p is a 3 digit no. and 1st digit is 3...and product of the digit is factor of that no..
how many no. are there who follow these ...n what is the sum of largest digit..satisfy above 3 condition

• there are only 3 such numbers those are 312,315,384
  the largest sum is:3+8+4=15
  Ex:
  the number start with 3
  the numbers are 300 to 399
  first take 300=3*0*0=0 no factors
  301=3*0*1=0 no factors
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  311=3*1*1=3 ,311 is not divisible by 3
  312=3*1*2=6,312 is divisible by 3
  313=3*1*3=9,313 is not divisible by 9
  314=3*1*4=12,314 is not divisible by 12
  315=3*1*5=15,315 is divisible by 15
  316=3*1*6=18, 316 is not divisible by 18
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  384=3*8*4=96,384 is divisible by 96
  there are only 3 such numbers
• 9 years agoHelpfull: Yes(14) No(4)
• updated... only 3 nos are there which are, 312,315 and 384.
  and sum of largest no is 3+8+4=15
  Though it is hit n trial ques, we can use some trick.
  As 3 is first digit of no, product of no is a factor of that no, hence 3 will be one of the factor of required no.
  now we will check all such no whose first digit is 3 and divisible by 3 excluding those no having 0 in them(as product of digit of that no will give 0,,eg 306 and 360).
• 9 years agoHelpfull: Yes(11) No(0)
• there are 5 such numbers: 306,312,315,360,384
  sum of largest no=15
• 9 years agoHelpfull: Yes(5) No(12)
• K S W
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• 9 years agoHelpfull: Yes(2) No(5)
• any one explane this question
  
• 9 years agoHelpfull: Yes(0) No(0)
• before submitting the solution,me doubt is that numbers like 306,360 or even 330 can be included or not because the product comes out to b 0 after multiplying the digits and 0 is factor of every number
• 9 years agoHelpfull: Yes(0) No(2)
• sorry the query i just submitted was wrong..that would result to infinity..answers would be 312 315 and 384
• 9 years agoHelpfull: Yes(0) No(2)