Consider the integers,
i. sum of the squares of the digits of the number = 50,
ii. every digit must be greater than the digits left to it(increasing order).

Find the product of the digits of the largest number, which satisfies those two conditions.

Ans: i got number as 1236, product is 36

• sum of squares of digits=50,
  so digits may be 1,2,3,4,5,6 for largest no.
  
  1^2+2^2+3^2+6^2=50
  
  no. is = 1236 => product of digit=1*2*3*6=36
• 10 years agoHelpfull: Yes(25) No(0)
• this type of question solve firsty check by option
• 10 years agoHelpfull: Yes(3) No(0)
• 345 product 60
• 10 years agoHelpfull: Yes(1) No(5)
• sum of the squares of the digits is
  3^2+4^2+5^2=50
  product
  3*4*5=60
• 10 years agoHelpfull: Yes(1) No(0)
• i got the number 17(1^2+7^2=1+49) 7 is grater than 1. so the number is 17 and product is 7
• 10 years agoHelpfull: Yes(0) No(6)
• number can be 345
  1st condition:3^2+4^2+5^2=9+16+25=50
  2nd condition:345(digit is greater than the digit left to it)
  product is 3*4*5=60
  
• 10 years agoHelpfull: Yes(0) No(1)