最大公约数（H.C.F）与最小公倍数（L.C.M.）—例题解析

答案 - B

解释

Prime numbers which are common to all the given numbers are 2,5 ,7.
∴ H.C.F = (22*5*73)= (4*5*343) = 6860

答案 - A

解释

108   = (22*33) , 360 = (23*32*5)  and 600 =  (23*52*3)
∴ H.C.F =  (22* 3) = (4* 3)=12

答案 - A

解释

Remainder of 185/148 = 37
Remainder of 148/37 = 0
∴ H.C.F. = 37

答案 - B

解释

Remainder of 1190/204 = 170
Remainder of 204/170 = 34
Remainder of 170/34 = 0

∴ H.C.F. of 204, 1190 = 34

Remainder of 1145/34 = 17
Remainder of 34/17 = 0

∴ H.C.F. of 204, 1190 and 1145 = 17

答案 - C

解释

First we find the H.C.F of 391 and 667.
Remainder of 667/391 = 276
Remainder of 391/276 =115
Remainder of 276/115 = 46
Remainder of 115/46 = 23
Remainder of 46/23= 0

∴ H.C.F. of 391, 667 = 23
∴    391/667   =( 391/23)/ (667/23)=  17/29

答案 - A

解释

We have  L.C.M = product of terms containing highest powers of (2,3,5,7,11)
= (2³* 32* 52*72*11) = (8*9*25*11*49)= 970200

答案 - D

解释

15 = 3 * 5
18 = 2* 3 * 3 = 2 * 3 2
24 = 2 *2 * 2 * 3   =  2 3 * 3
27 = 3 * 3 * 3 =  3 3
56 = 2 * 2 * 2 * 7=  2 3 * 7
L.C.M  = product of terms containing highest powers of (2,3,5,7) =2 3 * 3 3 * 5 * 7 = 7560

答案 - D

解释

H.C.F of 2,8,10,16     = 2
L.C.M of  3,9,27,81 = 81                      
H.C.f =  H.C.F of 2,8,10,16/L.C.M of 3,9,27,81 = 2/81          
L.C.M =  L.C.M of 2,8,10,16/H.C.F of  3,9, 27,81 = 80/3

答案 - B

解释

Let the numbers be 8x and 11 x. then, their H.C.F = x 
So, the numbers are (8*6), (11*6) i.e 48 and 66. 

答案 - B

解释

Let the Other number be X. then, 
Product of numbers = product of their H.C .F and L.C.M
35*x=   7* 210  ⇒ x=  7*210/35  = 42
Hence, the other number is 42.

答案 - A

解释

Required measure =   H.C.F  of 36 L, 45 L, and 72 L
= (32) liters = 9 liters
[As 36 = 22*32, 45 = 32*5 and 72 = 24* 34]

答案 - A

解释

Intervals of beeping 30  min, 60 min, 90 min, 105 min.
Interval of beeping together= L.C.M of 30 min. 60 min. 90 min. 105 min
= (3*5*2*2*3*7) min. = 1260 min = 21 hrs.
So, they will beep together again next morning at 9 am.

答案 - B

解释

Remainder of 783/513 = 270
Remainder of 513/270 =243
Remainder of 270/243 = 27
Remainder of 243/27 = 0

Remainder of 46/23= 0

∴ H.C.F. of 513, 783 = 23

Remainder of 1107/23 = 0

∴ H.C.F. of 513, 783 and 1107= 23

答案 - A

解释

Required no. = L.C.M of 12,15, 20 and 27
= (3*2*2*5*9) = 540

答案 - A

解释

Required number =  (L.C.M  of 6,7,8,9,12)+2   =  (2*3*2*7*2*3)+ 2 = (504+2)= 506.

答案 - B

解释

(1*2*3*4) =  24
∴ Required number = 24

答案 - A

解释

Here (35-18) =  17 , (45-28)=  17  and (55- 38) =  17
Required number =  (L.C.M of 35,45, 55)- 17  = (3465 -17) = 3448

答案 - A

解释

H.C.F =   H.C.F of 1,2,3,4/ L.C.M  of  2,3,4,5   = 1/120                                                 

答案 - C

解释

H.C.F  = H.C.F of 2, 8,10, 32/  L.C.M of 3,9, 27, 81   = 2/81

答案 - B

解释

H.C.F of  18 and 25 is 1.
∴  18 and 25 are co-primes.