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Problem 3 of 11 

3. Find the smallest no exactly divisible by .










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HCF_LCM 
This is sample question and answer for help purpose.
3. Find the smallest no exactly divisible by 12,15,20,27 .
`"Required number" = "LCM of " 12,15,20,27 = 540`.
2  12   15   20   27  2  6   15   10   27  3  3   15   5   27  3  1   5   5   9  3  1   5   5   3  5  1   5   5   1   1   1   1   1 
LCM of 12,15,20,27 = 2 × 2 × 3 × 3 × 3 × 5 = 540

Problem : 1 / 11
[ HCF_LCM ]
1.
The HCF of two nos is 14 and their LCM is 11592 . If one of the nos is 504 , find the other?




Solution:

`"Here, HCF"=14, "LCM"=11592, "No1"=504 and "No2"=?`
`"No1" * "No2" = "HCF" * "LCM"`
`"No2 "= ( "HCF" * "LCM" ) / ("No1") = ( 14 * 11592 ) / 504 = 322`.





Problem : 2 / 11
[ HCF_LCM ]
2.
Find the largest no which can exactly divide 513,783,1107 .




Solution:

`"Required number" = "HCF of " 513,783,1107 = 27`.
Find HCF of (513,783)
    
 513             
 
                    
 
                          
 
                              
          
HCF of (513, 783) = 27
Now find HCF of (27,1107)
HCF of (27, 1107) = 27
`:.` HCF of given numbers (513,783,1107) = 27





Problem : 3 / 11
[ HCF_LCM ]
3.
Find the smallest no exactly divisible by 12,15,20,27 .




Solution:

`"Required number" = "LCM of " 12,15,20,27 = 540`.
2  12   15   20   27  2  6   15   10   27  3  3   15   5   27  3  1   5   5   9  3  1   5   5   3  5  1   5   5   1   1   1   1   1 
LCM of 12,15,20,27 = 2 × 2 × 3 × 3 × 3 × 5 = 540





Problem : 4 / 11
[ HCF_LCM ]
4.
Find the least no which when divided by 6,7, 8, 9,12 leaves the same remainder 2 in each case.




Solution:

`"Required number" = ( "LCM of " 6,7,8,9,12 ) + 2 = 504 + 2 = 506`.
2  6   7   8   9   12  2  3   7   4   9   6  2  3   7   2   9   3  3  3   7   1   9   3  3  1   7   1   3   1  7  1   7   1   1   1   1   1   1   1   1 
LCM of 6,7,8,9,12 = 2 × 2 × 2 × 3 × 3 × 7 = 504





Problem : 5 / 11
[ HCF_LCM ]
5.
Find the largest no which divides 77 , 147 , 252 to leave the same remainder in each case.




Solution:

`"Required number" = "HCF of " ( 147  77 ), ( 252  147 ), ( 252  77 )`
`= "HCF of " 70, 105, 175 = 35`.
Factor of 70,105,175 are as follows...
70 = 2 × 5 × 7 105 = 3 × 5 × 7 175 = 5 × 5 × 7
HCF = 5 × 7 = 35
HCF of `70,105,175` is `35`





Problem : 6 / 11
[ HCF_LCM ]
6.
Greatest no which can divide 1354 , 1806 and 2762 leaving the same remainder 10 in each case.




Solution:

The number divides 1354 and leaves 10 as remainder `:.` The number exactly divides 1354  10 = 1344
The number divides 1806 and leaves 10 as remainder `:.` The number exactly divides 1806  10 = 1796
The number divides 2762 and leaves 10 as remainder `:.` The number exactly divides 2762  10 = 2752
Now, we have to find HCF of `1344, 1796, 2752`
Find HCF of (1344,1796)
    
 1344             
 
                    
 
                          
 
                              
                               
 
                                      
 
                                          
              
HCF of (1344, 1796) = 4
Now find HCF of (4,2752)
HCF of (4, 2752) = 4
`:.` HCF of given numbers (1344,1796,2752) = 4
`:.` Required number = HCF of `1344, 1796, 2752 = 4`.





Problem : 7 / 11
[ HCF_LCM ]
7.
The greatest no that will divide 1657 , 2772 leaving respectively 6 , 5 as remainder.




Solution:

The number divides 1657 and leaves 6 as remainder `:.` The number exactly divides 1657  6 = 1651
The number divides 2772 and leaves 5 as remainder `:.` The number exactly divides 2772  5 = 2767
Now, we have to find HCF of `1651, 2767`
Find HCF of (1651,2767)
    
 1651             
 
                    
 
                          
 
                              
                               
 
                                      
 
                                            
 
                                                  
 
                                                        
 
                                                              
 
                                                                  
                      
`:.` HCF of given numbers (1651,2767) = 1
`:.` Required number = HCF of `1651, 2767 = 1`.





Problem : 8 / 11
[ HCF_LCM ]
8.
The greatest no that will divide 290 , 460 , 552 leaving respectively 4 , 5 , 6 as remainder.




Solution:

The number divides 290 and leaves 4 as remainder `:.` The number exactly divides 290  4 = 286
The number divides 460 and leaves 5 as remainder `:.` The number exactly divides 460  5 = 455
The number divides 552 and leaves 6 as remainder `:.` The number exactly divides 552  6 = 546
Now, we have to find HCF of `286, 455, 546`
Find HCF of (286,455)
    
 286             
 
                    
 
                          
 
                                
 
                                    
            
HCF of (286, 455) = 13
Now find HCF of (13,546)
HCF of (13, 546) = 13
`:.` HCF of given numbers (286,455,546) = 13
`:.` Required number = HCF of `286, 455, 546 = 13`.





Problem : 9 / 11
[ HCF_LCM ]
9.
The product of HCF and LCM of 18 and 16 is




Solution:

Product of HCF and LCM = product of given numbers `= 18 * 16 = 288`.





Problem : 10 / 11
[ HCF_LCM ]
10.
LCM of two nos is 14 times their HCF. The sum of LCM and HCF is 600 . If one no is 280 , then find the other no ?




Solution:

`"LCM" = 14" HCF"`
`"LCM + HCF " = 600`
`=> 14 " HCF + HCF " = 600`
`=> 15 " HCF " = 600`
`=> "HCF " = 600 / 15 = 40`
Here, `"LCM" = 14 " HCF"`
`:. "LCM " = 14 * 40 = 560`.
`:. "Other No " = ( "HCF" * "LCM" ) / ("Given No"} = ( 40 * 560 ) / 280 = 80`.





Problem : 11 / 11
[ HCF_LCM ]
11.
Find the least number which is exactly divided by 28,36,45 when we add 19 to it




Solution:

`"Required number" = ( "LCM of " 28,36,45 )  19 = 1260  19 = 1241`.
2  28   36   45  2  14   18   45  3  7   9   45  3  7   3   15  5  7   1   5  7  7   1   1   1   1   1 
LCM of 28,36,45 = 2 × 2 × 3 × 3 × 5 × 7 = 1260










