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by harsh.champ » Mon Feb 08, 2010 3:10 pm
Suppose, the seed of any positive integer n is defined as follows:

seed(n) = n, if n < 10

= seed(s(n)), otherwise,

where s(n) indicates the sum of digits of n. For example,

seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5 etc.

How many positive integers n, such that n < 500, will have seed(n) = 9?

(A)39
(B)72
(C)81
(D)108
(E)55

The OA is E.

Solution approach needed!!
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Source: — Problem Solving |

by ldoolitt » Mon Feb 08, 2010 6:01 pm
Divisibility rule, right?

For 9's, the sum of the digits is divisible by 9 [rule]. So the question is straight up asking how many numbers between 1 and 500 are divisible by 9.

Each block of 100 has 11 multiples of 9 in it (until 500 at least) so 9*11 = 55
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by sumanr84 » Mon Feb 08, 2010 6:55 pm
ldoolitt wrote:Divisibility rule, right?

For 9's, the sum of the digits is divisible by 9 [rule]. So the question is straight up asking how many numbers between 1 and 500 are divisible by 9.

Each block of 100 has 11 multiples of 9 in it (until 500 at least) so 9*11 = 55
Bingo !! Right on target..
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by ajith » Mon Feb 08, 2010 9:46 pm
harsh.champ wrote:Suppose, the seed of any positive integer n is defined as follows:

seed(n) = n, if n < 10

= seed(s(n)), otherwise,

where s(n) indicates the sum of digits of n. For example,

seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5 etc.

How many positive integers n, such that n < 500, will have seed(n) = 9?

(A)39
(B)72
(C)81
(D)108
(E)55

The OA is E.

Solution approach needed!!
Seed(n)=9 only when n is a multiple of 9
the question is
How many positive integers n, such that n is divisible by 9

(495-9)/9 +1 = 55
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by shashank.ism » Wed Feb 10, 2010 5:44 am
harsh.champ wrote:Suppose, the seed of any positive integer n is defined as follows:

seed(n) = n, if n < 10

= seed(s(n)), otherwise,

where s(n) indicates the sum of digits of n. For example,

seed(7) = 7, seed(248) = seed(2 + 4 + 8) = seed(14) = seed(1 + 4) = seed(5) = 5 etc.

How many positive integers n, such that n < 500, will have seed(n) = 9?

(A)39
(B)72
(C)81
(D)108
(E)55

The OA is E.

Solution approach needed!!
We would go in cases
1st case n<10 , only 1 value (as seed(9)=9)
2nd case n>10 , so this case is possible only when n is divisble by 9 so no. of values = 500/9 -R = 55(which also includes 9)

so total value = (55-1)+1 =55
ans E
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