Source: NCTM Mathematics Teacher, September 2006

**Solution
**Let represent the length of one side of a square. The following equation is true

or (extraneous solution)

The length of one side of the square equals units.

**Answer**: units

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Source: NCTM Mathematics Teacher, September 2006

**Solution
**Let represent the number of small lemonades and the number of large lemonades sold. We have the following system of equations

Multiply Eqs. and by

Multiply Eq. by and add to Eq.

————————-

Substitute the value of into Eq.

John has sold small and large lemonades.

**Answer**: small and large

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Source: NCTM Mathematics Teacher, September 2006

**Solution**

In dollars the following equation depicts how you spent at the two stores

Collect like terms and simplify

At each store you purchase plums; in all you bought plums.

**Answer**:

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Source: NCTM Mathematics Teacher, September 2006

**Solution
**The square bracket in the sum expression contains 1003 pairs of consecutive powers which pairwise sum to zero

**Answer**:

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Source: NCTM Mathematics Teacher, September 2006

**Solution
**Areas of pizza slices

8-inch:

10-inch:

12-inch:

14-inch:

Since is the largest fraction, you want to take a slice from the 10-inch pizza.

**Answer**: 10-inch pizza

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Source: NCTM Mathematics Teacher, September 2006

**Solution
**A pin configuration is made up of ten pins each of which can be up or down. The number of possible configurations equals .

**Answer**:

**Alternative solution
**Let represent the number of pins left standing after a roll. In how many ways can they be left standing?

way to leave zero pin (bowl a strike)

ways to leave a single pin

ways to leave two pins

way to leave all ten pins (gutter ball)

Total number of possible ways

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Source: NCTM Mathematics Teacher, October 2006

**Solution
**Area in square meters per each year starting from year

In general in year , the area equals

The graph of this quadratic expresion is shown below

The area reaches a maximum of square meters in year .

**Answer**: square meters

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Source: NCTM Mathematics Teacher, October 2006

**Solution
** is such a large number that it is impractical to use modulo arithmetic to find the remainder. So we are going to find the remainders of a few powers of and hope to see a pattern emerge.

Remainders when are divided by

The pattern of remainders is . The remainder equals when the even exponent is the product of and an even number, for example, the remainder of equals because . The remainder equals when the even exponent is the product of and an odd number, for example, the remainder of equals because .

Since , the remainder of divided by equals .

**Answer**:

**Alternative solution
**We can group the powers according to their remainders as follows:

This is exactly what modulo arithmetic does, divide the whole numbers into four groups

Since divided by will have the same remainder as divided by .

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Source: NCTM Mathematics Teacher, October 2006

**Solution
**Sunday morning starts with a clean floor but in the night items appear.

Monday he cleans up but ends up with .

Tuesday

Wednesday

Thursday

Friday

Saturday

**Answer**:

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Source: NCTM Mathematics Teacher, October 2006

**Solution
**Is divisible by ? One way to find out is by determining if “behaves” like or or some other small multiples of . Modulo arithmetic will help us do that. In its most basic definition, modulo arithmetic is the arithmetic of remainder. We say “ is congruent to modulo ” and write , because and yield the same remainder when divided by .

Using , we will do the same calculation but with congruent numbers mod . For ease in presentation we drop the modulo notation.

Since ,

is divisible by .

Let be a 4-digit positive integer where the digits are , and . Suppose divides and . Show that divides .

Since divides and , there exist integers and such that

Subtract Eq. from Eq.

where

In other words, divides . Since does not divide , must divide .

Show that if divides , then divides .

Given for some integer ,

Since divides and , must divide . Hence, divides .

We have two cases to consider: and .

Case 1:

The integers are of the form where the numbers are divisible by . The possible numbers are: . The possible digits are: . There are possible .

Cases 2:

The integers are of the form where digits and are congruent. Interchanging congruent digits like and , and , and has no bearing on the remainder. The possible are: . There are possible .

Total number of integers equals .

List of some of the integers

**Answer**:

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