For a natural number n , let n^ denote the number obtained by deleting zero digits, if any. (For example, if n=260, n^ =26 ; if n=2020 , n^ =22 .) Find the number of 3 -digit numbers n for which n^ is a divisor of n , different from n .

Step-by-step solution

n^ n so atleast one digit of a must be zero Let n= a b c Case I. If only c=0 a b a b c a b 10 a b (always true) Number of numbers n=9 9=81 Case II. If only b=0 a c a b c aligned & 10 a+c 100 a+c & 10 a+c 10 a-8 c aligned So either 10 a=8 c or 8 c-10 a= (10 a+c) for some natural number . When 10 a=8 c 5 a=4 c (a, c)=(4,5) When c(8- )=10 a( +1) c=a [ 90 8- -10 ] for =1, c=a ( 20 7 ) (Not possible) for =2, c=5 a (a, c)=(1,5) for =3, c=8 a (a, c)=(1,8) for =4, c=a ( 25 2 ) (Not possible) and no further value of is need