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80 lines
1.4 KiB
Markdown
80 lines
1.4 KiB
Markdown
---
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id: 681cb1b0dab50c87ddb2e51b
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title: "Challenge 10: 3 Strikes"
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challengeType: 29
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dashedName: challenge-10
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---
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# --description--
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Given an integer between 1 and 10,000, return a count of how many numbers from 1 up to that integer whose square contains at least one digit 3.
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# --hints--
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`squares_with_three(1)` should return `0`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(squares_with_three(1), 0)`)
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}})
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```
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`squares_with_three(10)` should return `1`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(squares_with_three(10), 1)`)
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}})
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```
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`squares_with_three(100)` should return `19`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(squares_with_three(100), 19)`)
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}})
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```
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`squares_with_three(1000)` should return `326`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(squares_with_three(1000), 326)`)
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}})
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```
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`squares_with_three(10000)` should return `4531`.
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```js
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({test: () => { runPython(`
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from unittest import TestCase
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TestCase().assertEqual(squares_with_three(10000), 4531)`)
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}})
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```
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# --seed--
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## --seed-contents--
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```py
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def squares_with_three(n):
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return n
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```
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# --solutions--
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```py
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def squares_with_three(n):
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count = 0
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for i in range(1, n + 1):
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square = i * i
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if '3' in str(square):
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count += 1
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return count
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```
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