1) In how many ways can we write 16 as the sum of three distinct positive integers? [Note: Assume that 1 + 11 + 4 is the same as 4 + 1 + 11.] 

2. Two regular, fair dice are rolled. We add the two numbers on the top faces. How many different sums are possible? 

 3. A group of six strangers sat in a circle, and each one got acquainted only with the person to the left and the person to the right. Then all six people stood up and each one shook hands once with each of the others who was still a stranger. How many handshakes occurred?  

 4. How many odd, nonrepeating, three-digit numbers can we write using only the digits 0, 1, 2, an 3?

5. Until 1995, the rules for three-digit area codes in the U.S. were as follows: (i) The first digit could not be 0 or 1. (ii) The second digit had to be 0 or 1. (iii) The third digit had no restrictions. a) How many area codes were possible? b) In 1995, the restriction on the second digit of area codes was removed. How many area codes are possible? 

Answer:

I. 14 ways

II. 11 sums

III. 18 handshakes

IV. 12 numbers

V. 160 codes

Step-by-step explanation

I.

1+2+13

1+3+12

1+4+11

1+5+10

1+6+9

1+7+8

2+3+11

2+4+10

2+5+9

2+6+8

3+4+9

3+5+8

3+6+7

4+5+7

There are 14 ways

 

II.

The lowest possible number is 2 (1+1)

The highest possible number is 12 (6+6)

Anything in between 2 and 6 are possible sums. Therefore, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 are the only sums possible

There are 11 sums

 

III.

In a perspective of a person, he will not shake with himself, and the people on his left and right. Therefore, only three handshakes will occur per person. Since there are six people, then 3x6 is the number of handshakes that will occur. 18 handshakes will happen.

 

IV.

Every number which ends with 0 and 2 are even numbers, and they are not odd.

0123

1230 - eliminate

2301

3012 - eliminate

 

1023

0231

2310 -eliminate

3102 -eliminate

 

1203

2031

0312 -eliminate

3120 -eliminate

 

0213

2130 -eliminate

1302 -eliminate

3021

 

0132 -eliminate

1320 -eliminate

3201

2013

 

0321

3210 -eliminate

2103

1032- eliminate

 

This is also the same as 4P2 in the four numbers, only two of the digits can be selected because the remaining two are even numbers.

There are only 12 ways

 

V.

First digit - 2, 3, 4, 5, 6, 7, 8, 9 (8 digits possible)

Second digit- 0, 1 (2 digits possible)

Third digit, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 (10 digits possible)

 

200

201

202

203

204

205

206

207

208

209

210

211

212

213

214

215

216

217

218

219

There are 20 possible numbers if the first digit is set to 2

Since there are 8 possible numbers for the first digit, then 20 can be multiplied to 8 to determine the number of possible combinations.

This can also be determined by multiplying all of the possible numbers per digits. 8x2x10 = 160 codes