# Divisibility Tests and Factors Homework

Problem H1
The number abcabc (where each letter represents one particular digit) is divisible by 7, 11, and 13 for all one-digit values of a, b, and c. Why is that? Think about dividing abcabc by abc. Is the number you obtained divisible by 7, 11, and 13?

Problem H2
Use what you know about divisibility tests to find remainders for the following:
a.
7,252 3
b.
1,234 4
c.
3,457 9
d.
4,359 11

Problem H3
It is said that the mathematician Karl Gauss figured out how to find the sum of the first 100 counting numbers when he was in the second grade. Then he added the 101s to get

100 • 101 = 10,100.

But that number is twice the sum, so the actual sum is 10,100 2, or 5,050.

Examine Gauss’s process:
a. Are the missing totals 101? Why?
b. How many 101s are there in all? How do you know? Why did Gauss divide by 2?

Problem H4
Use Gauss’s method to find the sum of the first n counting numbers.

Problem H5
“Triangular numbers” describe the number of dots needed to make triangles like the ones below. The first triangular number is 1, the second is 3, and so on. Use the results of Problem H4 to write a rule for the number of dots in the nth triangular number. See Note 6 below. Think about how many dots the first triangle has and how many you need to add to make each new triangle.

Problem H6
Each * stands for any missing digit (i.e., they are not all the same digit). Decode these long-division problems:  Begin by looking for any additional digits that you can fill in, in order to have fewer unknowns. For example, the two asterisks below 5 have to be 5 as well, because there is no remainder.
We know that a two-digit number multiplied by a three-digit number produces a five-digit number. So think about the highest values you could have for those two numbers.

Also think about what numbers might work in the left-hand digit of the quotient and the right-hand digit of the divisor to produce a two-digit result that ends in 0.

### Solutions

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.

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.

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.

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.

Problem H5
The number of dots is measured as 1 + 2 + 3 + … + n, so the total number of dots is n • (n + 1) 2.

Problem H6
a.
13,095 45 = 291
b.
354,393 39 = 9,087