How many numbers divisible by 5 and lying between 5000 and 6000 can be formed from the digits 5, 6, 7, 8 and 9

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Q. How many numbers divisible by 5 and lying between 5000 and 6000 can be formed from the digits 5, 6, 7, 8 and 9 ?

Tamil Nadu DgeTamil Nadu HSC First Year (DGE) Board 2018Subjective· 2mImportance★★★★★

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The number must start and end in 5 (digits may repeat), leaving the two middle positions free to be any of the 5 digits each, giving $5\times5=25$ numbers.

A 4-digit number lying strictly between 5000 and 6000 must have its first (thousands) digit equal to 5.

For the number to be divisible by 5, its last (units) digit must be 0 or 5. Since 0 is not among the available digits $\{5,6,7,8,9\}$ , the units digit must be 5.

So the first digit is fixed as 5 and the last digit is fixed as 5 (digits may repeat across positions, since the set only offers one 5 and it's being reused in two positions).

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2025Set ANNUAL1 markMCQ
$Q. The number of 5 digit numbers, all digits of which are odd is:$
(a) $5^6$
(b) $25$
(c) $625$
(d) $5^5$
›Reveal solutionSolution

Each of the 5 digit-positions is filled independently from the 5 odd digits.

The odd digits available are $\{1,3,5,7,9\}$ , a set of 5 digits. Since none of them is $0$ , there is no restriction on the leading digit either. Each of the 5 positions of the number can be filled in 5 ways, independ …

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CBSE 2022Set ANNUAL1 markMCQ
$Q. The number of 5 digit numbers all digits of which are odd is:$
(a) $5^6$
(b) 25
(c) 625
(d) $5^5$
›Reveal solutionSolution

With 5 choices of odd digit for each of the 5 positions (no restriction, since all odd digits are already non-zero), the count is $5^5$ .

The odd digits are $1,3,5,7,9$ -- exactly 5 of them.

A 5-digit number has 5 positions, and unlike a general 5-digit number problem, there's no separate restriction on the leading digit here, because none of the allowed digits (odd digits) is zero anyway.

…

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CBSE 2019Set ANNUAL1 markMCQ
$Q. The number of five digit numbers in which all digits are even, is:$
(a) $4\times 5^4$
(b) $4\times 5^5$
(c) $5^5$
(d) $5\times 5$
›Reveal solutionSolution

With digits restricted to $\{0,2,4,6,8\}$ , the first digit excludes 0 (4 choices) and the remaining 4 digits are unrestricted (5 choices each), giving $4\times 5^4$ .

The even digits available are $0, 2, 4, 6, 8$ — 5 digits in total, and repetition is allowed (the question only restricts which digits may be used, not that they be distinct).

For a 5-digit number, the first (leftmost) digit cannot be 0, so it has $5-1=4$ choices.

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Q.How many numbers divisible by 5 and lying between 5000 and 6000 can be formed from the digits 5, 6, 7, 8 and 9 ?

Concept understanding — Fundamental Principles of Counting

The Fundamental Principle of Counting

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Q.How many numbers divisible by 5 and lying between 5000 and 6000 can be formed from the digits 5, 6, 7, 8 and 9 ?

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