Find the largest number, smaller than the smallest four-digit number, which when divided by 4,5,6 and 7 leaves a remainder 2 in each case.

• Take LCM of 4,5,6,7. It is 420
  BUt the no must leave remainder 2 in each case, so the no is of the form: 420k + 2.
  The smallest 4-digit no is 1000. So keeping k=0,1,2,3....
  We get that the largest no smaller than the smallest 4 -digit no is 842
• 10 years agoHelpfull: Yes(37) No(10)
• take LCM of all digit : 5,5,6,7 it should be =840
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  Than as per question reminder 2 in each case so add 2 in LCM= 840+2=842
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  means when we divide 842 with any on these number 4,5,6,7 than we get reminder as 2
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  So answer is 842
• 10 years agoHelpfull: Yes(13) No(28)
• LCM of 4,5,6,7= 840
  so that 842 leave the remainder 2 after divided by 4,5,6,7.
  
  
• 10 years agoHelpfull: Yes(4) No(23)
• take lcm of 4,5,6,7 =420
  remainder2 in each case. therefore 420k+2=1000, 1000 smallest 4 digit number
  k=2 840+2=842
  
• 9 years agoHelpfull: Yes(4) No(2)