ics (24/25) / Discrete Term 1 E-Assessment / PS1

Let $X$ be the set of all 7 digit numbers in ordinary decimal notation (i.e. leading zeros are not allowed, so the smallest member of $X$ is 1000000 ). Compute:
(a) $|X|=$ $\square$
(b) $\mid\{n: n \in X, 5$ exactly divides $n\} \mid=$
$\square$
(c) $\mid\{n: n \in X, n$ is a palindrome $\} \mid=$
$\square$
(a palindrome is a string that reads the same backwards as forwards; e.g. 'was it a cat I saw', ignoring the spaces).
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Question

ics (24/25) / Discrete Term 1 E-Assessment / PS1

Let $X$ be the set of all 7 digit numbers in ordinary decimal notation (i.e. leading zeros are not allowed, so the smallest member of $X$ is 1000000 ). Compute:
(a) $|X|=$ $\square$
(b) $\mid\{n: n \in X, 5$ exactly divides $n\} \mid=$
$\square$
(c) $\mid\{n: n \in X, n$ is a palindrome $\} \mid=$
$\square$
(a palindrome is a string that reads the same backwards as forwards; e.g. 'was it a cat I saw', ignoring the spaces).
Please answer all parts of the question.
Check
us page
Finish attempt ...
TACK syntax
Jump to...
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Accepted Answer

∻Solution by Steps∻ ‖ step 1 ⋮ To find the total number of 7-digit numbers, we note that the first digit can be any digit from 1 to 9 (9 options), and the remaining 6 digits can be any digit from 0 to 9 (10 options each). Thus, the total count is given by: $|X| = 9 imes 10^6$. ‖ ‖ step 2 ⋮ To find the count of 7-digit numbers divisible by 5, we note that a number is divisible by 5 if its last digit is either 0 or 5. For each case, we calculate the total: - If the last digit is 0, the first digit has 9 options and the remaining 5 digits have 10 options each: $9 imes 10^5$. - If the last digit is 5, the calculation is the same: $9 imes 10^5$. Thus, the total count is: $|{n: n \in X, 5 ext{ exactly divides } n}| = 9 imes 10^5 + 9 imes 10^5 = 18 imes 10^5$. ‖ ‖ step 3 ⋮ To find the count of 7-digit palindromes, we note that a 7-digit palindrome has the form $abcba$. The first digit (a) can be any digit from 1 to 9 (9 options), the second digit (b) can be any digit from 0 to 9 (10 options), and the third digit (c) can also be any digit from 0 to 9 (10 options). Thus, the total count is: $|{n: n \in X, n ext{ is a palindrome}}| = 9 imes 10 imes 10 = 900$. ‖ ∻Answer∻ ⚹ (a) 9000000, (b) 1800000, (c) 900 ⚹ ∻Key Concept∻ ⚹ Counting principles and properties of divisibility and palindromes in number theory ⚹ ∻Explanation∻ ⚹ The answers reflect the application of combinatorial counting for 7-digit numbers, considering divisibility by 5 and the structure of palindromes. ⚹◊What is the total number of 7-digit numbers that are palindromes?⍭ Generate me a similar question◊

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