Question

If n be the number of 4-digit numbers (in base 10) having

non-zero digits and which are divisible by 4 but not by 8.

Find [n/10], here [a] represents greatest integer less

than or equal to a.
please give the correct answer
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Answer 1

The 4 digit number N is divisible by 4 but not by 8.

So the number N should be in the form 8k + 4 for some natural number k.

• 8k + 4 leaves remainder 4 on division by 8.

• 8k + 4 = 4(2k + 1) is exactly divisible by 4.

It is clear that the number N = 8k + 4 is always an even number so the possible ones digit numbers are 0, 2, 4 and 6.

Since 6 - 0 = 6, if we consider four consecutive even numbers ending in 2, 4, 6 and 8 (0 is not possible), there exists exactly one number in the form 8k + 4.

Because, e.g. assume a number of the form 10n + 6 = 8k + 4, then the next possible number will be 10n + 6 + 8 = 10(n+1) + 4, but this number belongs to the set of next four consecutive even numbers, not the same.

So divide the 4 digit even numbers, having non - zero digits, into sets containing four consecutive even numbers each. And each should contain exactly one number in the form 8k + 4.

No. of 4 digit even numbers = 9 × 9 × 9 × 4.

[∵ Digits other than ones digit can occupy any among numbers from 1 to 9 but ones digit can occupy only any among 2, 4, 6 and 8.]

No. of possible sets = 9 × 9 × 9 × 4 / 4 = 729.

∴ No. of possible 4 - digit numbers, N = 729.

Now, [N/10] = [729/10] = 72.

Hence 72 is the answer.