Question

find the smallest number which when didvided by161,207,and 184 leaves remainder 21 in all the cases

Answer 1

Solution :-

To find the smallest number which when divided by 161, 207 and 184 leaves remainder 21 in all case, we have to compute the L.C.M. of 161, 207 and 184 and then we will add 21 in the L.C.M. of 161, 207 and 184. The number we have after adding will be the required number. 

  2  |  161, 207, 184
      |. ______________
  2  |  161, 207,  92
      |______________
  2  |  161, 207,  46
      |______________
  3  |  161, 207,  23
      |______________
  3  |  161,  69, 23
      |______________
  7  |  161,  23, 23
      |______________
  7  |   23,  23,  23
      |______________
 23 |     1,  1,   1
      |
      
      
L.C.M. of 161, 207 and 184 is 2*2*2*3*3*7*23

= 11592

So. L.C.M. of 161, 207 and 184 is 11592

Now, we will add 21 and 11592

11592 + 21 = 11613

The required smallest number is 11613, which when divided by 161, 207 and 184, leaves a remainder of 21 in each case.

Let us check our answer.

11613 ÷ 161
Quotient = 72 and Remainder = 21

11613 ÷ 207
Quotient = 56 and Remainder = 21

11613 ÷ 184
Quotient = 63 and Remainder = 21

Answer 2

Hi ,

First we have to write 161 , 207 and 184 into product of prime

161 = 7 × 23

207 = 3 × 3 × 23

184 = 2 × 2 × 2 × 23

Now find LCM of ( 161 , 207 , 184 ) = 23 × 7 × 3 ×3 × 2 × 2 ×2 = 11592

The required smallest number which when divided by 161 , 207 and

184 leaves remainder 21 = LCM ( 161, 207 , 184 ) + remainder

= 11592 + 21

= 11613

Verification :

1 ) 11613= 161 × 72 + 21

2) 11613 = 207 × 56 + 21

3 ) 11613 = 184 × 63 +21

I hope this will usful to you.

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