Discussion

Explanation:

‘0’ occurs exactly 3 times and 5 occurs exactly once while the number is an even number.

Case 1: First 3 digits are 305
‘0’ has to occur 2 more times and one ‘5’ has already been used.

Case 1(a): Last digit is ‘0’
Now, we have to use one more ‘0’ and other 3 digits can be any of 1, 2, 3, 4, 6, 7, 8, 9

3 0 5     __ __ __ _0_  

One ‘0’ can be placed in 4C1 = 4 ways and remaining 3 digits can be filled in 8 × 8 × 8 = 512 ways.

∴ Total number of ways = 4 × 512 = 2048.

Case 1(b): Last digit is not ‘0’
Now, we have to use two more ‘0s’ to be placed.

3 0 5     __ __ __ _2/4/6/8_  

One ‘0’ can be placed in 4C2 = 6 ways,
Last digit can be filled with either 2 or 4 or 6 or 8 i.e., in 4 ways and 
Remaining 2 digits can be filled in 8 × 8 = 64 ways.

∴ Total number of ways = 6 × 4 × 64 = 1536.

Case 2: First 3 digit are ‘936’

Case 2(a): Last digit is ‘0’
Now, we have to use two more ‘0s’ and one ‘5’. 

9 3 6     __ __ __ _0_  

Two ‘0’ and one 5 can be placed in 4C3 × 3 = 12 ways and remaining 1 digit can be filled in 8 ways.

∴ Total number of ways = 12 × 8 = 96.

Case 2(b): Last digit is not ‘0’
Now, we have to use three more ‘0s’ and one 5.

9 3 6     __ __ __ _2/4/6/8_  

Last digit can be filled with either 2 or 4 or 6 or 8 i.e., in 4 ways and 
Three ‘0s’ and a 5 can be placed in 4 ways,

∴ Total number of ways = 4 × 4 = 16.

⇒ Total number of combinations = 2048 + 1536 + 96 + 16 = 3696.

To make sure that Shubham reaches his friend, he will have to try a maximum of 3696 times.

Hence, 3696.

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