Problems On H.C.F And L.C.M - Aptitude MCQ Questions and Solutions with Explanations

HCF = 8
LCM = 48
One number = 24
Let other number be = y
∴ 24y = 48 × 8
⇔ y = 16

LCM of (4, 6, 8, 12, 16)
⇒ 16 × 3 = 48
∴ The number when divided by (4, 6, 8, 12, 16) leaves remainder 2 is
= 48 + 2
= 50

LCM = 495
HCF = 5       ( given )
∴ Let numbers are = 5x and 5y
∴ LCM = 5xy
⇒ 5xy = 495
⇒ xy = 99
∴ Possible co-prime factors are
1,       99
9,       11
∴ Possible numbers are 5x, 5y
45,       55
5,       495
Now given that sum of numbers = 100
So, required numbers are = (45, 55)
∴ Difference of numbers = 55 - 45 = 10

99 = 1 × 3 × 3 × 11
101 = 1 × 101
176 = 1 × 2 × 2 × 2 × 2 × 11
182 = 1 × 2 × 7 × 13
So, divisor of 99 are 1, 3, 9, 11, 33 and 99
Divisor of 101 are 1 and 101
Divisor of 176 are 1, 2, 4, 8, 11, 16, 22, 44, 88 and 176
Divisor of 182 are 1, 2, 7, 13, 14, 26, 91 and 182
Hence, 176 has the most number of divisors.

HCF = 8
Now LCM should have a factors 8
So check also the option we have only 60 which does not have a factor 8. So it will never be the LCM.

Numbers x = 28, y = 42
HCF (28 , 42)
Difference = 42 - 28 = 14
For HCF of any numbers take their difference. HCF will be either the factor of that difference or the difference itself.
Now, LCM of 28, 42
∴ 14 × 2 × 3 = 84
⇒ LCM : HCF
⇒ 84 : 14
⇒ 6 : 1

$$\eqalign{ & {\text{116690151}}\overline {){\text{427863887(}}} {\text{3}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{350070453}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {{\text{77793434)116690151(}}} 1 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\text{77793434}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\overline {{\text{38896717)77793434(}}} {\text{2}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{77793434}}{ \times } \cr & {\text{So, H}}{\text{.C}}{\text{.F of 116690151 and 427863887}} \cr & {\text{ = 38896717}} \cr & \therefore \frac{{116690151}}{{427863887}} = \frac{{116690151 \div 38896717}}{{427863887 \div 38896717}} \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \frac{3}{{11}} \cr} $$

$$\eqalign{ & {\text{LCM = 1820}} \cr & {\text{HCF = 26}} \cr & {{\text{1}}^{{\text{st}}}}{\text{ number = 130}} \cr & {\text{LCM}} \times {\text{HCF }} \cr & {\text{ = Product of numbers}} \cr & \Rightarrow {\text{Let the other number is }}x{\text{ }} \cr & 130 \times x = 1820 \times 26 \cr & x = \frac{{1820 \times 26}}{{130}} = 364 \cr} $$

HCF = 478
Number are = 12906 and 14818
LCM × HCF = 12906 × 14818
LCM × 478 = 12906 × 14818
LCM = 400086

LCM of (3, 5, 8, 12)
⇒ 3 × 5 × 4 × 2 = 120
⇒ Now greatest five digits number is 99999
On dividing 99999 by = 120 (LCM)
We get remainder = $$\frac{{99999}}{{120}},$$  remainder = 39
⇒ By subtracting remainder from 99999 we get the greatest five digits number which is completely divisible by given numbers (3, 5, 8, 12)
∴ 99999 - 39 = 99960
⇒ Now, we required greatest five digit number which when divided by (3, 5, 8, 12) leaves remainder 2 in each case.
⇒ add 2 in the 99960
⇒ 99960 + 2
⇒ 99962