Schaum's Outline of Discrete Mathematics, Third Edition (Schaum's Outlines)

CHAP. 11] PROPERTIES OF THE INTEGERS 283

We now substitute into the formula (11.9) to obtain the following solution of our original system:

x 0 = 35 · 2 · 2 + 21 · 1 · 4 + 15 · 1 · 6 = 314

Dividing this solution by the modulusM=105, we obtain the remainder

x= 104

which is the unique solution of the riddle between 0 and 105.

Remark:The above solutionss 1 =2,s 2 =1,s 3 =1 were obtained by inspection. If the moduli are large, we

can always use the Euclidean algorithm to find such solutions as in Example 11.17.

SolvedProblems

INEQUALITIES, ABSOLUTE VALUE

11.1. Insert the correct symbol,<, >,or=, between each pair of integers:

(a) 4 ____−7; (b) −2 _____−9; (c)(− 3 )^2 ____ 9; (d)−8 ____ 3,

For each pair of integers, sayaandb, determine their relative positions on the number lineR; or, alternatively,

computeb−a, and writea<b,a>b,ora=baccording asb−ais positive, negative, or zero. Hence:

(a) 4 >− 7 ; (b)− 2 >− 9 ; (c) (− 3 )^2 = 9 ; (d)− 8 < 3.

11.2. Evaluate: (a)| 2 − 5 |,|− 2 + 5 |,|− 2 − 5 |;(b)| 5 − 8 |+| 2 − 4 |,| 4 − 3 |−| 3 − 9 |.
Evaluate inside the absolute value sign first:

(a) | 2 − 5 |=|− 3 |= 3 , |− 2 + 5 |=| 3 |= 3 , |− 2 − 5 |=|− 7 |= 7

(b) | 5 − 8 |+| 2 − 4 |=|− 3 |+|− 2 |= 3 + 2 = 5 ;| 4 − 3 |−| 3 − 9 |=| 1 |−|− 6 |= 1 − 6 =− 5

11.3. Find the distancedbetween each pair of integers:
(a) 3 and –7; (b) –4 and 2; (c) 1 and 9; (d) –8 and –3; (e) –5 and –8.
The distancedbetweenaandbis given byd=|a−b|=|b−a|.Alternatively, as indicated by Fig. 11-5,
d=|a|+|b|whenaandbhave different signs, andd=|a|−|b|whenaandbhave the same sign and|a|>|b|.
Thus:(a) d= 3 + 7 = 10 ;(b) d= 4 + 2 = 6 ;(c) d= 9 − 1 = 8 ;(d) d= 8 − 3 = 5 ;(e) d= 8 − 5 = 3.

Fig. 11-5

11.4. Find all integersnsuch that: (a)1< 2 n− 6 <14; (b)2< 8 − 3 n<18.

(a) Add 6 to the “three sides” to get 7< 2 n<20. Then divide all sides by 2 (or multiply by 1/2) to get 3. 5 <n<10.

Hencen= 4 , 5 , 6 , 7 , 8 ,9.

(b) Add –8 to the three sides to get− 6 <− 3 n<10. Divide by –3 (or multiply by –1/3) and, since –3 is negative,

change the direction of the inequality to get

2 >n>− 3 .3or− 3. 3 <n< 2

Hencen=− 3 ,− 2 ,− 1 , 0 ,1.