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Learning Math Home
Number and Operations Session 5: Solutions
 
Session 5 Part A Part B Part C Homework
 
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Solutions for Session 5 Homework

See solutions for Problems: H1 | H2 | H3 | H4 | H5 | H6


Problem H1

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

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

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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Problem H6

a. 

13,09545 = 291

b. 

354,39339 = 9,087

<< back to Problem H6


 

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