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find the number of $4$-digit numbers where the second digit is even, and the fourth digit is at least twice the second digit. (note that digits are read from the left, so the first digit is the leftmost digit, and so on.)

Answer :

There will be a total of 1620 four digit numbers having an even second digit and a fourth digit that is atleast twice of the second digit.

Calculation:

Possible second digits allowed = 0,2,4

digits 6 and 8 are also even but cant be included as their twice yeild double digits.

For each of the allowed digits, the possible fourth digit ranges from 0-9(if second digit is zero), 4-9(if second digit is considered as 2) and 8-9(if second digit is 4). Hence the total number of fourth digits allowed            = 10+6+2=18.

The first digit can be any non zero number whereas the third digit can be from 0-9.

Hence the total number of four digits allowed: 9×10×18 = 1620.

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