisors of 90 (i.e. 2*5*32) = 2*2*3 = 12.
So, 1 + b = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90
=> (a,b) = (89,0), (44,1), (29,2), (17,4), (14,5), (9,8), (8,9), (5,14), (4,17), (2,29), (1,44) or (0,89)
Now a and b are non zero b'coz they are natural numbers.
Also HCF(a,b) = 1.
So final values for (a,b) = (1,44) , (2,29) , (4,17) , (5,14) , (8,9) i.e. 5 pairs.
0
22/02/2018 7:57 am
Set A consist of all n positive integers less than 100 such that no two numbers are co prime to each other and no number is multiple of any other number in the group.
I. what is the maximum value of n?
II. What are the possible values?
TG Team 04/07/2018 4:06 pm
Hello Apoorva ,
Select all even numbers from 50 - 98
{ 50 , 52 , 54 , ......., 98 }
Hence , maximum value of n = 25
0
03/03/2018 11:50 pm
Set A consists of all positive integers les than 100 such that no two numbers are co prime to each other and no. number is a multiple of any other number.
What is the maximum value of n?
TG Team 04/07/2018 4:07 pm
Hello Yamin!
Select all even numbers from 50 - 98
{ 50 , 52 , 54 , ......., 98 }
Hence , maximum value of n = 25
0
18/05/2018 11:55 pm
A number when divided successively by 5, 8 and 11. It leaves the respectively remainder of 2, 5 and 7. What will be the remainder when such a least possible number is divided by 14?
AmitK 19/05/2018 5:04 pm
Hello Surabhi,
Minimum such number = 5 x {8 x (11 x 0+7) + 5} + 2 = 307
307 when divided by 14, remainder = 13
0
18/05/2018 11:56 pm
Bhima creates 5 types of sweets, 1452, 1188, 1716, 528 and 792, respectively, in her wedding party. If he packed in cartons to give sweets to all guests in equal number, so that the sweets of all kinds in each carton, then how many maximum cartons to be needed.
AmitK 19/05/2018 7:54 pm
Hello Surabhi,
in this case, the maximum number of cartons will be the HCF of the given numbers.
1452 = 22 * 3 * 112
1188 = 22 * 33 * 11
1716 = 22 * 3 * 11 * 13
528 = 24 * 3 * 11
792 = 23 * 32 * 11
Their HCF = 22 * 3 * 11 = 132, So Maximum number of cartons = 132
0
06/07/2018 2:10 pm
Find the pairs of natural numbers whose LCM is 78 and GCD is 13.
TG Team 06/07/2018 2:35 pm
Hi Nilesh .
Let the numbers be 13a and 13 b.
( Where a and b are coprime )
LCM of 13a and 13b = 13ab
13ab = 78
ab = 6
a = 1 , b = 6 or a = 2 , b = 3
Hence , two possible pairs .
0
06/07/2018 2:14 pm
The least common multiple of 2^6-1 and 2^9-1.
TG Team 06/07/2018 3:15 pm
Hi Nilesh !
A = (26 - 1) = ((23)3 -1) = ( 23 - 1 ) ( 23 + 1 )
B = (29 - 1) = (23 -1) ( 23 + 1 + 26 )
HCF (A, B) = ( 23 - 1 )
We know,
Product of two numbers = HCF x LCM
Hence, LCM (A, B) = [(26 - 1) (29 -1)]/ [23 -1]
LCM (A, B) = (23 +1) (29 -1)
= 212 - 23+ 29 -1
= 212 - 23 (26-1) -1
= 212 + 23 × 63 - 1.
0
20/10/2018 11:57 pm
TG Team 21/10/2018 9:40 am
B is multiplied by 25, LCM (A,B) remains unchanged so A has 5^3 in its prime factorization .
When A is multiplied by 8 ,LCM (A,B) remains unchanged so B has 2^4 in its prime factorization .
And A can have 2^0 or 2^1 => 2 cases
for each case exponent of 7 can range from 0 to 4 => 5 cases
Total : 2 x 5 = 10 cases
0
28/10/2018 9:36 am
H.C.F of how many distinct pairs of factors of 18000 is 75?
TG Team 09/11/2018 5:43 pm
75k and 75n are the factors , where k and n are coprime to each other .
=> k,n are the coprime factors of 18000/75= 240.
240 = 24 x 3 x 5
k = 2a1 3b1 5c1
n = 2a2 3b2 5c2
one of a1 , a2 has to be 0
=> ( 5 x 5) – ( 4 x 4) = 9 cases
one of b1 , b2 has to be 0
=> ( 2 x 2 ) – ( 1x 1 )= 3 cases
one of c1 , c2 has to be 0
=> ( 2 x 2 ) – ( 1 x 1 )= 3 cases
Hence , total number of ordered pair solution : 9 x 3 x 3 = 81 cases
Unordered pairs 82/2 = 41
0
09/11/2018 3:22 pm
What is the HCF of 3n+23 and n+8? (Please explain with long division method)
0
09/11/2018 3:24 pm
Find the largest 3 digit number which gives same remainder when divided by 13 or 17?
A.998
B.996
C.896
D.884
How can we do this without options?
TG Team 09/11/2018 5:28 pm
Largest possible remainder among 13 and 17 is 12 .since remainder has to be lesser than divisor
So , n [ LCM (13,17)] + 12
221n + 12 < 1000
largest possible n : 4
Hence 221 x 4 + 12 = 896.
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