HCF and LCM
Practice aptitude questions
Q1. Which of the following has most number of divisors?
Solution
99 = 1 * 3 * 3 * 11; 101 = 1 * 101;
176 = 1 * 2 * 2 * 2 * 2 * 11; 182 = 1 * 2 * 7 * 13
So, divisors of 99 are 1, 3, 9, 11, 33 and 99;
divisors of 101 are 1 and 101;
divisors of 176 are 1, 2, 4, 8, 16, 22, 44, 88 and 176;
divisors of 182 are 1, 2, 7, 13, 14, 26, 91 and 182.
Hence, 176 has the most number of divisors.
Q2. 1095/1168 when expressed in simplest form is:
Solution
1095)1168(1
1095
-----------
73)1095(15
1095
----------
0
So, H.C.F of 1095 and 1168 = 73.
1095/1168 = (1095/73) / (1168/73) = 15/16.
Q3. H.C.F of 4 * 27 * 3125, 8 * 9 * 25 * 7 and 16 * 81 * 5 * 11 * 49 is:
Solution
4 * 27 * 3125 = 22 * 33 * 55;
8 * 9 * 25 * 7 = 23 * 32 * 52 * 7;
16 * 81 * 5 * 11 * 49 = 24 * 34 * 5 * 72 * 11
H.C.F = 22 * 32 * 5 = 180.
Q4. Find the highest common factor of 36 and 84.
Solution
Explanation:
36 = 22 * 32; 84 = 22 * 3 * 7
H.C.F = 22 * 3 = 12
Q5. Find the lowest common multiple of 24, 36 and 40.
Solution
2 24 - 36 - 40
--------------------
2 12 - 18 - 20
--------------------
2 6 - 9 - 10
-------------------
3 3 - 9 - 5
--------------------
1 - 3 - 5
L.C.M = 2 * 2 * 2 * 3 * 3 * 5 = 360.
Q6. The G.C.D of 1.08, 0.36 and 0.9 is
Solution
Given numbers are 1.08, 0.36 and 0.90.
H.C.F of 108, 36 and 90 is 18.
H.C.F of a given numbers = 0.18
Q7. H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 24 * 35 * 52 * 72. The third number is:
Solution
3240 = 23 * 34 * 5; 3600 = 24 * 32 * 52
H.C.F = 36 = 22 * 32
Since H.C.F is the product of lowest powers of common factors, so the third number must have (22 * 32 ) as its factor.
Since L.C.M is the product of highest powers of common prime factors, so the third number must have 35 and 72 as its factors.
Third number = 22 * 35 * 72
Q8. Three numbers are in the ratio 1:2:3 and their H.C.F is 12. The numbers are:
Solution
Let the required numbers be x, 2x and 3x. Then, their H.C.F = x. So, x = 12.
The numbers are 12, 24, 36.
Q9. The ratio of numbers is 3:4 and their H.C.F is 4. Their L.C.M is:
Solution
Let the numbers be 3x and 4x.
Then their H.C.F = x. So, x = 4.
So, the numbers are 12 and 16.
L.C.M of 12 and 16 = 48.
Q10. The sum of two numbers is 528 and their H.C.F is 33. The number of pairs of numbers satisfying the above conditions is:
Solution
Let the required numbers be 33a and 33b.
Then, 33a + 33b = 528 => a + b = 16.
Now, co-primes with sum 16 are (1, 15), (3, 13), (5, 11) and (7, 9).
Required numbers are (33 * 1, 33 * 15), (33 * 3, 33 * 13), (33 * 5, 33 * 11), (33 * 7, 33 * 9).
The number of such pairs is 4.