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Solutions for Session 5 Homework
See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6
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Problem H1 | |
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This is true because abcabc divided by abc is 1,001. (Break up abcabc, the number, into [abc • 1,000] + abc, or abc • 1,001.) Since 1,001 is a multiple of 7, 11, and 13 (7 • 11 • 13 = 1,001), the number abcabc must be divisible by 7, 11, and 13.
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Problem H2 | |
a. | Add the digits: 7 + 2 + 5 + 2 = 16. This is not a multiple of 3, but it is one more than a multiple of 3. Therefore, the remainder is 1. |
b. | Use the last two digits of the number: 34. Thirty-four is two more than a multiple of 4, so the remainder is 2. |
c. | Add the digits: 3 + 4 + 5 + 7 = 19. This is not a multiple of 9, but it is one more than a multiple of 9. Therefore, the remainder is 1. |
d. | Add the even power digits: 3 + 9 = 12. Add the odd power digits: 4 + 5 = 9. In order for these to be equal (and produce a multiple of 11), we would need to subtract 3 from the units digit, so the number 4,356 is a multiple of 11. This means that 4,359 is three more than a multiple of 11, so the remainder is 3. |
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Problem H3 | |
a. | Yes, each is formed by adding 1 to the first sequence and subtracting 1 from the second. The sum must stay constant, and this constant is 101. |
b. | There are 100 of the 101s, since there are 100 numbers in each sequence. Gauss divided his sum by 2 because each number appears twice in the sequence -- once in the top row and once in the bottom row. |
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Problem H4 | |
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The sum can be written two ways:
The value (n + 1) appears n times. The total is n • (n + 1)  2. Testing this value for n = 100 gives the sum 100 • (101) 2 = 5,050, which is correct.
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Problem H5 | |
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The number of dots is measured as 1 + 2 + 3 + ... + n, so the total number of dots is n • (n + 1)2.
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