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# Which is the least number that must be subtracted from 1856 so that the remainder, when divided by 7, 12, and 16, is 4? 

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Which is the least number that must be subtracted from 1856 so that the remainder, when divided by 7, 12, and 16, is 4? 

a. 137       b. 1361         C. 140          d. 157          e. 172

#### Solution:

Again the same method used in the question the smallest number which when divided by 4, 6 and 7 will be adopted but now from the smallest to 1856.

(This is just another method to find the LCM)

7 x 12=84 (two numbers)

=84 x 16

$\frac { 84\times 16 }{ 2 } =42\times8$      (reduce it to find the LCM)

42×8=336; we have to check here whether this number is divisible by all the given numbers or not. If this is divisible, this means it is the LCM. If not, increase the number by multiplying it by 2 or decrease the number by dividing it by 2. In all cases, the obtained number has to be divisible by all the given numbers.

The LCM of 7, 12, 16 is 336 because it is divisible by all the numbers.

Now we have to find multiple of 336 until 1856.

336×5= 1680  ( we cannot go beyond 1680 because then the multiple will exceed 1856 ).

As given in the question, add 4 to it to have a remainder 4.

It becomes 1680+4 = 1684.

Now subtract 1856-1680 = 172.

This means to have a remainder of 4, we have to subtract 172 from 1856.

Hence, the option (e) is correct.

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