Robert Bosch LCM & HCF Quiz-2

Required number of students
= LCM of 6, 8, 10 = 120

Here, Divisor – remainder = 1
e.g., 10 – 9 = 1, 9 – 8 = 1,8 – 7 = 1
\therefore Required number
= (L.C.M. of 10, 9, 8) –1
= 360 – 1 = 359

We find LCM of 5, 6 and 8
5=5
6= 3\times2
8= 2^{3}
= 23 \times3\times5 = 8\times 15 = 120
Required number = 120K + 3
\therefore when K = 2, 120 \times 2 + 3 = 243 is required no
It is completely divisible by 9

 

LCM of 16, 18, 20 and 25= 3600
\therefore Required number = 3600K +4 which is exactly divisible by 7 for certain value of K.
When K = 5,
number = 3600 \times 5 + 4
= 18004 which is exactly divisible by 7.

 

LCM of 3, 5, 6, 8, 10 and 12
= 120
\therefore Required number
= 120x + 2, which is exactly divisible by 13.
120x + 2 = 13 \times 9x + 3x + 2
Clearly 3x + 2 should be divisible by 13.
For x=8,3x + 2 is divisible by 13.

We will find the LCM of 16, 24,30 and 36.
\therefore LCM = 2 \times 2 \times 2 \times 3 \times 2 \times 5 \times 3
The largest number of five digits
= 99999
On dividing 99999 by 720, the remainder = 639
\therefore The largest five-digit number divisible by 720
= 99999 – 639 = 99360
\therefore Required number = 99360 + 10
= 99370

LCM of 5, 10, 12, 15 = 2 \times 3 \times 5 \times 2 \times = 60
\therefore Number = 60k + 2
Now, the required number should be divisible by 7.
Now, 60k + 2 = 7 \times 8k + 4k + 2
If we put k = 3, (4k + 2) is equal to 14 which is exactly divisible by 7.
\therefore Required number = 60 \times 3 + 2
= 182

LCM of 9, 10 and 15 = 90
=> The multiple of 90 are also divisible by 9, 10 or 15.
\therefore 21\times 90 = 1890 will be divisible by them.
\therefore Now, 1897 will be the number that will give the remainder 7.
Required number
= 1936 – 1897 = 39

 

Here,12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10
Now, LCM of 12, 16 and 24 = 48
\therefore The greatest 4–digit number
exactly divisible by 48 = 9984
\therefore Required number
= 9984 – 10 = 9974

LCM of 15, 20 and 35 = 420
\therefore Required least number
= 420 + 8 = 428