Please wait...
/
-
Let the numbers be where are co-primes.
Possible and acceptable pairs of x and y satisfying the condition are: (1, 7) and (3, 5)
Numbers are: 38 × 1 = 38 and 38 × 7 = 266
& 38 × 3 = 114 and 38 × 5 = 190
Required difference = 266 – 38 = 228
Option C is correct.
LCM of 21, 25 and 36 = 6300
Required number = 6300K + 6 which is exactly divisible by 11 for certain value of K.
When K = 2
Number = 6300 × 2 + 6 = 12606 which is exactly divisible by 11.
So, option D is correct.
Let us consider any value of a, b such that a and b are positive integer.
For a=21 and b = 12 H.C.F (a, b)= H.C.F(21, 12) = 3.
Now H.C.F (a/H.C.F.(a,b),b/H.C.F.(a,b) =H.C.F.(21/3,12/3) = H.C.F. (7, 4) = 1 For any value of positive integer a and b H.C.F. (a/H.C.F.(a,b),b/H.C.F.(a,b)) will always give '1' as answer.
For any integer n.
H.C.F. (22n + 7, 33n + 10)
Now for n = 0, H.C.F. (7, 10) = 1
for n = 1, H.C.F.(29, 43) = 1
Hence for any integer n,
H.C.F. (22n + 7, 33n + 10) equal to 1
If HCF=21
then let first number= 21a
second number=21b
Now LCM= 21ab= 4641
⇒ ab =221
If ab= 221
Then possible pairs of a & b is (221,1) & (13,17)
But as one number is in between 200 and 300 then
First number = 21 × 13=273
Second number= 21 × 17 = 357
Correct (-)
Wrong (-)
Skipped (-)